Chapter 8 Rational Numbers

Given:

The rational number $\frac{2}{3}$ and four options.

To Find:

Which of the given rational numbers is equivalent to $\frac{2}{3}$.

Solution:

Two rational numbers are equivalent if they represent the same value. We can check for equivalence by simplifying the given options or by cross-multiplication.

The given rational number is $\frac{2}{3}$. This is already in its simplest form.

Let's examine each option:

(a) $\frac{3}{2}$: This fraction is in simplest form. $\frac{3}{2} \neq \frac{2}{3}$.

(b) $\frac{4}{9}$: This fraction is in simplest form. $\frac{4}{9} \neq \frac{2}{3}$.

(c) $\frac{4}{6}$: We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2.

$\frac{4}{6} = \frac{4 \div 2}{6 \div 2} = \frac{2}{3}$

Since the simplified form of $\frac{4}{6}$ is $\frac{2}{3}$, these two rational numbers are equivalent.

(d) $\frac{9}{4}$: This fraction is in simplest form. $\frac{9}{4} \neq \frac{2}{3}$.

Alternatively, using cross-multiplication, we check which option $\frac{a}{b}$ satisfies $2 \times b = 3 \times a$ when compared with $\frac{2}{3}$.

For option (c) $\frac{4}{6}$, we check if $2 \times 6 = 3 \times 4$.

$2 \times 6 = 12$

$3 \times 4 = 12$

Since $12 = 12$, the ratios are proportional (or the fractions are equivalent).

Answer:

The rational number equivalent to $\frac{2}{3}$ is $\frac{4}{6}$.

The correct option is (c).

Given:

Four rational numbers are provided as options.

To Find:

Which of the given rational numbers is in standard form.

Solution:

A rational number $\frac{p}{q}$ is in standard form if:

1. $p$ and $q$ are integers with no common factors other than 1 (i.e., $\text{gcd}(p, q) = 1$).

2. The denominator $q$ is a positive integer.

Let's examine each option:

(a) $\frac{20}{30}$: Both 20 and 30 have common factors other than 1 (e.g., 10). $\frac{20}{30} = \frac{2}{3}$. So, it is not in standard form.

(b) $\frac{10}{4}$: Both 10 and 4 have common factors other than 1 (e.g., 2). $\frac{10}{4} = \frac{5}{2}$. So, it is not in standard form.

(c) $\frac{1}{2}$: The numerator 1 and the denominator 2 have no common factors other than 1. The denominator 2 is positive. So, this rational number is in standard form.

(d) $\frac{1}{-3}$: The denominator is a negative integer ($-3$). To write this in standard form, the negative sign is usually moved to the numerator or placed in front of the fraction: $\frac{-1}{3}$ or $-\frac{1}{3}$. So, it is not in standard form as given.

Based on the definition of standard form, only option (c) satisfies both conditions.

Answer:

The rational number in standard form is $\frac{1}{2}$.

The correct option is (c).

Given:

Two rational numbers: $\frac{-3}{2}$ and $\frac{1}{2}$.

To Find:

The sum of the two rational numbers.

Solution:

To find the sum of two rational numbers with the same denominator, we add the numerators and keep the denominator the same.

Sum = $\frac{-3}{2} + \frac{1}{2}$

Sum = $\frac{-3 + 1}{2}$

Sum = $\frac{-2}{2}$

Simplify the fraction:

Sum = $-1$

Answer:

The sum of $\frac{-3}{2}$ and $\frac{1}{2}$ is $-1$.

The correct option is (a).

Given:

The expression to evaluate: $-\frac{4}{3} - \frac{-1}{3}$.

To Find:

The value of the given expression.

Solution:

The given expression is the subtraction of two rational numbers.

$-\frac{4}{3} - \frac{-1}{3}$

Subtracting a negative number is equivalent to adding the corresponding positive number. So, $-\frac{-1}{3}$ is the same as $+\frac{1}{3}$.

The expression becomes:

$-\frac{4}{3} + \frac{1}{3}$

Since the two rational numbers have the same denominator (3), we can add the numerators directly and keep the common denominator:

$\frac{-4 + 1}{3}$

Perform the addition in the numerator:

$-4 + 1 = -3$

So the expression simplifies to:

$\frac{-3}{3}$

Now, simplify the fraction by dividing the numerator and the denominator by their greatest common divisor, which is 3:

$\frac{-3}{3} = -1$

The value of the expression is $-1$.

Answer:

The value of $-\frac{4}{3} - \frac{-1}{3}$ is $-1$.

The correct option is (d).

Given:

A statement with a blank regarding the number of rational numbers between two given rational numbers.

To Find:

The term that correctly fills the blank to make the statement true.

Solution:

The set of rational numbers is dense. This means that between any two distinct rational numbers, no matter how close they are, there exists at least one other rational number. In fact, there are infinitely many rational numbers between any two distinct rational numbers.

Answer:

There are infinitely many number of rational numbers between two rational numbers.

Given:

A statement with a blank identifying a rational number that is neither positive nor negative.

To Find:

The rational number that correctly fills the blank to make the statement true.

Solution:

A number is considered positive if it is greater than zero, and negative if it is less than zero. The only number that is neither greater than zero nor less than zero is zero itself.

Zero can be expressed as a rational number in the form $\frac{p}{q}$, where $p=0$ and $q$ is any non-zero integer (e.g., $\frac{0}{1}, \frac{0}{2}, \frac{0}{-5}$, etc.). Therefore, zero is a rational number.

Zero is neither positive nor negative.

Answer:

The rational number $0$ (or zero) is neither positive nor negative.

Given:

The statement: "In any rational number $\frac{p}{q}$, denominator is always a non-zero integer."

To Determine:

Whether the given statement is True or False.

Solution:

A rational number is defined as any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is not equal to zero ($q \neq 0$).

The denominator of the rational number $\frac{p}{q}$ is $q$. The definition explicitly states that $q$ must be a non-zero integer.

Therefore, the statement that the denominator is always a non-zero integer is consistent with the definition of a rational number.

Answer:

The statement is True.

Given:

The statement: "To reduce the rational number to its standard form, we divide its numerator and denominator by their HCF".

To Determine:

Whether the given statement is True or False.

Solution:

Reducing a rational number to its standard form involves two main steps:

1. Making the denominator positive (if it is negative).

2. Simplifying the fraction by dividing both the numerator and the denominator by their greatest common factor (HCF) to make the numerator and denominator coprime (having no common factors other than 1).

The given statement focuses on the second step, which is crucial for reducing the fraction to its simplest form where the numerator and denominator are coprime. Dividing both the numerator and denominator by their HCF ensures that there are no common factors left, making the fraction irreducible.

The statement correctly describes a key process in reducing a rational number to its standard form, specifically regarding the simplification of the fraction part.

Answer:

The statement is True.

Given:

The statement: "All rational numbers are integers".

To Determine:

Whether the given statement is True or False.

Solution:

An integer is a whole number (positive, negative, or zero). Examples of integers are ..., -3, -2, -1, 0, 1, 2, 3, ...

A rational number is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Consider some examples of rational numbers:

$\frac{1}{2}$ is a rational number, but it is not an integer.

$\frac{3}{4}$ is a rational number, but it is not an integer.

$- \frac{5}{2}$ is a rational number, but it is not an integer.

While all integers can be expressed as rational numbers (e.g., $3 = \frac{3}{1}$, $-2 = \frac{-2}{1}$), not all rational numbers are integers.

The statement claims that *all* rational numbers are integers, which is false because there exist rational numbers (like $\frac{1}{2}$) that are not integers.

Answer:

The statement is False.

Given:

Two rational numbers: $\frac{4}{5}$ and $\frac{5}{6}$.

To Find:

Three rational numbers between $\frac{4}{5}$ and $\frac{5}{6}$.

Solution:

To find rational numbers between two given rational numbers, we first need to express them with a common denominator. The least common multiple (LCM) of the denominators 5 and 6 is 30.

Convert $\frac{4}{5}$ to an equivalent fraction with a denominator of 30:

$\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$

Convert $\frac{5}{6}$ to an equivalent fraction with a denominator of 30:

$\frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30}$

Now we need to find three rational numbers between $\frac{24}{30}$ and $\frac{25}{30}$. There are no integers between the numerators 24 and 25. To find numbers between them, we can multiply both fractions by an integer greater than the number of rational numbers we want to find (which is 3). Let's multiply by $10$.

Convert $\frac{24}{30}$ to an equivalent fraction by multiplying the numerator and denominator by 10:

$\frac{24}{30} = \frac{24 \times 10}{30 \times 10} = \frac{240}{300}$

Convert $\frac{25}{30}$ to an equivalent fraction by multiplying the numerator and denominator by 10:

$\frac{25}{30} = \frac{25 \times 10}{30 \times 10} = \frac{250}{300}$

Now we need to find three rational numbers between $\frac{240}{300}$ and $\frac{250}{300}$. We can choose any three fractions with a denominator of 300 and a numerator between 240 and 250 (exclusive of 240 and 250).

Three possible numerators are 241, 242, 243.

The corresponding rational numbers are $\frac{241}{300}$, $\frac{242}{300}$, $\frac{243}{300}$.

These are three rational numbers between $\frac{240}{300}$ and $\frac{250}{300}$, which are equivalent to $\frac{4}{5}$ and $\frac{5}{6}$ respectively.

Note: We can simplify $\frac{242}{300}$ to $\frac{121}{150}$ and $\frac{243}{300}$ to $\frac{81}{100}$. The simplified forms are also between the original numbers.

Answer:

Three rational numbers between $\frac{4}{5}$ and $\frac{5}{6}$ are $\frac{241}{300}$, $\frac{242}{300}$ (or $\frac{121}{150}$), and $\frac{243}{300}$ (or $\frac{81}{100}$). Many other answers are possible.

Given:

Two pairs of rational numbers.

To Find:

Which pair(s) represent equivalent rational numbers.

Solution:

Two rational numbers are equivalent if they represent the same value. We can check for equivalence by simplifying both fractions in a pair or by cross-multiplication.

Pair (i): $\frac{7}{12}$ and $\frac{28}{48}$

Check by simplifying the second fraction:

$\frac{28}{48}$ - The greatest common divisor (GCD) of 28 and 48 is 4.

$\frac{28}{48} = \frac{28 \div 4}{48 \div 4} = \frac{7}{12}$

Since the simplified form of $\frac{28}{48}$ is $\frac{7}{12}$, the two fractions are equivalent.

Alternatively, check by cross-multiplication:

Is $7 \times 48$ equal to $12 \times 28$?

$7 \times 48 = 336$

$12 \times 28 = 336$

Since $336 = 336$, the fractions are equivalent.

Pair (ii): $\frac{-2}{-3}$ and $\frac{-16}{24}$

Simplify the first fraction:

$\frac{-2}{-3} = \frac{2}{3}$ (A negative divided by a negative is positive).

Simplify the second fraction:

$\frac{-16}{24}$ - The greatest common divisor (GCD) of 16 and 24 is 8.

$\frac{-16}{24} = \frac{-16 \div 8}{24 \div 8} = \frac{-2}{3}$

Now compare the simplified forms: $\frac{2}{3}$ and $\frac{-2}{3}$.

These two fractions are not equal ($\frac{2}{3} \neq \frac{-2}{3}$). Therefore, the original fractions are not equivalent.

Alternatively, check by cross-multiplication:

Is $(-2) \times 24$ equal to $(-3) \times (-16)$?

$(-2) \times 24 = -48$

$(-3) \times (-16) = 48$

Since $-48 \neq 48$, the fractions are not equivalent.

Answer:

Only pair (i) represents equivalent rational numbers.

Given:

A sequence of rational numbers: $\frac{-1}{3}$, $\frac{-2}{6}$, $\frac{-3}{9}$, and four blanks.

To Find:

The next four rational numbers in the pattern.

Solution:

Let's examine the given rational numbers to identify the pattern.

The first number is $\frac{-1}{3}$.

The second number is $\frac{-2}{6}$. We can see that this is obtained by multiplying the numerator and denominator of $\frac{-1}{3}$ by 2: $\frac{-1 \times 2}{3 \times 2} = \frac{-2}{6}$.

The third number is $\frac{-3}{9}$. This is obtained by multiplying the numerator and denominator of $\frac{-1}{3}$ by 3: $\frac{-1 \times 3}{3 \times 3} = \frac{-3}{9}$.

The pattern appears to be generating equivalent fractions of $\frac{-1}{3}$ by multiplying the numerator and the denominator by consecutive integers starting from 1.

The sequence is $\frac{-1 \times 1}{3 \times 1}$, $\frac{-1 \times 2}{3 \times 2}$, $\frac{-1 \times 3}{3 \times 3}$, ...

To find the next four numbers, we should multiply the numerator and denominator of $\frac{-1}{3}$ by the next consecutive integers, which are 4, 5, 6, and 7.

The fourth number (multiply by 4): $\frac{-1 \times 4}{3 \times 4} = \frac{-4}{12}$.

The fifth number (multiply by 5): $\frac{-1 \times 5}{3 \times 5} = \frac{-5}{15}$.

The sixth number (multiply by 6): $\frac{-1 \times 6}{3 \times 6} = \frac{-6}{18}$.

The seventh number (multiply by 7): $\frac{-1 \times 7}{3 \times 7} = \frac{-7}{21}$.

Answer:

The next four rational numbers in the pattern are $\frac{-4}{12}$, $\frac{-5}{15}$, $\frac{-6}{18}$, $\frac{-7}{21}$.

The completed pattern is:

$\frac{-1}{3}$ , $\frac{-2}{6}$ , $\frac{-3}{9}$ , $\frac{-4}{12}$, $\frac{-5}{15}$, $\frac{-6}{18}$, $\frac{-7}{21}$.

Given:

Two mixed numbers: $-4\frac{5}{6}$ and $-7\frac{3}{4}$.

To Find:

The sum of the two mixed numbers.

Solution:

First, convert the mixed numbers into improper fractions.

$-4\frac{5}{6} = -\left(\frac{(4 \times 6) + 5}{6}\right) = -\left(\frac{24 + 5}{6}\right) = -\frac{29}{6}$

$-7\frac{3}{4} = -\left(\frac{(7 \times 4) + 3}{4}\right) = -\left(\frac{28 + 3}{4}\right) = -\frac{31}{4}$

Now we need to find the sum: $-\frac{29}{6} + \left(-\frac{31}{4}\right) = -\frac{29}{6} - \frac{31}{4}$.

To add or subtract fractions, they must have a common denominator. The least common multiple (LCM) of the denominators 6 and 4 is 12.

Convert $-\frac{29}{6}$ to an equivalent fraction with a denominator of 12:

$-\frac{29}{6} = -\frac{29 \times 2}{6 \times 2} = -\frac{58}{12}$

Convert $-\frac{31}{4}$ to an equivalent fraction with a denominator of 12:

$-\frac{31}{4} = -\frac{31 \times 3}{4 \times 3} = -\frac{93}{12}$

Now, add the equivalent fractions:

$-\frac{58}{12} - \frac{93}{12} = \frac{-58 - 93}{12}$

Add the numbers in the numerator:

$-58 - 93 = -151$

The sum is $\frac{-151}{12}$.

We can convert this improper fraction back to a mixed number by dividing 151 by 12.

$151 \div 12$. $12 \times 12 = 144$. The remainder is $151 - 144 = 7$.

So, $\frac{151}{12} = 12 \frac{7}{12}$.

Since the fraction is negative, the result is $-12\frac{7}{12}$.

Answer:

The sum of $-4\frac{5}{6}$ and $-7\frac{3}{4}$ is $-12\frac{7}{12}$ (or $\frac{-151}{12}$).

Given:

Two mixed numbers: $-2\frac{3}{4}$ and $5\frac{6}{7}$.

To Find:

The product of the two mixed numbers.

Solution:

First, convert the mixed numbers into improper fractions.

$-2\frac{3}{4} = -\left(\frac{(2 \times 4) + 3}{4}\right) = -\left(\frac{8 + 3}{4}\right) = -\frac{11}{4}$

$5\frac{6}{7} = \frac{(5 \times 7) + 6}{7} = \frac{35 + 6}{7} = \frac{41}{7}$

Now, multiply the improper fractions:

Product = $\left(-\frac{11}{4}\right) \times \left(\frac{41}{7}\right)$

To multiply fractions, multiply the numerators together and multiply the denominators together.

Product = $\frac{-11 \times 41}{4 \times 7}$

Multiply the numerators: $-11 \times 41 = -451$

Multiply the denominators: $4 \times 7 = 28$

Product = $\frac{-451}{28}$

This is an improper fraction. We can convert it back to a mixed number by dividing 451 by 28.

$451 \div 28$. $28 \times 10 = 280$. $451 - 280 = 171$.

$28 \times 6 = 168$. $171 - 168 = 3$.

So, $451 = 16 \times 28 + 3$.

$\frac{451}{28} = 16 \frac{3}{28}$.

Since the original product was negative, the result is $-16\frac{3}{28}$.

Answer:

The product of $-2\frac{3}{4}$ and $5\frac{6}{7}$ is $-16\frac{3}{28}$ (or $\frac{-451}{28}$).

Given:

Mathematical expressions in Column I and values in Column II.

To Find:

Match each expression in Column I with its correct value in Column II.

Solution:

We will evaluate each expression in Column I and determine its value to find the corresponding entry in Column II.

Recall that dividing by a fraction is the same as multiplying by its reciprocal.

(i) $\frac{3}{4}$ ÷ $\frac{3}{4}$

$\frac{3}{4} \div \frac{3}{4} = \frac{3}{4} \times \frac{4}{3} = \frac{3 \times 4}{4 \times 3} = \frac{12}{12} = 1$

This matches Column II (e).

(ii) $\frac{1}{2}$ ÷ $\frac{4}{3}$

$\frac{1}{2} \div \frac{4}{3} = \frac{1}{2} \times \frac{3}{4} = \frac{1 \times 3}{2 \times 4} = \frac{3}{8}$

This matches Column II (d).

(iii) $\frac{2}{3}$ ÷ (-1)

Dividing by -1 is the same as multiplying by $\frac{1}{-1} = -1$.

$\frac{2}{3} \div (-1) = \frac{2}{3} \times (-1) = -\frac{2}{3}$

This matches Column II (b).

(iv) $\frac{3}{4}$ ÷ $\frac{1}{2}$

$\frac{3}{4} \div \frac{1}{2} = \frac{3}{4} \times \frac{2}{1} = \frac{3 \times 2}{4 \times 1} = \frac{6}{4}$

Simplify the fraction: $\frac{6}{4} = \frac{6 \div 2}{4 \div 2} = \frac{3}{2}$.

This matches Column II (c).

(v) $\frac{5}{7}$ ÷ $\left( \frac{-5}{7} \right)$

$\frac{5}{7} \div \left( \frac{-5}{7} \right) = \frac{5}{7} \times \left( \frac{7}{-5} \right) = \frac{5 \times 7}{7 \times (-5)} = \frac{35}{-35} = -1$

This matches Column II (a).

Answer:

The correct matches are:

(i) - (e)

(ii) - (d)

(iii) - (b)

(iv) - (c)

(v) - (a)

Given:

The expression $\frac{2}{11}$ ÷ $-\frac{5}{55}$.

To Find:

The reciprocal of the value of the given expression.

Solution:

First, evaluate the given expression $\frac{2}{11}$ ÷ $-\frac{5}{55}$.

Dividing by a fraction is equivalent to multiplying by its reciprocal.

Reciprocal of $-\frac{5}{55}$ is $\frac{55}{-5}$.

Expression = $\frac{2}{11} \times \frac{55}{-5}$

We can simplify before multiplying. Notice that 55 is divisible by 11 and by 5.

$\frac{2}{\cancel{11}_{1}} \times \frac{\cancel{55}^{5}}{\cancel{-5}_{-1}}$

Expression = $\frac{2}{1} \times \frac{5}{-1}$

Multiply the numerators and the denominators:

Expression = $\frac{2 \times 5}{1 \times (-1)} = \frac{10}{-1} = -10$

The value of the expression $\frac{2}{11}$ ÷ $-\frac{5}{55}$ is $-10$.

Now, we need to find the reciprocal of $-10$.

The reciprocal of a number $x$ is $\frac{1}{x}$. For a non-zero number, the reciprocal is found by inverting the fraction (writing $x$ as $\frac{x}{1}$).

Reciprocal of $-10$ (or $\frac{-10}{1}$) is $\frac{1}{-10}$.

In standard form, the reciprocal is $-\frac{1}{10}$.

Answer:

The reciprocal of $\frac{2}{11}$ ÷ $-\frac{5}{55}$ is $-\frac{1}{10}$.

To Find: The correct condition for $q$ in the definition of a rational number.

Solution:

A rational number is formally defined as any number that can be expressed as the quotient or fraction $\frac{p}{q}$ of two integers, a numerator $p$ and a non-zero denominator $q$.

That is, for a number to be rational, it must satisfy the form:

$\frac{p}{q}$, where $p \in \mathbb{Z}$, $q \in \mathbb{Z}$, and $q \neq 0$.

The condition that the denominator $q$ cannot be zero is crucial because division by zero is undefined in mathematics.

Let's examine the given options:

• (a) $q = 0$: This is incorrect because the denominator of a fraction cannot be zero.
• (b) $q = 1$: This is not the only condition. While numbers like $5 = \frac{5}{1}$ are rational, $q$ can be any non-zero integer (e.g., $\frac{3}{4}$, $\frac{-2}{7}$).
• (c) $q \neq 1$: This is partially correct as $q$ can be values other than 1 (like 2, -3, etc.), but it is not the fundamental restriction. $q$ is allowed to be 1.
• (d) $q \neq 0$: This is the fundamental and necessary condition for the expression $\frac{p}{q}$ to be a well-defined number that can represent a rational number.

Therefore, the correct condition on $q$ is that $q$ must not be equal to zero.

The correct option is (d).

To Find: The positive rational number among the given options.

Solution:

A rational number $\frac{p}{q}$ is positive if the numerator $p$ and the denominator $q$ have the same sign (both positive or both negative).

Let's check each option:

• (a) $\frac{-8}{7}$: The numerator is $-8$ (negative) and the denominator is $7$ (positive). The signs are different, so the rational number is negative.
• (b) $\frac{19}{-13}$: The numerator is $19$ (positive) and the denominator is $-13$ (negative). The signs are different, so the rational number is negative. This can be written as $\frac{-19}{13}$.
• (c) $\frac{-3}{-4}$: The numerator is $-3$ (negative) and the denominator is $-4$ (negative). The signs are the same (both negative), so the rational number is positive. This simplifies to $\frac{3}{4}$.
• (d) $\frac{-21}{13}$: The numerator is $-21$ (negative) and the denominator is $13$ (positive). The signs are different, so the rational number is negative.

Based on the analysis, only option (c) has a positive value.

The correct option is (c).

To Find: The negative rational number among the given options.

Solution:

A rational number $\frac{p}{q}$ is negative if the numerator $p$ and the denominator $q$ have different signs (one positive and one negative).

Let's examine each option:

• (a) $-\left( \frac{-3}{7} \right)$: The fraction inside the parenthesis $\frac{-3}{7}$ has a negative numerator ($-3$) and a positive denominator ($7$). Thus, $\frac{-3}{7}$ is negative. The expression is the negative of $\frac{-3}{7}$, which means it is $-(\text{negative number})$. This results in a positive number. $-\left( \frac{-3}{7} \right) = \frac{3}{7}$, which is positive.
• (b) $\frac{-5}{-8}$: The numerator is $-5$ (negative) and the denominator is $-8$ (negative). Since both signs are the same, the rational number is positive. $\frac{-5}{-8} = \frac{5}{8}$, which is positive.
• (c) $\frac{9}{8}$: The numerator is $9$ (positive) and the denominator is $8$ (positive). Since both signs are the same, the rational number is positive.
• (d) $\frac{3}{-7}$: The numerator is $3$ (positive) and the denominator is $-7$ (negative). Since the signs are different, the rational number is negative. This can also be written as $\frac{-3}{7}$.

Therefore, the only rational number that is negative is $\frac{3}{-7}$.

The correct option is (d).

To Find: The common factor of the numerator and denominator in the standard form of a rational number.

Solution:

A rational number $\frac{p}{q}$ is said to be in its standard form if it satisfies two conditions:

1. The denominator $q$ is a positive integer.

2. The numerator $p$ and the denominator $q$ have no common factor other than 1. This means that the greatest common divisor (GCD) of $|p|$ and $|q|$ is 1.

The second condition directly answers the question. If the only common factor between the numerator and the denominator is 1, then their common factor in the standard form is always 1.

Let's consider an example: The rational number $\frac{6}{9}$ is not in standard form. To convert it to standard form, we find the GCD of 6 and 9, which is 3. We divide both the numerator and the denominator by 3:

$\frac{\cancel{6}^2}{\cancel{9}_3} = \frac{2}{3}$

In the standard form $\frac{2}{3}$, the numerator 2 and the denominator 3 have no common factor other than 1.

Reviewing the options:

• (a) 0: Not a common factor for non-zero numbers.
• (b) 1: This matches the definition of the standard form where the numerator and denominator are relatively prime.
• (c) – 2: If -2 is a common factor, then 2 is also a common factor, meaning the number is not in standard form.
• (d) 2: If 2 is a common factor, the fraction can be simplified further, so it is not in standard form.

Therefore, the common factor of the numerator and denominator in the standard form of a rational number is always 1.

The correct option is (b).

To Find: The rational number among the options that is equal to its reciprocal.

Solution:

Let the rational number be $x$. The reciprocal of $x$ is $\frac{1}{x}$, provided $x \neq 0$.

We are looking for a number $x$ such that $x$ is equal to its reciprocal. So we set up the equation:

$x = \frac{1}{x}$

To solve for $x$, multiply both sides by $x$ (we must assume $x \neq 0$, which is required for the reciprocal to be defined):

$x \times x = \frac{1}{x} \times x$

$x^2 = 1$

Taking the square root of both sides gives:

$x = \pm \sqrt{1}$

$x = 1$ or $x = -1$

Thus, the rational numbers that are equal to their reciprocals are 1 and -1.

Now, let's check the given options to see which one is either 1 or -1:

• (a) 1: This is one of the values we found. The reciprocal of 1 is $\frac{1}{1} = 1$. So, $1 = 1$.
• (b) 2: The reciprocal of 2 is $\frac{1}{2}$. $2 \neq \frac{1}{2}$.
• (c) $\frac{1}{2}$: The reciprocal of $\frac{1}{2}$ is $2$. $\frac{1}{2} \neq 2$.
• (d) 0: The reciprocal of 0 is $\frac{1}{0}$, which is undefined. 0 is not equal to an undefined value.

Among the given options, only 1 is equal to its reciprocal.

The correct option is (a).

To Find: The reciprocal of the rational number $\frac{1}{2}$.

Solution:

The reciprocal of a non-zero rational number $\frac{p}{q}$ is defined as $\frac{q}{p}$. This is the number that, when multiplied by $\frac{p}{q}$, gives a product of 1.

In this question, the given rational number is $\frac{1}{2}$. Here, the numerator is $p=1$ and the denominator is $q=2$.

According to the definition, the reciprocal of $\frac{1}{2}$ is obtained by swapping the numerator and the denominator.

Reciprocal of $\frac{1}{2} = \frac{2}{1}$

$\frac{2}{1} = 2$

Let's verify this by multiplication:

$\frac{1}{2} \times 2 = \frac{1}{2} \times \frac{2}{1} = \frac{1 \times 2}{2 \times 1} = \frac{2}{2} = 1$

Since the product is 1, our calculation for the reciprocal is correct.

Now, let's compare our result with the given options:

• (a) 3: Incorrect.
• (b) 2: Correct.
• (c) – 1: Incorrect.
• (d) 0: Incorrect (reciprocal of 0 is undefined).

The reciprocal of $\frac{1}{2}$ is 2.

The correct option is (b).

To Find: The standard form of the rational number $\frac{-48}{60}$.

Solution:

To convert a rational number to its standard form, we need to:

1. Ensure the denominator is positive.

2. Divide both the numerator and the denominator by their greatest common divisor (GCD) to make them relatively prime (their only common factor is 1).

The given rational number is $\frac{-48}{60}$.

Step 1: The denominator is 60, which is already positive. So, this step is satisfied.

Step 2: Find the GCD of the absolute values of the numerator and the denominator, i.e., GCD of $|-48|$ and $|60|$, which is GCD(48, 60).

We can find the GCD using prime factorization or by listing factors.

Prime factorization of 48: $\begin{array}{c|cc} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$ So, $48 = 2^4 \times 3$.

Prime factorization of 60: $\begin{array}{c|cc} 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$ So, $60 = 2^2 \times 3 \times 5$.

The common prime factors are $2^2$ and $3^1$.

GCD(48, 60) = $2^2 \times 3 = 4 \times 3 = 12$.

Now, divide both the numerator and the denominator of $\frac{-48}{60}$ by 12:

$\frac{-48 \div 12}{60 \div 12} = \frac{-4}{5}$

The resulting fraction is $\frac{-4}{5}$. The denominator 5 is positive, and the numerator -4 and denominator 5 have no common factors other than 1 (GCD(|-4|, |5|) = GCD(4, 5) = 1). Therefore, $\frac{-4}{5}$ is the standard form.

Let's check the given options:

• (a) $\frac{48}{60}$: Not simplified, and the sign is incorrect for the standard form based on the original number.
• (b) $\frac{-60}{48}$: Incorrect fraction and not simplified.
• (c) $\frac{-4}{5}$: Denominator is positive, and GCD(|-4|, 5) = 1. This is the standard form.
• (d) $\frac{-4}{-5}$: Denominator is negative. This is equivalent to $\frac{4}{5}$, which is positive, whereas the original number $\frac{-48}{60}$ is negative.

The standard form of $\frac{-48}{60}$ is $\frac{-4}{5}$.

The correct option is (c).

To Find: The rational number among the options that is equivalent to $\frac{4}{5}$.

Solution:

Two rational numbers are equivalent if one can be obtained from the other by multiplying or dividing both the numerator and the denominator by the same non-zero integer. In simpler terms, they represent the same value when simplified to their standard form.

The given rational number is $\frac{4}{5}$. This is already in its standard form because the GCD of 4 and 5 is 1, and the denominator is positive.

To find an equivalent fraction, we can multiply the numerator and denominator by the same integer. For example, multiplying by 2 gives $\frac{4 \times 2}{5 \times 2} = \frac{8}{10}$, multiplying by 3 gives $\frac{4 \times 3}{5 \times 3} = \frac{12}{15}$, and multiplying by 4 gives $\frac{4 \times 4}{5 \times 4} = \frac{16}{20}$.

Alternatively, we can check each option and simplify it to its standard form to see if it is equal to $\frac{4}{5}$.

• (a) $\frac{5}{4}$: This is the reciprocal of $\frac{4}{5}$, not equivalent unless $\frac{4}{5} = \frac{5}{4}$, which implies $4 \times 4 = 5 \times 5$, or $16 = 25$, which is false.
• (b) $\frac{16}{25}$: The GCD of 16 and 25 is 1. This fraction is already in standard form. $\frac{16}{25} \neq \frac{4}{5}$.
• (c) $\frac{16}{20}$: Find the GCD of 16 and 20. Factors of 16 are 1, 2, 4, 8, 16. Factors of 20 are 1, 2, 4, 5, 10, 20. The GCD is 4. Divide the numerator and denominator by 4: $\frac{16 \div 4}{20 \div 4} = \frac{4}{5}$. This is equal to the given rational number.
• (d) $\frac{15}{25}$: Find the GCD of 15 and 25. Factors of 15 are 1, 3, 5, 15. Factors of 25 are 1, 5, 25. The GCD is 5. Divide the numerator and denominator by 5: $\frac{15 \div 5}{25 \div 5} = \frac{3}{5}$. $\frac{3}{5} \neq \frac{4}{5}$.

The only option equivalent to $\frac{4}{5}$ is $\frac{16}{20}$.

The correct option is (c).

To Find: The number of rational numbers between any two distinct rational numbers.

Solution:

This question relates to the property of rational numbers called density. The set of rational numbers is dense, which means that between any two distinct rational numbers, there exists another rational number.

Let $r_1$ and $r_2$ be two distinct rational numbers such that $r_1 < r_2$.

One way to find a rational number between $r_1$ and $r_2$ is to calculate their average:

$r_{avg} = \frac{r_1 + r_2}{2}$

Since $r_1$ and $r_2$ are rational, their sum $(r_1 + r_2)$ is rational, and dividing a rational number by a non-zero rational number (2) results in a rational number. So, $r_{avg}$ is a rational number.

Furthermore, because $r_1 < r_2$, we have:

$r_1 + r_1 < r_1 + r_2 < r_2 + r_2$

$2r_1 < r_1 + r_2 < 2r_2$

Dividing by 2 (a positive number), the inequalities are preserved:

$\frac{2r_1}{2} < \frac{r_1 + r_2}{2} < \frac{2r_2}{2}$

$r_1 < r_{avg} < r_2$

So, $\frac{r_1 + r_2}{2}$ is a rational number strictly between $r_1$ and $r_2$.

Now consider the numbers $r_1$ and $r_{avg}$. Since they are distinct rational numbers, there must be another rational number between them, for instance, $\frac{r_1 + r_{avg}}{2}$. Similarly, there is a rational number between $r_{avg}$ and $r_2$. This process can be repeated infinitely many times, generating infinitely many distinct rational numbers between the original two rational numbers $r_1$ and $r_2$.

Therefore, there is an unlimited number of rational numbers between any two distinct rational numbers.

Let's consider the options:

• (a) 1: Incorrect, there is always at least one, but more than one.
• (b) 0: Incorrect, there is always at least one rational number between two distinct rational numbers.
• (c) unlimited: Correct, due to the density property.
• (d) 100: Incorrect, the number is not fixed and is much larger.

The correct option is (c).

To Find: The nature of the denominator in the standard form of a rational number.

Solution:

As discussed in a previous question (Question 4), the standard form of a rational number $\frac{p}{q}$ requires two main conditions:

1. The denominator $q$ must be a positive integer.

2. The numerator $p$ and the denominator $q$ must be coprime (their greatest common divisor must be 1).

The first condition explicitly states that the denominator in the standard form is always a positive integer.

For example, the standard form of $\frac{3}{-4}$ is obtained by making the denominator positive. We can multiply both the numerator and the denominator by -1:

$\frac{3 \times (-1)}{-4 \times (-1)} = \frac{-3}{4}$

In the standard form $\frac{-3}{4}$, the denominator is 4, which is a positive integer.

Let's look at the options:

• (a) 0: The denominator of any rational number cannot be zero.
• (b) negative integer: The standard form requires the denominator to be positive.
• (c) positive integer: This matches the definition of the standard form.
• (d) 1: The denominator can be 1 (e.g., $\frac{5}{1} = 5$), but it can also be any other positive integer (e.g., $\frac{2}{3}$, $\frac{-4}{7}$).

The most general and correct description among the options is that the denominator is always a positive integer.

The correct option is (c).

To Find: The quantity by which the numerator and denominator of a rational number are divided to reduce it to its standard form.

Solution:

Reducing a rational number to its standard form means expressing it as a fraction $\frac{p}{q}$ where the denominator $q$ is positive and the numerator $p$ and the denominator $q$ have no common factors other than 1 (i.e., they are relatively prime).

To achieve this, we first ensure the denominator is positive. If it's negative, we multiply both numerator and denominator by -1.

Next, we need to eliminate any common factors between the numerator and the denominator. The most efficient way to do this is to divide both the numerator and the denominator by the largest possible common factor. This largest common factor is known as the Greatest Common Divisor (GCD) or Highest Common Factor (HCF) of the absolute values of the numerator and the denominator.

For example, consider the rational number $\frac{12}{18}$.

The numerator is 12 and the denominator is 18. Both are positive.

We find the HCF of 12 and 18. The factors of 12 are 1, 2, 3, 4, 6, 12. The factors of 18 are 1, 2, 3, 6, 9, 18.

The common factors are 1, 2, 3, and 6. The greatest common factor (HCF) is 6.

To reduce $\frac{12}{18}$ to its standard form, we divide both the numerator and the denominator by their HCF, which is 6:

$\frac{12 \div 6}{18 \div 6} = \frac{2}{3}$

The resulting fraction $\frac{2}{3}$ is in standard form because the denominator 3 is positive and the GCD of 2 and 3 is 1.

Let's review the options:

• (a) LCM: Used for finding common denominators.
• (b) HCF: Used for simplifying fractions to their lowest terms (standard form).
• (c) product: Dividing by the product would oversimplify or change the value incorrectly.
• (d) multiple: Dividing by a multiple might simplify, but dividing by the HCF guarantees the standard form in one step.

Therefore, to reduce a rational number to its standard form, we divide its numerator and denominator by their HCF.

The correct option is (b).

To Find: The greatest number among the given options.

Solution:

To compare numbers, especially rational numbers and integers, it's helpful to visualize them on a number line or convert them to decimal form for easier comparison.

Let's evaluate the value of each option:

• (a) $\frac{-1}{2} = -0.5$
• (b) $0$
• (c) $\frac{1}{2} = 0.5$
• (d) – 2

Now, let's compare these values: -0.5, 0, 0.5, -2.

On a number line, numbers increase as you move from left to right.

Ordering the numbers from smallest to largest:

-2 (most negative) < -0.5 < 0 < 0.5 (most positive)

In the given options, the values are:

(a) -0.5

(b) 0

(c) 0.5

(d) -2

Comparing these values directly: -2 is the smallest, followed by -0.5, then 0, and finally 0.5 is the largest.

-2 < -0.5 < 0 < 0.5

Thus, the greatest number among the options is 0.5, which is represented by $\frac{1}{2}$.

Let's check the options again:

• (a) $\frac{-1}{2}$ (equals -0.5)
• (b) 0
• (c) $\frac{1}{2}$ (equals 0.5)
• (d) – 2

Comparing the values -0.5, 0, 0.5, and -2, the largest value is 0.5.

The correct option is (c).

To Fill in the Blank: Identify the type of rational number $-\frac{3}{8}$.

Solution:

A rational number $\frac{p}{q}$ is classified as:

• Positive if $p$ and $q$ have the same sign (both positive or both negative).
• Negative if $p$ and $q$ have different signs (one positive and one negative).

In the given rational number $-\frac{3}{8}$, the numerator is $-3$ and the denominator is $8$.

The numerator $-3$ is negative.

The denominator $8$ is positive.

Since the numerator and the denominator have different signs, the rational number $-\frac{3}{8}$ is a negative rational number.

The completed statement is: $-\frac{3}{8}$ is a negative rational number.

To Fill in the Blank: Identify the type of rational number 1.

Solution:

The number 1 can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. For example, $1 = \frac{1}{1}$. Therefore, 1 is a rational number.

To determine if it is positive or negative, we look at the signs of the numerator and denominator in its standard form, which is $\frac{1}{1}$.

The numerator is 1 (positive).

The denominator is 1 (positive).

Since the numerator and the denominator have the same sign (both positive), the rational number 1 is a positive rational number.

The completed statement is: 1 is a positive rational number.

To Fill in the Blank: Find the standard form of the rational number $\frac{-8}{-36}$.

Solution:

To find the standard form of a rational number $\frac{p}{q}$, we need to:

1. Make the denominator positive.

2. Divide both the numerator and the denominator by their Greatest Common Divisor (GCD).

The given rational number is $\frac{-8}{-36}$.

Step 1: Make the denominator positive. The denominator is -36. We multiply both the numerator and the denominator by -1:

$\frac{-8 \times (-1)}{-36 \times (-1)} = \frac{8}{36}$

Step 2: Find the GCD of the new numerator and denominator, which are 8 and 36. We find the GCD(8, 36).

Prime factorization of 8: $8 = 2^3$.

Prime factorization of 36: $36 = 2^2 \times 3^2$.

The common prime factor is $2^2$.

GCD(8, 36) = $2^2 = 4$.

Now, divide both the numerator and the denominator of $\frac{8}{36}$ by their GCD, 4:

$\frac{8 \div 4}{36 \div 4} = \frac{2}{9}$

The resulting fraction $\frac{2}{9}$ has a positive denominator (9) and the numerator (2) and denominator (9) have no common factors other than 1 (GCD(2, 9) = 1). So, this is the standard form.

Alternatively, we could first simplify the fraction $\frac{-8}{-36}$ by canceling the negative signs, giving $\frac{8}{36}$. Then proceed to find the GCD and simplify as above.

$\frac{\cancel{-}8}{\cancel{-}36} = \frac{8}{36}$

$\frac{\cancel{8}^2}{\cancel{36}_9} = \frac{2}{9}$ (dividing numerator and denominator by 4, the GCD)

The completed statement is: The standard form of $\frac{-8}{-36}$ is $\frac{2}{9}$.

To Fill in the Blank: Find the standard form of the rational number $\frac{8}{-24}$.

Solution:

To find the standard form of $\frac{8}{-24}$, we follow the two steps:

1. Make the denominator positive.

2. Divide the numerator and denominator by their GCD.

The given rational number is $\frac{8}{-24}$.

Step 1: Make the denominator positive. The denominator is -24. Multiply both parts by -1:

$\frac{8 \times (-1)}{-24 \times (-1)} = \frac{-8}{24}$

Step 2: Find the GCD of the absolute values of the numerator and denominator, which are $|-8|$ and $|24|$, so GCD(8, 24).

Prime factorization of 8: $8 = 2^3$.

Prime factorization of 24: $24 = 2^3 \times 3$.

The common prime factor is $2^3$.

GCD(8, 24) = $2^3 = 8$.

Now, divide both the numerator and the denominator of $\frac{-8}{24}$ by their GCD, 8:

$\frac{-8 \div 8}{24 \div 8} = \frac{-1}{3}$

The resulting fraction $\frac{-1}{3}$ has a positive denominator (3) and the numerator (-1) and denominator (3) have no common factors other than 1 (GCD(|-1|, 3) = GCD(1, 3) = 1). This is the standard form.

Alternatively, we could simplify the original fraction $\frac{8}{-24}$ directly by dividing by 8, the GCD of 8 and 24, and placing the negative sign in the numerator or in front of the fraction:

$\frac{\cancel{8}^1}{\cancel{-24}_{-3}} = \frac{1}{-3} = -\frac{1}{3} = \frac{-1}{3}$

The completed statement is: The standard form of $\frac{8}{-24}$ is $\frac{-1}{3}$.

To Fill in the Blank: Determine the position of $\frac{-1}{2}$ relative to zero on a number line.

Solution:

On a standard horizontal number line:

• Positive numbers are located to the right of zero.
• Negative numbers are located to the left of zero.
• Zero is the point of origin.

The given rational number is $\frac{-1}{2}$.

Let's evaluate its value: $\frac{-1}{2} = -0.5$.

Since -0.5 is a negative number (it is less than zero), it is located to the left of zero on the number line.

The completed statement is: On a number line, $\frac{-1}{2}$ is to the left of zero (0).

To Fill in the Blank: Determine the position of $\frac{4}{3}$ relative to zero on a number line.

Solution:

As established previously, on a number line, positive numbers are to the right of zero, and negative numbers are to the left of zero.

The given rational number is $\frac{4}{3}$.

To determine its sign, we look at the numerator and the denominator. The numerator is 4 (positive) and the denominator is 3 (positive). Since both have the same sign, the rational number $\frac{4}{3}$ is positive.

Its decimal value is approximately $1.333...$, which is greater than 0.

Since $\frac{4}{3} > 0$, it is located to the right of zero on the number line.

The completed statement is: On a number line, $\frac{4}{3}$ is to the right of zero (0).

To Fill in the Blank: Compare the two rational numbers $-\frac{1}{2}$ and $\frac{1}{5}$.

Solution:

We need to compare $-\frac{1}{2}$ and $\frac{1}{5}$.

One number is negative ($-\frac{1}{2}$) and the other number is positive ($\frac{1}{5}$).

On the number line, all negative numbers are to the left of all positive numbers. Therefore, any negative number is less than any positive number.

So, $-\frac{1}{2}$ is less than $\frac{1}{5}$.

In terms of inequality: $-\frac{1}{2} < \frac{1}{5}$.

Alternatively, we can convert them to decimals:

$-\frac{1}{2} = -0.5$

$\frac{1}{5} = 0.2$

Comparing -0.5 and 0.2, we see that $-0.5 < 0.2$.

So, $-\frac{1}{2}$ is less than $\frac{1}{5}$.

The completed statement is: $-\frac{1}{2}$ is less than $\frac{1}{5}$.

To Fill in the Blank: Compare the rational number $-\frac{3}{5}$ with 0.

Solution:

We need to compare $-\frac{3}{5}$ and 0.

The number $-\frac{3}{5}$ is a negative rational number because the numerator -3 is negative and the denominator 5 is positive (they have different signs).

On the number line, all negative numbers are located to the left of zero.

Therefore, any negative number is less than 0.

So, $-\frac{3}{5}$ is less than 0.

In terms of inequality: $-\frac{3}{5} < 0$.

Alternatively, we can convert $-\frac{3}{5}$ to decimal form: $-\frac{3}{5} = -0.6$.

Comparing -0.6 and 0, we see that $-0.6 < 0$.

So, $-\frac{3}{5}$ is less than 0.

The completed statement is: $-\frac{3}{5}$ is less than 0.

To Fill in the Blank: Determine the relationship between the rational numbers $\frac{-16}{24}$ and $\frac{20}{-16}$.

Solution:

We need to determine if the two rational numbers are equivalent or different.

Let's find the standard form for each rational number.

For the first rational number, $\frac{-16}{24}$:

The denominator 24 is positive. We need to simplify the fraction by dividing the numerator and denominator by their GCD. Find GCD(16, 24).

Prime factorization of 16: $16 = 2^4$.

Prime factorization of 24: $24 = 2^3 \times 3$.

GCD(16, 24) = $2^3 = 8$.

Divide the numerator and denominator of $\frac{-16}{24}$ by 8:

$\frac{-16 \div 8}{24 \div 8} = \frac{-2}{3}$

The standard form of $\frac{-16}{24}$ is $\frac{-2}{3}$.

For the second rational number, $\frac{20}{-16}$:

Step 1: Make the denominator positive. Multiply numerator and denominator by -1:

$\frac{20 \times (-1)}{-16 \times (-1)} = \frac{-20}{16}$

Step 2: Find the GCD of the new numerator and denominator, $|-20|$ and $|16|$, so GCD(20, 16).

Prime factorization of 20: $20 = 2^2 \times 5$.

Prime factorization of 16: $16 = 2^4$.

GCD(20, 16) = $2^2 = 4$.

Divide the numerator and denominator of $\frac{-20}{16}$ by 4:

$\frac{-20 \div 4}{16 \div 4} = \frac{-5}{4}$

The standard form of $\frac{20}{-16}$ is $\frac{-5}{4}$.

Comparing the standard forms, we have $\frac{-2}{3}$ and $\frac{-5}{4}$.

These two standard forms are different ($\frac{-2}{3} \neq \frac{-5}{4}$). Therefore, the original rational numbers are not equivalent.

The completed statement is: $\frac{-16}{24}$ and $\frac{20}{-16}$ represent different rational numbers.

To Fill in the Blank: Determine the relationship between the rational numbers $\frac{-27}{45}$ and $\frac{-3}{5}$.

Solution:

We need to check if the two rational numbers are equivalent.

The second rational number, $\frac{-3}{5}$, is already in standard form: the denominator 5 is positive, and the GCD of |-3| and 5 (which is GCD(3, 5)) is 1.

Now, let's reduce the first rational number, $\frac{-27}{45}$, to its standard form.

The denominator 45 is positive. We need to find the GCD of |-27| and 45, which is GCD(27, 45).

Prime factorization of 27: $27 = 3^3$.

Prime factorization of 45: $45 = 3^2 \times 5$.

The common prime factor is $3^2$.

GCD(27, 45) = $3^2 = 9$.

Now, divide the numerator and the denominator of $\frac{-27}{45}$ by their GCD, 9:

$\frac{-27 \div 9}{45 \div 9} = \frac{-3}{5}$

The standard form of $\frac{-27}{45}$ is $\frac{-3}{5}$.

Comparing the standard form of $\frac{-27}{45}$ ($\frac{-3}{5}$) with the second given rational number ($\frac{-3}{5}$), we see that they are the same.

Therefore, the rational numbers $\frac{-27}{45}$ and $\frac{-3}{5}$ are equivalent.

The completed statement is: $\frac{-27}{45}$ and $\frac{-3}{5}$ represent equivalent rational numbers.

To Fill in the Blank: Find the additive inverse of the rational number $\frac{2}{3}$.

Solution:

The additive inverse of a number $x$ is the number that, when added to $x$, results in a sum of 0. If the number is $x$, its additive inverse is $-x$.

The given rational number is $\frac{2}{3}$.

To find its additive inverse, we simply negate the number.

Additive inverse of $\frac{2}{3} = - \left( \frac{2}{3} \right) = \frac{-2}{3}$.

Let's verify this by adding the number and its additive inverse:

$\frac{2}{3} + \left( -\frac{2}{3} \right) = \frac{2}{3} - \frac{2}{3} = 0$

Since the sum is 0, $\frac{-2}{3}$ is indeed the additive inverse of $\frac{2}{3}$.

The completed statement is: Additive inverse of $\frac{2}{3}$ is $\frac{-2}{3}$.

To Fill in the Blank: Calculate the sum of the rational numbers $\frac{-3}{5}$ and $\frac{2}{5}$.

Solution:

We are asked to add two rational numbers with the same denominator:

$\frac{-3}{5} + \frac{2}{5}$

When adding rational numbers with a common denominator, we add the numerators and keep the denominator the same.

Sum = $\frac{\text{Numerator 1} + \text{Numerator 2}}{\text{Common Denominator}}$

Sum = $\frac{-3 + 2}{5}$

Now, perform the addition in the numerator:

$-3 + 2 = -1$

So, the sum is $\frac{-1}{5}$.

The completed statement is: $\frac{-3}{5}$ + $\frac{2}{5}$ = $\frac{-1}{5}$.

To Fill in the Blank: Calculate the sum of the rational numbers $\frac{-5}{6}$ and $\frac{-1}{6}$.

Solution:

We need to add two rational numbers with a common denominator:

$\frac{-5}{6} + \frac{-1}{6}$

Since the denominators are the same, we add the numerators and keep the common denominator.

Sum = $\frac{\text{Numerator 1} + \text{Numerator 2}}{\text{Common Denominator}}$

Sum = $\frac{-5 + (-1)}{6}$

Adding the numerators: $-5 + (-1) = -5 - 1 = -6$.

So, the sum is $\frac{-6}{6}$.

This fraction can be simplified:

$\frac{-6}{6} = -1$

The completed statement is: $\frac{-5}{6}$ + $\frac{-1}{6}$ = -1.

To Fill in the Blank: Calculate the product of the rational numbers $\frac{3}{4}$ and $\frac{-2}{3}$.

Solution:

To multiply rational numbers, we multiply the numerators together and multiply the denominators together.

Product = $\frac{\text{Numerator 1} \times \text{Numerator 2}}{\text{Denominator 1} \times \text{Denominator 2}}$

Product = $\frac{3 \times (-2)}{4 \times 3}$

Perform the multiplications:

Numerator product = $3 \times (-2) = -6$

Denominator product = $4 \times 3 = 12$

So, the product is $\frac{-6}{12}$.

This fraction can be simplified to its standard form. Find the GCD of |-6| and 12, which is GCD(6, 12). The GCD is 6.

Divide the numerator and denominator by 6:

$\frac{-6 \div 6}{12 \div 6} = \frac{-1}{2}$

Alternatively, we can cancel common factors before multiplying:

$\frac{\cancel{3}^1}{4} \times \frac{-2}{\cancel{3}^1} = \frac{1 \times (-2)}{4 \times 1} = \frac{-2}{4}$

Now, simplify $\frac{-2}{4}$. The GCD of |-2| and 4 is GCD(2, 4), which is 2.

Divide the numerator and denominator by 2:

$\frac{-2 \div 2}{4 \div 2} = \frac{-1}{2}$

Both methods yield the same result.

The completed statement is: $\frac{3}{4}$ × $\left( \frac{-2}{3} \right)$ = $\frac{-1}{2}$.

To Fill in the Blank: Calculate the product of the rational numbers $\frac{-5}{3}$ and $\frac{-3}{5}$.

Solution:

We are asked to find the product of $\frac{-5}{3}$ and $\frac{-3}{5}$.

Product = $\frac{\text{Numerator 1} \times \text{Numerator 2}}{\text{Denominator 1} \times \text{Denominator 2}}$

Product = $\frac{-5 \times (-3)}{3 \times 5}$

Perform the multiplications:

Numerator product = $(-5) \times (-3) = 15$ (since negative times negative is positive)

Denominator product = $3 \times 5 = 15$

So, the product is $\frac{15}{15}$.

This fraction simplifies to 1.

Alternatively, we can cancel common factors before multiplying:

$\frac{\cancel{-5}^{-1}}{\cancel{3}^1} \times \frac{\cancel{-3}^{-1}}{\cancel{5}^1} = \frac{-1 \times (-1)}{1 \times 1} = \frac{1}{1} = 1$

Notice that $\frac{-3}{5}$ is the reciprocal of $\frac{-5}{3}$ (or vice versa). The product of a non-zero number and its reciprocal is always 1.

The completed statement is: $\frac{-5}{3}$ × $\left( \frac{-3}{5} \right)$ = 1.

To Fill in the Blank: Find the missing numerator in the equivalent rational number $\frac{-}{42}$, given that it is equal to $\frac{-6}{7}$.

Solution:

We are given the equation $\frac{-6}{7} = \frac{x}{42}$, where $x$ is the missing numerator.

To find $x$, we can observe how the denominator has changed from 7 to 42. The denominator has been multiplied by some factor.

$7 \times \text{factor} = 42$

Factor = $\frac{42}{7} = 6$

Since the denominator was multiplied by 6, the numerator must also be multiplied by the same factor to maintain equivalence.

$x = -6 \times \text{factor}$

$x = -6 \times 6$

$x = -36$

So, the missing numerator is -36.

The equivalent fraction is $\frac{-36}{42}$.

Alternatively, we can use cross-multiplication:

$\frac{-6}{7} = \frac{x}{42}$

$-6 \times 42 = 7 \times x$

$-252 = 7x$

Divide both sides by 7:

$x = \frac{-252}{7}$

$x = -36$

Both methods give the same missing numerator.

The completed statement is: $\frac{-6}{7}$ = $\frac{\mathbf{-36}}{42}$.

To Fill in the Blank: Find the missing denominator in the equivalent rational number $\frac{6}{-}$, given that it is equal to $\frac{1}{2}$.

Solution:

We are given the equation $\frac{1}{2} = \frac{6}{y}$, where $y$ is the missing denominator.

We can observe how the numerator has changed from 1 to 6. The numerator has been multiplied by some factor.

$1 \times \text{factor} = 6$

Factor = $\frac{6}{1} = 6$

Since the numerator was multiplied by 6, the denominator must also be multiplied by the same factor to maintain equivalence.

$y = 2 \times \text{factor}$

$y = 2 \times 6$

$y = 12$

So, the missing denominator is 12.

The equivalent fraction is $\frac{6}{12}$.

Alternatively, we can use cross-multiplication:

$\frac{1}{2} = \frac{6}{y}$

$1 \times y = 2 \times 6$

$y = 12$

Both methods yield the same missing denominator.

The completed statement is: $\frac{1}{2}$ = $\frac{6}{\mathbf{12}}$.

To Fill in the Blank: Calculate the difference between the rational numbers $\frac{-2}{9}$ and $\frac{7}{9}$.

Solution:

We need to subtract one rational number from another, and they have the same denominator:

$\frac{-2}{9} - \frac{7}{9}$

When subtracting rational numbers with a common denominator, we subtract the numerators and keep the denominator the same.

Difference = $\frac{\text{Numerator 1} - \text{Numerator 2}}{\text{Common Denominator}}$

Difference = $\frac{-2 - 7}{9}$

Now, perform the subtraction in the numerator:

$-2 - 7 = -9$

So, the difference is $\frac{-9}{9}$.

This fraction can be simplified:

$\frac{-9}{9} = -1$

The completed statement is: $\frac{-2}{9}$ - $\frac{7}{9}$ = -1.

To Fill in the Box: Compare the rational numbers $\frac{7}{-8}$ and $\frac{8}{9}$ using $>, <,$ or $=$.

Solution:

We are comparing $\frac{7}{-8}$ and $\frac{8}{9}$.

First, let's write $\frac{7}{-8}$ in its standard form. Make the denominator positive by multiplying the numerator and denominator by -1:

$\frac{7}{-8} = \frac{7 \times (-1)}{-8 \times (-1)} = \frac{-7}{8}$

Now we need to compare $\frac{-7}{8}$ and $\frac{8}{9}$.

The rational number $\frac{-7}{8}$ has a negative numerator and a positive denominator, so it is a negative rational number.

The rational number $\frac{8}{9}$ has a positive numerator and a positive denominator, so it is a positive rational number.

Any negative number is always less than any positive number.

Therefore, $\frac{-7}{8} < \frac{8}{9}$.

This means $\frac{7}{-8} < \frac{8}{9}$.

Alternatively, we can find a common denominator or convert to decimals, but comparing signs is the quickest method here.

Common denominator method (LCM of 8 and 9 is 72):

$\frac{-7}{8} = \frac{-7 \times 9}{8 \times 9} = \frac{-63}{72}$

$\frac{8}{9} = \frac{8 \times 8}{9 \times 8} = \frac{64}{72}$

Comparing $\frac{-63}{72}$ and $\frac{64}{72}$, since $-63 < 64$, we have $\frac{-63}{72} < \frac{64}{72}$.

Thus, $\frac{-7}{8} < \frac{8}{9}$.

The completed comparison is: $\frac{7}{-8} < \frac{8}{9}$.

To Fill in the Box: Compare the rational numbers $\frac{3}{7}$ and $\frac{-5}{6}$ using $>, <,$ or $=$.

Solution:

We are comparing $\frac{3}{7}$ and $\frac{-5}{6}$.

Let's determine the sign of each rational number.

For $\frac{3}{7}$: The numerator is 3 (positive) and the denominator is 7 (positive). Both signs are the same, so $\frac{3}{7}$ is a positive rational number.

For $\frac{-5}{6}$: The numerator is -5 (negative) and the denominator is 6 (positive). The signs are different, so $\frac{-5}{6}$ is a negative rational number.

Since $\frac{3}{7}$ is a positive number and $\frac{-5}{6}$ is a negative number, any positive number is always greater than any negative number.

Therefore, $\frac{3}{7} > \frac{-5}{6}$.

The completed comparison is: $\frac{3}{7} > \frac{-5}{6}$.

To Fill in the Box: Compare the rational numbers $\frac{5}{6}$ and $\frac{8}{4}$ using $>, <,$ or $=$.

Solution:

We need to compare $\frac{5}{6}$ and $\frac{8}{4}$.

Let's first simplify the second rational number $\frac{8}{4}$.

$\frac{8}{4} = 2$

Now we compare $\frac{5}{6}$ with 2.

We can convert $\frac{5}{6}$ to a decimal or compare it to a number relative to 2.

$\frac{5}{6}$ is a proper fraction, meaning the numerator (5) is less than the denominator (6). Therefore, $\frac{5}{6}$ is less than 1.

Since $\frac{5}{6} < 1$ and $2 > 1$, it follows that $\frac{5}{6} < 2$.

Therefore, $\frac{5}{6} < \frac{8}{4}$.

Alternatively, using a common denominator (the denominator of 2 can be considered 1, LCM of 6 and 1 is 6):

$\frac{5}{6}$

$2 = \frac{2}{1} = \frac{2 \times 6}{1 \times 6} = \frac{12}{6}$

Comparing $\frac{5}{6}$ and $\frac{12}{6}$, since $5 < 12$, we have $\frac{5}{6} < \frac{12}{6}$.

Thus, $\frac{5}{6} < \frac{8}{4}$.

The completed comparison is: $\frac{5}{6} < \frac{8}{4}$.

To Fill in the Box: Compare the rational numbers $\frac{-9}{7}$ and $\frac{4}{-7}$ using $>, <,$ or $=$.

Solution:

We need to compare $\frac{-9}{7}$ and $\frac{4}{-7}$.

First, let's write the second rational number in its standard form by making the denominator positive. Multiply the numerator and denominator by -1:

$\frac{4}{-7} = \frac{4 \times (-1)}{-7 \times (-1)} = \frac{-4}{7}$

Now we need to compare $\frac{-9}{7}$ and $\frac{-4}{7}$.

Both rational numbers have the same positive denominator (7). To compare fractions with the same positive denominator, we just need to compare their numerators.

We are comparing -9 and -4.

On the number line, -9 is to the left of -4.

Therefore, $-9 < -4$.

Since the numerator of the first fraction (-9) is less than the numerator of the second fraction (-4), and the denominators are positive and equal, the first fraction is less than the second fraction.

So, $\frac{-9}{7} < \frac{-4}{7}$.

This means $\frac{-9}{7} < \frac{4}{-7}$.

The completed comparison is: $\frac{-9}{7} < \frac{4}{-7}$.

To Fill in the Box: Compare the rational numbers $\frac{8}{8}$ and $\frac{2}{2}$ using $>, <,$ or $=$.

Solution:

We need to compare $\frac{8}{8}$ and $\frac{2}{2}$.

Let's simplify each rational number:

$\frac{8}{8}$: The numerator and denominator are equal and non-zero. $\frac{8}{8} = 1$.

$\frac{2}{2}$: The numerator and denominator are equal and non-zero. $\frac{2}{2} = 1$.

Now we are comparing 1 and 1.

Since both values are equal to 1, they are equal.

$1 = 1$

Therefore, $\frac{8}{8} = \frac{2}{2}$.

The completed comparison is: $\frac{8}{8} = \frac{2}{2}$.

To Fill in the Blank: Identify the number whose reciprocal does not exist.

Solution:

The reciprocal of a non-zero rational number $\frac{p}{q}$ is defined as $\frac{q}{p}$. This is the number that, when multiplied by the original number, gives a product of 1.

If the original number is $x$, its reciprocal is $\frac{1}{x}$. For the reciprocal to exist, the denominator of $\frac{1}{x}$ cannot be zero.

The reciprocal is $\frac{1}{x}$. The denominator is $x$.

The reciprocal does not exist if the denominator is zero, i.e., if $x = 0$.

Let's consider the number 0. Can we find a number $y$ such that $0 \times y = 1$? There is no such real number $y$. Multiplication by 0 always results in 0.

Therefore, the reciprocal of 0 is undefined or does not exist.

The completed statement is: The reciprocal of 0 does not exist.

To Fill in the Blank: Find the reciprocal of the number 1.

Solution:

The reciprocal of a non-zero number $x$ is $\frac{1}{x}$.

We need to find the reciprocal of 1.

Reciprocal of $1 = \frac{1}{1}$

$\frac{1}{1} = 1$

Alternatively, we can think of the definition of reciprocal: The reciprocal of a number is the number that, when multiplied by the original number, gives 1.

We are looking for a number $y$ such that $1 \times y = 1$.

The only number that satisfies this equation is $y = 1$.

Thus, the reciprocal of 1 is 1.

The completed statement is: The reciprocal of 1 is 1.

To Fill in the Blank: Calculate the result of the division $\frac{-3}{7}$ ÷ $\left( \frac{-7}{3} \right)$.

Solution:

Dividing by a rational number is the same as multiplying by its reciprocal.

The expression is $\frac{-3}{7}$ ÷ $\left( \frac{-7}{3} \right)$.

The number we are dividing by is $\frac{-7}{3}$.

The reciprocal of $\frac{-7}{3}$ is obtained by flipping the numerator and the denominator, keeping the sign with the number it belongs to (usually the numerator for standard form):

Reciprocal of $\frac{-7}{3}$ is $\frac{3}{-7}$, which is equivalent to $\frac{-3}{7}$.

Now, we rewrite the division as multiplication by the reciprocal:

$\frac{-3}{7}$ ÷ $\left( \frac{-7}{3} \right) = \frac{-3}{7} \times \left( \text{Reciprocal of } \frac{-7}{3} \right)$

$\frac{-3}{7}$ ÷ $\left( \frac{-7}{3} \right) = \frac{-3}{7} \times \frac{-3}{7}$

Now, perform the multiplication:

Product = $\frac{\text{Numerator 1} \times \text{Numerator 2}}{\text{Denominator 1} \times \text{Denominator 2}}$

Product = $\frac{(-3) \times (-3)}{7 \times 7}$

Numerator product = $(-3) \times (-3) = 9$

Denominator product = $7 \times 7 = 49$

So, the result is $\frac{9}{49}$.

The completed statement is: $\frac{-3}{7}$ ÷ $\left( \frac{-7}{3} \right)$ = $\frac{9}{49}$.

To Fill in the Blank: Calculate the result of the division 0 ÷ $\left( \frac{-5}{6} \right)$.

Solution:

We are asked to divide 0 by a non-zero rational number $\frac{-5}{6}$.

Division by a non-zero number is defined as multiplication by the reciprocal of the divisor.

The divisor is $\frac{-5}{6}$. Its reciprocal is $\frac{6}{-5}$, which is equivalent to $\frac{-6}{5}$.

So, $0 \div \left( \frac{-5}{6} \right) = 0 \times \left( \text{Reciprocal of } \frac{-5}{6} \right)$

$0 \div \left( \frac{-5}{6} \right) = 0 \times \left( \frac{-6}{5} \right)$

When we multiply any number (including a rational number) by zero, the result is always zero.

$0 \times \frac{-6}{5} = 0$

Therefore, 0 divided by any non-zero number is 0.

The completed statement is: 0 ÷ $\left( \frac{-5}{6} \right)$ = 0.

To Fill in the Blank: Calculate the result of the multiplication 0 × $\left( \frac{-5}{6} \right)$.

Solution:

We are asked to multiply 0 by the rational number $\frac{-5}{6}$.

In mathematics, the product of any number and zero is always zero.

This property holds true for all types of numbers, including rational numbers.

So, $0 \times \frac{-5}{6} = 0$.

The completed statement is: 0 × $\left( \frac{-5}{6} \right)$ = 0.

To Fill in the Blank: Find the missing number that, when multiplied by $\frac{-2}{5}$, gives a product of 1.

Solution:

The problem asks for a number $x$ such that $x \times \left( \frac{-2}{5} \right) = 1$.

By definition, the number that, when multiplied by a given non-zero number, results in a product of 1 is called the reciprocal or multiplicative inverse of the given number.

The given number is $\frac{-2}{5}$.

To find the reciprocal of $\frac{-2}{5}$, we swap the numerator and the denominator. The sign stays with the numerator (or the number it was originally associated with).

Reciprocal of $\frac{-2}{5}$ is $\frac{5}{-2}$.

To write this in standard form, we make the denominator positive by multiplying the numerator and denominator by -1:

$\frac{5}{-2} = \frac{5 \times (-1)}{-2 \times (-1)} = \frac{-5}{2}$

Let's verify the multiplication:

$\frac{-5}{2} \times \frac{-2}{5} = \frac{(-5) \times (-2)}{2 \times 5} = \frac{10}{10} = 1$

The product is 1, so the missing number is the reciprocal of $\frac{-2}{5}$, which is $\frac{-5}{2}$.

The completed statement is: $\frac{-5}{2}$ × $\left( \frac{-2}{5} \right)$ = 1.

To Fill in the Blank: Find the standard form of the rational number -1.

Solution:

The standard form of a rational number $\frac{p}{q}$ requires two conditions:

1. The denominator $q$ is a positive integer.

2. The numerator $p$ and the denominator $q$ have no common factor other than 1 (GCD(|p|, |q|) = 1).

The number given is -1.

We can express -1 as a rational number in the form $\frac{p}{q}$. The simplest way to do this is $\frac{-1}{1}$.

Let's check if $\frac{-1}{1}$ is in standard form:

1. Is the denominator positive? Yes, the denominator is 1, which is a positive integer.

2. Is the GCD of the absolute values of the numerator and denominator equal to 1? The absolute value of the numerator is |-1| = 1. The absolute value of the denominator is |1| = 1. GCD(1, 1) = 1. Yes, they have no common factors other than 1.

Since both conditions are met, $\frac{-1}{1}$ is the standard form of -1 as a rational number.

Other forms like $\frac{-2}{2}$ or $\frac{-3}{3}$ are equal to -1, but they are not in standard form because the GCD of the numerator and denominator is greater than 1 (GCD(2, 2) = 2, GCD(3, 3) = 3, etc.).

The completed statement is: The standard form of rational number –1 is $\frac{-1}{1}$.

To Fill in the Blank: Complete the equivalent fraction when the numerator is divided by a common divisor $m$.

Solution:

The property of equivalent rational numbers states that if you multiply or divide both the numerator and the denominator of a rational number by the same non-zero integer, the value of the rational number remains unchanged.

We are given that $m$ is a common divisor of $a$ and $b$. This means that $a$ is divisible by $m$ and $b$ is divisible by $m$. Since $m$ is a common divisor, $m$ must be non-zero (otherwise, the rational number $\frac{a}{b}$ itself might not be well-defined if $b=0$, and division by zero is undefined).

The expression is $\frac{a}{b} = \frac{a \;÷\; m}{----}$.

To maintain the equivalence of the fraction, if the numerator $a$ is divided by a non-zero number $m$, the denominator $b$ must also be divided by the same non-zero number $m$.

So, the missing part in the denominator is $b \;÷\; m$.

Thus, $\frac{a}{b} = \frac{a \;÷\; m}{b \;÷\; m}$.

This is the fundamental principle used to simplify rational numbers by dividing out common factors.

The completed statement is: If m is a common divisor of a and b, then $\frac{a}{b}$ = $\frac{a \;÷\; m}{\mathbf{b \;÷\; m}}$.

To Fill in the Blanks: Determine the type (positive or negative) of the rational numbers $\frac{p}{q}$ and $\frac{p}{-q}$, given that $p$ and $q$ are positive integers.

Solution:

We are given that $p$ and $q$ are positive integers.

Consider the first rational number, $\frac{p}{q}$.

The numerator is $p$, which is a positive integer.

The denominator is $q$, which is a positive integer.

Since both the numerator and the denominator are positive, they have the same sign. A rational number with a numerator and denominator of the same sign is a positive rational number.

Consider the second rational number, $\frac{p}{-q}$.

The numerator is $p$, which is a positive integer.

The denominator is $-q$. Since $q$ is a positive integer, $-q$ is a negative integer.

The numerator ($p$) is positive, and the denominator ($-q$) is negative. Since the numerator and the denominator have different signs, the rational number $\frac{p}{-q}$ is a negative rational number. Note that $\frac{p}{-q} = \frac{-p}{q}$ by multiplying the numerator and denominator by -1.

The completed statement is: If p and q are positive integers, then $\frac{p}{q}$ is a positive rational number and $\frac{p}{-q}$ is a negative rational number.

To Fill in the Blank: Identify the form that determines if two rational numbers are equivalent.

Solution:

Two rational numbers are equivalent if they represent the same value. For example, $\frac{1}{2}$, $\frac{2}{4}$, $\frac{3}{6}$, and $\frac{-5}{-10}$ are all equivalent rational numbers.

The most reliable way to determine if two rational numbers are equivalent is to reduce both of them to their standard form.

The standard form of a rational number is the form $\frac{p}{q}$ where $q$ is a positive integer and the greatest common divisor (GCD) of $|p|$ and $q$ is 1.

Let's check the examples:

• $\frac{1}{2}$ is in standard form (denominator is positive, GCD(1, 2) = 1).
• $\frac{2}{4}$: GCD(2, 4) = 2. $\frac{2 \div 2}{4 \div 2} = \frac{1}{2}$.
• $\frac{3}{6}$: GCD(3, 6) = 3. $\frac{3 \div 3}{6 \div 3} = \frac{1}{2}$.
• $\frac{-5}{-10}$: Denominator is negative. $\frac{-5 \times (-1)}{-10 \times (-1)} = \frac{5}{10}$. GCD(5, 10) = 5. $\frac{5 \div 5}{10 \div 5} = \frac{1}{2}$.

All equivalent rational numbers reduce to the same standard form.

Therefore, two rational numbers are equivalent if they have the same standard form.

The completed statement is: Two rational numbers are said to be equivalent or equal, if they have the same standard form.

To Fill in the Blank: Identify the value that the denominator $q$ cannot be in the definition of a rational number $\frac{p}{q}$.

Solution:

The definition of a rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, and the denominator $q$ is not equal to zero.

This condition $q \neq 0$ is fundamental because division by zero is an undefined operation in mathematics.

The question directly asks for the value that $q$ cannot be.

Based on the definition of a rational number, $q$ cannot be 0.

The completed statement is: If $\frac{p}{q}$ is a rational number, then q cannot be 0.

To Determine: Whether the statement "Every natural number is a rational number but every rational number need not be a natural number" is True or False.

Solution:

Let's analyze the two parts of the statement separately.

Part 1: "Every natural number is a rational number."

Natural numbers are the counting numbers: 1, 2, 3, 4, ... The set of natural numbers is often denoted by $\mathbb{N}$.

A rational number is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Consider any natural number, say $n$. We can write $n$ as a fraction $\frac{n}{1}$. Since $n$ is an integer and 1 is a non-zero integer, $\frac{n}{1}$ is a rational number by definition.

Thus, every natural number can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers ($p=n$, $q=1$) and $q \neq 0$. This part of the statement is True.

Part 2: "Every rational number need not be a natural number."

This means there exists at least one rational number that is not a natural number.

Consider the rational number $\frac{1}{2}$. It is in the form $\frac{p}{q}$ where $p=1$ and $q=2$ are integers and $q \neq 0$. So, $\frac{1}{2}$ is a rational number.

Is $\frac{1}{2}$ a natural number? Natural numbers are 1, 2, 3, ... The number $\frac{1}{2}$ (or 0.5) is not in this set.

Other examples of rational numbers that are not natural numbers include $\frac{-3}{4}$, 0, etc.

Thus, it is true that every rational number need not be a natural number. This part of the statement is True.

Since both parts of the statement are True, the entire statement is True.

The statement is True.

To Determine: Whether the statement "Zero is a rational number" is True or False.

Solution:

A rational number is defined as a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

We need to check if the number zero can be expressed in this form.

Consider expressing 0 as a fraction. We can write 0 as $\frac{0}{1}$.

Here, $p = 0$ and $q = 1$. Both $p$ and $q$ are integers. The denominator $q=1$ is not equal to zero.

Since 0 can be expressed in the form $\frac{p}{q}$ where $p=0$ and $q=1$ (which are integers and $q \neq 0$), 0 fits the definition of a rational number.

We can also express 0 as $\frac{0}{2}$, $\frac{0}{-5}$, etc. In all these cases, the numerator is an integer (0), the denominator is a non-zero integer, satisfying the definition.

The statement "Zero is a rational number" is True.

The statement is True.

To Determine: Whether the statement "Every integer is a rational number but every rational number need not be an integer" is True or False.

Solution:

Let's break down the statement into two parts.

Part 1: "Every integer is a rational number."

Integers are whole numbers and their negatives: ..., -3, -2, -1, 0, 1, 2, 3, ... The set of integers is denoted by $\mathbb{Z}$.

A rational number is defined as $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$.

Consider any integer, say $z$. We can express any integer $z$ as a fraction $\frac{z}{1}$. Here, the numerator $p=z$ is an integer, and the denominator $q=1$ is a non-zero integer. Thus, every integer can be written in the form $\frac{p}{q}$ satisfying the conditions for a rational number.

This part of the statement is True.

Part 2: "Every rational number need not be an integer."

This means there exists at least one rational number that is not an integer.

Consider the rational number $\frac{1}{2}$. It is in the form $\frac{p}{q}$ where $p=1$ and $q=2$ are integers and $q \neq 0$. So, $\frac{1}{2}$ is a rational number.

Is $\frac{1}{2}$ an integer? Integers are whole numbers and their negatives. $\frac{1}{2}$ (or 0.5) is not in the set of integers.

Other examples include $\frac{-3}{4}$, $\frac{5}{3}$, etc. These are rational numbers but not integers.

Thus, it is true that every rational number need not be an integer. This part of the statement is True.

Since both parts of the statement are True, the entire statement is True.

The statement is True.

To Determine: Whether the statement "Every negative integer is not a negative rational number" is True or False.

Solution:

Let's analyze the statement carefully. It claims that every negative integer is *not* a negative rational number. This is equivalent to saying that no negative integer is a negative rational number.

Consider a negative integer, for example, -3. According to the definition of rational numbers, -3 can be written as $\frac{-3}{1}$. Here, the numerator is -3 (negative) and the denominator is 1 (positive).

A rational number $\frac{p}{q}$ is negative if the numerator $p$ and the denominator $q$ have different signs.

In the case of $\frac{-3}{1}$, the numerator (-3) is negative and the denominator (1) is positive. They have different signs. Therefore, $\frac{-3}{1}$ is a negative rational number.

So, the negative integer -3 is a negative rational number.

This contradicts the statement "Every negative integer is not a negative rational number".

Let's take another negative integer, say -5. It can be written as $\frac{-5}{1}$. Numerator is -5 (negative), denominator is 1 (positive). Different signs, so $\frac{-5}{1}$ is a negative rational number.

In general, any negative integer $-n$ (where $n$ is a positive integer) can be written as $\frac{-n}{1}$. Here, the numerator $-n$ is negative and the denominator 1 is positive. Thus, $\frac{-n}{1}$ is a negative rational number.

So, every negative integer is indeed a negative rational number (as it can be written with a negative numerator and a positive denominator). The statement claims the opposite.

The statement "Every negative integer is not a negative rational number" is False.

The statement is False.

To Determine: Whether the statement "If $\frac{p}{q}$ is a rational number and m is a non-zero integer, then $\frac{p}{q} = \frac{p \times m}{ p \times m}$" is True or False.

Solution:

The property of equivalent rational numbers states that if you multiply both the numerator and the denominator of a rational number by the same non-zero integer, the value of the rational number remains unchanged.

So, if $\frac{p}{q}$ is a rational number and $m$ is a non-zero integer, the equivalent fraction is obtained by multiplying the numerator by $m$ and the denominator by $m$:

$\frac{p}{q} = \frac{p \times m}{q \times m}$

The statement given is $\frac{p}{q} = \frac{p \times m}{ p \times m}$. This implies that the denominator is also being multiplied by $p$, not $q$.

Let's consider an example. Let $\frac{p}{q} = \frac{1}{2}$ and $m = 3$.

According to the property of equivalent fractions, $\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$.

According to the statement, $\frac{1}{2} = \frac{1 \times 3}{1 \times 3} = \frac{3}{3} = 1$.

Since $\frac{1}{2} \neq 1$, the statement is incorrect.

The correct statement for creating equivalent fractions is $\frac{p}{q} = \frac{p \times m}{q \times m}$. The given statement has the denominator incorrectly written as $p \times m$ instead of $q \times m$.

The statement "If $\frac{p}{q}$ is a rational number and m is a non-zero integer, then $\frac{p}{q}$ = $\frac{p × m}{ p × m}$" is False.

The statement is False.

To Determine: Whether the statement "If $\frac{p}{q}$ is a rational number and m is a non-zero common divisor of p and q, then $\frac{p}{q} = \frac{p \div m}{q \div m}$" is True or False.

Solution:

The property of equivalent rational numbers states that if you divide both the numerator and the denominator of a rational number by the same non-zero integer, the value of the rational number remains unchanged.

We are given that $\frac{p}{q}$ is a rational number, which means $p$ and $q$ are integers and $q \neq 0$.

We are also given that $m$ is a non-zero common divisor of $p$ and $q$. This means that $p$ is divisible by $m$ and $q$ is divisible by $m$, and $m \neq 0$.

The statement says that $\frac{p}{q}$ is equal to the fraction obtained by dividing the numerator $p$ by $m$ and the denominator $q$ by $m$, i.e., $\frac{p \div m}{q \div m}$.

This is precisely how we simplify fractions and find equivalent rational numbers. As long as $m$ is a non-zero number that divides both the numerator and the denominator, dividing both by $m$ yields an equivalent fraction.

For example, let $\frac{p}{q} = \frac{6}{10}$. Here $p=6$ and $q=10$. Let $m=2$, which is a non-zero common divisor of 6 and 10.

According to the statement: $\frac{6}{10} = \frac{6 \div 2}{10 \div 2} = \frac{3}{5}$.

The rational numbers $\frac{6}{10}$ and $\frac{3}{5}$ are indeed equivalent.

This property is fundamental for reducing rational numbers to their simplest or standard form (when $m$ is the GCD).

The statement "If $\frac{p}{q}$ is a rational number and m is a non-zero common divisor of p and q, then $\frac{p}{q} = \frac{p \div m}{q \div m}$" is True.

The statement is True.

To Determine: Whether the statement "In a rational number, denominator always has to be a non-zero integer" is True or False.

Solution:

The definition of a rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

The definition itself explicitly states the condition on the denominator $q$. The denominator must be an integer and it must not be equal to zero.

The phrase "non-zero integer" means an integer that is not 0 (i.e., any integer except 0, such as ..., -3, -2, -1, 1, 2, 3, ...).

The statement says the denominator always has to be a non-zero integer. This directly matches the condition $q \in \mathbb{Z}$ and $q \neq 0$ from the definition of a rational number $\frac{p}{q}$.

If the denominator were zero, the expression $\frac{p}{q}$ would be undefined, and thus would not represent a number, rational or otherwise.

The statement "In a rational number, denominator always has to be a non-zero integer" is True.

The statement is True.

To Determine: Whether the statement "If $\frac{p}{q}$ is a rational number and m is a non-zero integer, then $\frac{p \times m}{q \times m}$ is a rational number not equivalent to $\frac{p}{q}$" is True or False.

Solution:

We are given that $\frac{p}{q}$ is a rational number, meaning $p$ and $q$ are integers and $q \neq 0$. We are also given that $m$ is a non-zero integer, meaning $m \neq 0$.

Consider the expression $\frac{p \times m}{q \times m}$.

Since $p$ and $m$ are integers, their product $p \times m$ is an integer.

Since $q$ and $m$ are non-zero integers, their product $q \times m$ is a non-zero integer (because the product of two non-zero numbers is non-zero).

Therefore, $\frac{p \times m}{q \times m}$ is in the form $\frac{\text{integer}}{\text{non-zero integer}}$, so it is a rational number.

The core of the statement is the claim that $\frac{p \times m}{q \times m}$ is not equivalent to $\frac{p}{q}$.

However, a fundamental property of fractions and rational numbers is that multiplying both the numerator and the denominator by the same non-zero number results in an equivalent fraction.

$\frac{p \times m}{q \times m} = \frac{p}{q}$ (provided $m \neq 0$)

This is because $\frac{m}{m} = 1$ (since $m$ is non-zero), and multiplying any number by 1 does not change its value: $\frac{p}{q} \times \frac{m}{m} = \frac{p \times m}{q \times m}$.

Therefore, the rational number $\frac{p \times m}{q \times m}$ is always equivalent to $\frac{p}{q}$, as long as $m$ is a non-zero integer.

The statement claims the opposite, that they are *not* equivalent.

The statement "If $\frac{p}{q}$ is a rational number and m is a non-zero integer, then $\frac{p × m}{q × m}$ is a rational number not equivalent to $\frac{p}{q}$" is False.

The statement is False.

To Determine: Whether the statement "Sum of two rational numbers is always a rational number" is True or False.

Solution:

This statement is about the closure property of the set of rational numbers under addition. A set of numbers is closed under an operation if performing that operation on any two numbers in the set always results in a number that is also in the set.

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers, where $a, b, c, d$ are integers and $b \neq 0$, $d \neq 0$.

Their sum is $\frac{a}{b} + \frac{c}{d}$.

To add these fractions, we find a common denominator, which is typically the LCM of $b$ and $d$, or simply the product $bd$. Let's use $bd$ as the common denominator.

$\frac{a}{b} + \frac{c}{d} = \frac{a \times d}{b \times d} + \frac{c \times b}{d \times b} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad + bc}{bd}$

Let's examine the numerator and the denominator of the resulting sum $\frac{ad + bc}{bd}$:

Numerator: $ad + bc$. Since $a, b, c, d$ are integers, $ad$ is an integer and $bc$ is an integer (by the closure property of integers under multiplication). The sum of two integers ($ad + bc$) is also an integer (by the closure property of integers under addition). So, the numerator is an integer.

Denominator: $bd$. Since $b$ and $d$ are non-zero integers, their product $bd$ is a non-zero integer (because the product of two non-zero numbers is non-zero). So, the denominator is a non-zero integer.

Since the sum can be expressed in the form $\frac{\text{integer}}{\text{non-zero integer}}$, the sum $\frac{ad + bc}{bd}$ is a rational number.

This holds true for any two rational numbers.

Examples:

$\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$ (which is rational)

$\frac{-1}{4} + \frac{3}{4} = \frac{-1+3}{4} = \frac{2}{4} = \frac{1}{2}$ (which is rational)

$\frac{-2}{3} + \frac{2}{3} = \frac{-2+2}{3} = \frac{0}{3} = 0$ (which is rational)

The sum of two rational numbers is always a rational number.

The statement "Sum of two rational numbers is always a rational number" is True.

The statement is True.

To Determine: Whether the statement "All decimal numbers are also rational numbers" is True or False.

Solution:

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Decimal numbers can be broadly classified into two types: terminating decimals and non-terminating decimals.

1. Terminating decimals: These are decimals that have a finite number of digits after the decimal point. Examples: 0.5, 2.75, -1.345.

Any terminating decimal can be written as a fraction with a power of 10 in the denominator. For example:

• $0.5 = \frac{5}{10} = \frac{1}{2}$ (which is rational)
• $2.75 = \frac{275}{100} = \frac{11}{4}$ (which is rational)
• $-1.345 = \frac{-1345}{1000} = \frac{-269}{200}$ (which is rational)

Terminating decimals can always be expressed in the form $\frac{p}{q}$ where $p$ is an integer and $q$ is a power of 10 (a non-zero integer). So, terminating decimals are rational numbers.

2. Non-terminating decimals: These decimals have an infinite number of digits after the decimal point. Non-terminating decimals can be further classified into:

• Repeating (or recurring) decimals: These have a pattern of digits that repeats infinitely. Examples: $0.333... = 0.\overline{3}$, $1.272727... = 1.\overline{27}$.

Any repeating decimal can also be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. For example:

$0.\overline{3} = \frac{1}{3}$ (which is rational)

$1.\overline{27} = 1 + 0.\overline{27} = 1 + \frac{27}{99} = 1 + \frac{3}{11} = \frac{11+3}{11} = \frac{14}{11}$ (which is rational)

So, repeating decimals are rational numbers.

• Non-repeating decimals: These have an infinite number of digits after the decimal point with no repeating pattern. Examples: $\pi = 3.14159265...$, $\sqrt{2} = 1.41421356...$.

Numbers with non-terminating, non-repeating decimal expansions cannot be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. These numbers are called irrational numbers.

The statement "All decimal numbers are also rational numbers" claims that every number with a decimal representation is rational.

However, as shown above, irrational numbers also have decimal representations, and these decimal representations are non-terminating and non-repeating.

For example, $\sqrt{2}$ is a decimal number ($1.41421356...$), but it is an irrational number, not a rational number.

Therefore, not all decimal numbers are rational numbers.

The statement "All decimal numbers are also rational numbers" is False.

The statement is False.

To Determine: Whether the statement "The quotient of two rationals is always a rational number" is True or False.

Solution:

The statement is about the closure property of the set of rational numbers under division. A set is closed under division if dividing any number in the set by any other number in the set always results in a number that is also in the set, with the exception of division by zero.

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers, where $a, b, c, d$ are integers, $b \neq 0$, and $d \neq 0$.

Their quotient is $\frac{a}{b} \div \frac{c}{d}$.

Dividing by a fraction is equivalent to multiplying by its reciprocal. The reciprocal of $\frac{c}{d}$ is $\frac{d}{c}$, provided $\frac{c}{d}$ is not zero. This means $c \neq 0$.

So, $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{a \times d}{b \times c}$.

Let's examine the numerator and the denominator of the resulting quotient $\frac{ad}{bc}$:

Numerator: $ad$. Since $a$ and $d$ are integers, their product $ad$ is an integer.

Denominator: $bc$. Since $b$ and $c$ are integers, their product $bc$ is an integer.

However, for the quotient to be a rational number, the denominator $bc$ must be non-zero. This requires both $b \neq 0$ and $c \neq 0$.

We were initially given that $b \neq 0$ and $d \neq 0$. But the denominator of the *divisor* is $c$, and the numerator of the *divisor* is also $c$. The reciprocal $\frac{d}{c}$ is only defined if $c \neq 0$.

If the second rational number $\frac{c}{d}$ is 0, which happens when $c=0$, then division by $\frac{c}{d} = \frac{0}{d} = 0$ is undefined.

The statement says "The quotient of two rationals is always a rational number." This implies it holds true for *any* two rational numbers. However, if the second rational number (the divisor) is 0, the quotient is undefined, and an undefined value is not a rational number.

Therefore, the statement is only true if the divisor is non-zero. Since the statement claims it is *always* a rational number (without the condition that the divisor is non-zero), the statement is False.

If the statement were "The quotient of a rational number by a non-zero rational number is always a rational number," that would be True.

The statement "The quotient of two rationals is always a rational number" is False.

The statement is False.

To Determine: Whether the statement "Every fraction is a rational number" is True or False.

Solution:

Let's consider the definitions:

A rational number is a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

A fraction is typically understood as a number expressed in the form $\frac{a}{b}$, where $a$ is usually an integer or a whole number (the numerator) and $b$ is usually a non-zero whole number or a positive integer (the denominator).

Let's take a standard definition of a fraction as $\frac{a}{b}$, where $a$ is an integer and $b$ is a positive integer.

If $a$ is an integer and $b$ is a positive integer ($b \neq 0$), then a fraction $\frac{a}{b}$ fits the definition of a rational number because the numerator $p=a$ is an integer, and the denominator $q=b$ is a non-zero integer.

Examples of fractions:

• $\frac{3}{4}$: $p=3$ (integer), $q=4$ (non-zero integer). Rational.
• $\frac{-5}{7}$: $p=-5$ (integer), $q=7$ (non-zero integer). Rational.
• $\frac{0}{2}$: $p=0$ (integer), $q=2$ (non-zero integer). Rational (equals 0).
• $\frac{6}{1}$: $p=6$ (integer), $q=1$ (non-zero integer). Rational (equals 6).

What if the definition of a fraction allows the denominator to be any non-zero integer (positive or negative)? For instance, $\frac{3}{-4}$. This is often considered a fraction. $p=3$ (integer), $q=-4$ (non-zero integer). This still fits the definition of a rational number.

What if the definition of a fraction allows the numerator to be a non-integer? In some contexts (like algebraic fractions), the numerator and denominator can be expressions involving variables or even irrational numbers, e.g., $\frac{\sqrt{2}}{3}$ or $\frac{x+1}{y-2}$. However, when discussing basic number systems, a fraction usually refers to a ratio of integers.

Assuming the common understanding of a fraction in this context refers to the ratio of an integer (numerator) to a non-zero integer (denominator, often restricted to positive integers), then every such fraction meets the definition of a rational number.

Let's consider if there's any number commonly called a "fraction" that is not rational. Expressions like $\frac{\pi}{2}$ or $\frac{1}{\sqrt{2}}$ might be called fractions in a broader sense (representing a part of a whole or a division), but they do not fit the standard definition of a rational number because the numerator or denominator (or both) are not integers.

In elementary and middle school mathematics, "fraction" is generally used for numbers like $\frac{a}{b}$ where $a$ is a whole number and $b$ is a positive integer. This subset is certainly rational. Expanding the definition slightly to integer numerator and non-zero integer denominator still fits the rational number definition.

Unless a broader, non-standard definition of "fraction" is implied (where numerator or denominator can be non-integers), the statement holds true under the standard definition of a fraction as a ratio of integers.

Given the context of rational numbers (defined as $\frac{p}{q}$ with $p,q$ integers, $q \neq 0$), the statement "Every fraction is a rational number" is generally considered true, where "fraction" means a ratio of integers.

The statement "Every fraction is a rational number" is True (assuming "fraction" means a ratio of integers).

To Determine: Whether the statement "Two rationals with different numerators can never be equal" is True or False.

Solution:

The statement claims that if two rational numbers $\frac{p_1}{q_1}$ and $\frac{p_2}{q_2}$ have different numerators (i.e., $p_1 \neq p_2$), then they must be unequal ($\frac{p_1}{q_1} \neq \frac{p_2}{q_2}$).

To check if this is true, let's look for a counterexample: Can we find two equal rational numbers that have different numerators?

Consider the rational number $\frac{1}{2}$. Its numerator is 1.

Can we find another rational number equal to $\frac{1}{2}$ but with a different numerator? Yes, we can create equivalent fractions.

Multiply the numerator and denominator of $\frac{1}{2}$ by 2: $\frac{1 \times 2}{2 \times 2} = \frac{2}{4}$. The numerator is 2, which is different from 1. The two rational numbers are $\frac{1}{2}$ and $\frac{2}{4}$. They have different numerators (1 and 2), but they are equal ($\frac{1}{2} = \frac{2}{4}$).

Multiply the numerator and denominator of $\frac{1}{2}$ by 3: $\frac{1 \times 3}{2 \times 3} = \frac{3}{6}$. The numerator is 3, different from 1. $\frac{1}{2} = \frac{3}{6}$.

Multiply the numerator and denominator of $\frac{1}{2}$ by -1: $\frac{1 \times (-1)}{2 \times (-1)} = \frac{-1}{-2}$. The numerator is -1, different from 1. $\frac{1}{2} = \frac{-1}{-2}$.

In all these cases, we found pairs of equal rational numbers with different numerators.

Therefore, the statement "Two rationals with different numerators can never be equal" is incorrect.

The statement is False.

To Determine: Whether the statement "8 can be written as a rational number with any integer as denominator" is True or False.

Solution:

A rational number is defined as a number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

The number 8 can be written as a rational number in many ways, such as $\frac{8}{1}$, $\frac{16}{2}$, $\frac{24}{3}$, $\frac{-8}{-1}$, etc.

The statement says that 8 can be written as a rational number with "any integer" as the denominator.

Let the rational number be $\frac{p}{q}$. We want $\frac{p}{q} = 8$, where $q$ is "any integer".

From the definition of a rational number, the denominator $q$ must be a non-zero integer. This means $q$ can be any integer except 0.

The statement says $q$ can be "any integer", which includes 0. If we were to use 0 as the denominator, the expression $\frac{p}{0}$ would be undefined, and thus cannot be equal to 8.

So, 8 cannot be written with 0 as the denominator. Therefore, 8 cannot be written as a rational number with *any* integer as the denominator.

The statement would be true if it said "with any non-zero integer as denominator".

The statement "8 can be written as a rational number with any integer as denominator" is False.

The statement is False.

To Determine: Whether the statement "$\frac{4}{6}$ is equivalent to $\frac{2}{3}$" is True or False.

Solution:

Two rational numbers are equivalent if they represent the same value. We can check for equivalence by reducing both fractions to their standard form or by cross-multiplication.

Method 1: Reducing to Standard Form

Consider the fraction $\frac{4}{6}$. The denominator is positive. Find the GCD of 4 and 6.

Factors of 4: 1, 2, 4

Factors of 6: 1, 2, 3, 6

The GCD(4, 6) = 2.

Divide the numerator and denominator by the GCD:

$\frac{4 \div 2}{6 \div 2} = \frac{2}{3}$

The standard form of $\frac{4}{6}$ is $\frac{2}{3}$.

The second fraction given is $\frac{2}{3}$. This fraction is already in standard form because the denominator 3 is positive and GCD(2, 3) = 1.

Since the standard form of $\frac{4}{6}$ is $\frac{2}{3}$, and the second fraction is also $\frac{2}{3}$, they have the same standard form.

Method 2: Cross-Multiplication

Two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ are equivalent if and only if $a \times d = b \times c$.

We are comparing $\frac{4}{6}$ and $\frac{2}{3}$. Here, $a=4, b=6, c=2, d=3$.

Calculate the cross-products:

$a \times d = 4 \times 3 = 12$

$b \times c = 6 \times 2 = 12$

Since the cross-products are equal ($12 = 12$), the rational numbers are equivalent.

The statement "$\frac{4}{6}$ is equivalent to $\frac{2}{3}$" is True.

The statement is True.

To Determine: Whether the statement "The rational number $\frac{-3}{4}$ lies to the right of zero on the number line" is True or False.

Solution:

On a standard horizontal number line, zero is the point of origin. Numbers to the right of zero are positive, and numbers to the left of zero are negative.

We need to determine the sign of the rational number $\frac{-3}{4}$.

The numerator is -3 (negative).

The denominator is 4 (positive).

Since the numerator and the denominator have different signs, the rational number $\frac{-3}{4}$ is a negative number.

Its decimal value is $-0.75$.

Since $\frac{-3}{4} = -0.75$ is less than 0, it is located to the left of zero on the number line.

The statement claims that $\frac{-3}{4}$ lies to the right of zero. This is the opposite of its actual position.

The statement "The rational number $\frac{-3}{4}$ lies to the right of zero on the number line" is False.

The statement is False.

To Determine: Whether the statement "The rational numbers $\frac{-12}{-5}$ and $\frac{-7}{17}$ are on the opposite sides of zero on the number line" is True or False.

Solution:

To determine the position of a number relative to zero on the number line, we look at its sign.

• Positive numbers are to the right of zero.
• Negative numbers are to the left of zero.

Consider the first rational number, $\frac{-12}{-5}$.

The numerator is -12 (negative).

The denominator is -5 (negative).

Since the numerator and the denominator have the same sign (both negative), the rational number is positive. $\frac{-12}{-5} = \frac{12}{5}$.

Since $\frac{-12}{-5}$ is a positive number, it lies to the right of zero on the number line.

Consider the second rational number, $\frac{-7}{17}$.

The numerator is -7 (negative).

The denominator is 17 (positive).

Since the numerator and the denominator have different signs, the rational number is negative.

Since $\frac{-7}{17}$ is a negative number, it lies to the left of zero on the number line.

One number ($\frac{-12}{-5}$) is to the right of zero, and the other number ($\frac{-7}{17}$) is to the left of zero. This means they are on opposite sides of zero on the number line.

The statement correctly claims that these two rational numbers are on the opposite sides of zero on the number line.

The statement "The rational numbers $\frac{-12}{-5}$ and $\frac{-7}{17}$ are on the opposite sides of zero on the number line" is True.

The statement is True.

To Determine: Whether the statement "Every rational number is a whole number" is True or False.

Solution:

Let's recall the definitions of rational numbers and whole numbers.

A rational number is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Examples include $\frac{1}{2}$, $\frac{-3}{4}$, 0, 5, $\frac{-7}{1}$.

A whole number is a non-negative integer. The set of whole numbers is $\{0, 1, 2, 3, ...\}$.

The statement "Every rational number is a whole number" implies that the set of rational numbers is a subset of the set of whole numbers. This would mean that any number belonging to the set of rational numbers must also belong to the set of whole numbers.

Let's consider some rational numbers and see if they are whole numbers:

• Consider the rational number $\frac{1}{2}$. It is in the form $\frac{p}{q}$ ($p=1$, $q=2$). It is a rational number. Is $\frac{1}{2}$ a whole number? The whole numbers are 0, 1, 2, 3, ... The number $\frac{1}{2}$ (or 0.5) is not in this set.
• Consider the rational number $\frac{-3}{5}$. It is in the form $\frac{p}{q}$ ($p=-3$, $q=5$). It is a rational number. Is $\frac{-3}{5}$ a whole number? Whole numbers are non-negative, and $\frac{-3}{5}$ is negative. So, it is not a whole number.
• Consider the rational number $-2$. It is in the form $\frac{-2}{1}$. It is a rational number. Is $-2$ a whole number? No, whole numbers are non-negative.

Since we have found rational numbers (like $\frac{1}{2}$, $\frac{-3}{5}$, and -2) that are not whole numbers, the statement "Every rational number is a whole number" is false.

Note that the reverse statement, "Every whole number is a rational number," is true, because any whole number $w$ can be written as $\frac{w}{1}$, which fits the definition of a rational number.

The statement "Every rational number is a whole number" is False.

The statement is False.

To Determine: Whether the statement "Zero is the smallest rational number" is True or False.

Solution:

A rational number is any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Rational numbers include positive numbers, negative numbers, and zero.

The statement claims that 0 is the smallest rational number. This would mean that for any rational number $r$, $r \geq 0$.

Let's consider rational numbers that are less than zero.

Take the rational number $\frac{-1}{2}$. This is a rational number because -1 and 2 are integers and $2 \neq 0$.

On the number line, negative numbers are located to the left of zero, while positive numbers are to the right. Zero is the boundary between positive and negative numbers.

Any negative number is less than zero. The number $\frac{-1}{2} = -0.5$, which is a negative number.

So, $\frac{-1}{2} < 0$.

Since we have found a rational number ($\frac{-1}{2}$) that is smaller than 0, the statement that 0 is the smallest rational number is false.

In fact, there are infinitely many negative rational numbers (e.g., -1, $\frac{-10}{3}$, $\frac{-1}{1000}$) which are all less than 0. The set of rational numbers extends infinitely in both the positive and negative directions, so there is no smallest (or largest) rational number.

The statement "Zero is the smallest rational number" is False.

The statement is False.

To Match: Match the expressions in Column I with their equivalent simplified forms in Column II.

Solution:

We evaluate each expression in Column I, assuming $a, b, c, d$ are non-zero integers where they appear in denominators or as divisors.

(i) $\frac{a}{b}$ ÷ $\frac{a}{b}$

Dividing a non-zero number by itself always results in 1.

$\frac{a}{b} \div \frac{a}{b} = \frac{a}{b} \times \frac{b}{a} = \frac{a \times b}{b \times a} = \frac{ab}{ab} = 1$ (assuming $a \neq 0$).

This matches option (c).

(ii) $\frac{a}{b}$ ÷ $\frac{c}{d}$

Division by a fraction is multiplication by its reciprocal ($\frac{d}{c}$, assuming $c \neq 0$).

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{a \times d}{b \times c} = \frac{ad}{bc}$ (assuming $c \neq 0$).

This matches option (e).

(iii) $\frac{a}{b}$ ÷ (–1)

Dividing by -1 is equivalent to multiplying by the reciprocal of -1, which is -1.

$\frac{a}{b} \div (-1) = \frac{a}{b} \times \frac{1}{-1} = \frac{a}{b} \times (-1) = \frac{a \times (-1)}{b} = \frac{-a}{b}$.

This matches option (a).

(iv) $\frac{a}{b}$ ÷ $\frac{-a}{b}$

Division by a fraction is multiplication by its reciprocal ($\frac{b}{-a}$, assuming $a \neq 0$).

$\frac{a}{b} \div \frac{-a}{b} = \frac{a}{b} \times \frac{b}{-a} = \frac{a \times b}{b \times (-a)} = \frac{ab}{-ab}$ (assuming $a \neq 0$).

Assuming $a \neq 0$ and $b \neq 0$, $ab \neq 0$. The expression simplifies to -1.

$\frac{ab}{-ab} = -1$.

This matches option (b).

(v) $\frac{b}{a}$ ÷ $\left( \frac{d}{c} \right)$

Division by a fraction is multiplication by its reciprocal ($\frac{c}{d}$, assuming $a \neq 0$ and $d \neq 0$).

$\frac{b}{a} \div \frac{d}{c} = \frac{b}{a} \times \frac{c}{d} = \frac{b \times c}{a \times d} = \frac{bc}{ad}$ (assuming $a \neq 0, d \neq 0$).

This matches option (d).

Matching results:

(i) - (c)

(ii) - (e)

(iii) - (a)

(iv) - (b)

(v) - (d)

Solution:

To write a rational number with a positive denominator, we multiply both the numerator and the denominator by $-1$. This is equivalent to multiplying by $\frac{-1}{-1}$, which is equal to $1$, and hence does not change the value of the rational number.

Let's consider the first rational number: $\frac{5}{-8}$.

To make the denominator positive, multiply the numerator and denominator by $-1$:

$\frac{5}{-8} = \frac{5 \times (-1)}{-8 \times (-1)}$

$\frac{5}{-8} = \frac{-5}{8}$

Thus, $\frac{5}{-8}$ with a positive denominator is $\frac{-5}{8}$.

Let's consider the second rational number: $\frac{15}{-28}$.

To make the denominator positive, multiply the numerator and denominator by $-1$:

$\frac{15}{-28} = \frac{15 \times (-1)}{-28 \times (-1)}$

$\frac{15}{-28} = \frac{-15}{28}$

Thus, $\frac{15}{-28}$ with a positive denominator is $\frac{-15}{28}$.

Let's consider the third rational number: $\frac{-17}{-13}$.

To make the denominator positive, multiply the numerator and denominator by $-1$:

$\frac{-17}{-13} = \frac{-17 \times (-1)}{-13 \times (-1)}$

$\frac{-17}{-13} = \frac{17}{13}$

Thus, $\frac{-17}{-13}$ with a positive denominator is $\frac{17}{13}$.

Solution:

To express a given rational number with a specific denominator, we find the factor by which the original denominator must be multiplied to get the new denominator. Then, we multiply both the numerator and the denominator by this factor.

(i) Denominator 36

The given rational number is $\frac{3}{4}$. We want to express it with a denominator of 36.

Let the required rational number be $\frac{3 \times k}{4 \times k}$ such that $4 \times k = 36$.

To find the value of $k$, we divide 36 by 4:

$k = \frac{36}{4} = 9$

Now, we multiply both the numerator and the denominator of $\frac{3}{4}$ by 9:

$\frac{3}{4} = \frac{3 \times 9}{4 \times 9} = \frac{27}{36}$

Thus, $\frac{3}{4}$ expressed as a rational number with denominator 36 is $\frac{27}{36}$.

(ii) Denominator – 80

The given rational number is $\frac{3}{4}$. We want to express it with a denominator of – 80.

Let the required rational number be $\frac{3 \times m}{4 \times m}$ such that $4 \times m = -80$.

To find the value of $m$, we divide – 80 by 4:

$m = \frac{-80}{4} = -20$

Now, we multiply both the numerator and the denominator of $\frac{3}{4}$ by – 20:

$\frac{3}{4} = \frac{3 \times (-20)}{4 \times (-20)} = \frac{-60}{-80}$

Thus, $\frac{3}{4}$ expressed as a rational number with denominator – 80 is $\frac{-60}{-80}$.

Solution:

To reduce a rational number to its lowest form, we divide both the numerator and the denominator by their Greatest Common Divisor (GCD). The rational number $\frac{p}{q}$ is in its lowest form if the GCD of $|p|$ and $|q|$ is 1.

(i) Reduce $\frac{-60}{72}$ to its lowest form.

We need to find the GCD of $|-60| = 60$ and $|72| = 72$.

Prime factorization of 60 is $2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5$.

Prime factorization of 72 is $2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.

The common prime factors are $2$ and $3$.

The lowest power of $2$ is $2^2$. The lowest power of $3$ is $3^1$.

So, GCD$(60, 72) = 2^2 \times 3^1 = 4 \times 3 = 12$.

Now, we divide both the numerator and the denominator of $\frac{-60}{72}$ by 12:

$\frac{-60}{72} = \frac{-60 \div 12}{72 \div 12} = \frac{-5}{6}$

The rational number $\frac{-5}{6}$ is in its lowest form because the GCD of $|-5| = 5$ and $|6| = 6$ is 1.

Thus, the lowest form of $\frac{-60}{72}$ is $\frac{-5}{6}$.

(ii) Reduce $\frac{91}{-364}$ to its lowest form.

We need to find the GCD of $|91| = 91$ and $|-364| = 364$.

Prime factorization of 91 is $7 \times 13$.

To find the prime factorization of 364, we can divide by small prime numbers:

$\begin{array}{c|cc} 2 & 364 \\ \hline 2 & 182 \\ \hline 7 & 91 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

Prime factorization of 364 is $2 \times 2 \times 7 \times 13 = 2^2 \times 7 \times 13$.

The common prime factors are $7$ and $13$.

The lowest power of $7$ is $7^1$. The lowest power of $13$ is $13^1$.

So, GCD$(91, 364) = 7^1 \times 13^1 = 7 \times 13 = 91$.

Now, we divide both the numerator and the denominator of $\frac{91}{-364}$ by 91:

$\frac{91}{-364} = \frac{91 \div 91}{-364 \div 91} = \frac{1}{-4}$

To express this with a positive denominator, we multiply the numerator and denominator by $-1$:

$\frac{1}{-4} = \frac{1 \times (-1)}{-4 \times (-1)} = \frac{-1}{4}$

The rational number $\frac{-1}{4}$ is in its lowest form because the GCD of $|-1| = 1$ and $|4| = 4$ is 1.

Thus, the lowest form of $\frac{91}{-364}$ is $\frac{-1}{4}$.

Solution:

A rational number $\frac{p}{q}$ is said to be in its standard form if:

1. The denominator $q$ is a positive integer.

2. The Greatest Common Divisor (GCD) of $|p|$ and $|q|$ is 1 (i.e., $p$ and $q$ are coprime).

To reduce a rational number to its standard form, first make the denominator positive (if it is negative) by multiplying both numerator and denominator by $-1$. Then, divide both the numerator and the denominator by their GCD.

(i) $\frac{-12}{-30}$

The denominator is $-30$, which is negative. Multiply both the numerator and the denominator by $-1$ to make the denominator positive:

$\frac{-12}{-30} = \frac{-12 \times (-1)}{-30 \times (-1)} = \frac{12}{30}$

Now, find the GCD of 12 and 30.

Prime factorization of 12 is $2 \times 2 \times 3 = 2^2 \times 3^1$.

Prime factorization of 30 is $2 \times 3 \times 5 = 2^1 \times 3^1 \times 5^1$.

GCD$(12, 30) = 2^1 \times 3^1 = 6$.

Divide both the numerator and the denominator by the GCD, 6:

$\frac{12}{30} = \frac{12 \div 6}{30 \div 6} = \frac{2}{5}$

The standard form of $\frac{-12}{-30}$ is $\frac{2}{5}$. The denominator 5 is positive, and GCD(2, 5) = 1.

(ii) $\frac{14}{-49}$

The denominator is $-49$, which is negative. Multiply both the numerator and the denominator by $-1$:

$\frac{14}{-49} = \frac{14 \times (-1)}{-49 \times (-1)} = \frac{-14}{49}$

Now, find the GCD of $|-14| = 14$ and 49.

Prime factorization of 14 is $2 \times 7 = 2^1 \times 7^1$.

Prime factorization of 49 is $7 \times 7 = 7^2$.

GCD$(14, 49) = 7^1 = 7$.

Divide both the numerator and the denominator by the GCD, 7:

$\frac{-14}{49} = \frac{-14 \div 7}{49 \div 7} = \frac{-2}{7}$

The standard form of $\frac{14}{-49}$ is $\frac{-2}{7}$. The denominator 7 is positive, and GCD($|-2|$, 7) = GCD(2, 7) = 1.

(iii) $\frac{-15}{35}$

The denominator is 35, which is positive. Find the GCD of $|-15| = 15$ and 35.

Prime factorization of 15 is $3 \times 5 = 3^1 \times 5^1$.

Prime factorization of 35 is $5 \times 7 = 5^1 \times 7^1$.

GCD$(15, 35) = 5^1 = 5$.

Divide both the numerator and the denominator by the GCD, 5:

$\frac{-15}{35} = \frac{-15 \div 5}{35 \div 5} = \frac{-3}{7}$

The standard form of $\frac{-15}{35}$ is $\frac{-3}{7}$. The denominator 7 is positive, and GCD($|-3|$, 7) = GCD(3, 7) = 1.

(iv) $\frac{299}{-161}$

The denominator is $-161$, which is negative. Multiply both the numerator and the denominator by $-1$:

$\frac{299}{-161} = \frac{299 \times (-1)}{-161 \times (-1)} = \frac{-299}{161}$

Now, find the GCD of $|-299| = 299$ and 161.

Find the prime factors of 299 and 161:

For 161:

$\begin{array}{c|cc} 7 & 161 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$

So, $161 = 7 \times 23$.

For 299:

$\begin{array}{c|cc} 13 & 299 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$

So, $299 = 13 \times 23$.

The common prime factor is 23.

GCD$(299, 161) = 23$.

Divide both the numerator and the denominator by the GCD, 23:

$\frac{-299}{161} = \frac{-299 \div 23}{161 \div 23} = \frac{-13}{7}$

The standard form of $\frac{299}{-161}$ is $\frac{-13}{7}$. The denominator 7 is positive, and GCD($|-13|$, 7) = GCD(13, 7) = 1.

Solution:

To check if two rational numbers are equivalent, we can reduce them to their lowest forms. If their lowest forms are the same, the rational numbers are equivalent.

Consider the first rational number: $\frac{-8}{28}$.

We can divide both the numerator and the denominator by their greatest common divisor. The greatest common divisor of 8 and 28 is 4.

$\frac{-8}{28} = \frac{-8 \div 4}{28 \div 4} = \frac{-2}{7}$

The lowest form of $\frac{-8}{28}$ is $\frac{-2}{7}$.

Consider the second rational number: $\frac{32}{-112}$.

First, make the denominator positive by multiplying the numerator and denominator by $-1$.

$\frac{32}{-112} = \frac{32 \times (-1)}{-112 \times (-1)} = \frac{-32}{112}$

Now, divide both the numerator and the denominator by their greatest common divisor. The greatest common divisor of 32 and 112 is 16.

$\frac{-32}{112} = \frac{-32 \div 16}{112 \div 16} = \frac{-2}{7}$

The lowest form of $\frac{32}{-112}$ is $\frac{-2}{7}$.

Since the lowest forms of both rational numbers are the same ($\frac{-2}{7}$), the rational numbers are equivalent.

Conclusion:

Yes, the rational numbers $\frac{-8}{28}$ and $\frac{32}{-112}$ are equivalent.

Reason: Both rational numbers reduce to the same lowest form, $\frac{-2}{7}$.

Solution:

To arrange the rational numbers in ascending order, we first need to express them with a common denominator. It is helpful to make all denominators positive first.

The given rational numbers are:

$\frac{7}{10}$

$\frac{5}{-8} = \frac{5 \times (-1)}{-8 \times (-1)} = \frac{-5}{8}$

$\frac{2}{-3} = \frac{2 \times (-1)}{-3 \times (-1)} = \frac{-2}{3}$

$\frac{-1}{4}$

$\frac{-3}{5}$

The rational numbers with positive denominators are: $\frac{7}{10}$, $\frac{-5}{8}$, $\frac{-2}{3}$, $\frac{-1}{4}$, $\frac{-3}{5}$.

The denominators are 10, 8, 3, 4, and 5.

We find the Least Common Multiple (LCM) of these denominators.

$\begin{array}{c|cc} 2 & 10 \;, & 8 \;, & 3 \;, & 4 \;, & 5 \\ \hline 2 & 5 \; , & 4 \; , & 3 \; , & 2 \; , & 5 \\ \hline 2 & 5 \; , & 2 \; , & 3 \; , & 1 \; , & 5 \\ \hline 3 & 5 \; , & 1 \; , & 3 \; , & 1 \; , & 5 \\ \hline 5 & 1 \; , & 1 \; , & 1 \; , & 1 \; , & 1 \\ \hline & & & & & \end{array}$

LCM$(10, 8, 3, 4, 5) = 2 \times 2 \times 2 \times 3 \times 5 = 8 \times 15 = 120$.

The LCM of the denominators is 120.

Now, we express each rational number with a denominator of 120:

$\frac{7}{10} = \frac{7 \times 12}{10 \times 12} = \frac{84}{120}$

$\frac{-5}{8} = \frac{-5 \times 15}{8 \times 15} = \frac{-75}{120}$

$\frac{-2}{3} = \frac{-2 \times 40}{3 \times 40} = \frac{-80}{120}$

$\frac{-1}{4} = \frac{-1 \times 30}{4 \times 30} = \frac{-30}{120}$

$\frac{-3}{5} = \frac{-3 \times 24}{5 \times 24} = \frac{-72}{120}$

The rational numbers with the common denominator are: $\frac{84}{120}$, $\frac{-75}{120}$, $\frac{-80}{120}$, $\frac{-30}{120}$, $\frac{-72}{120}$.

To arrange these in ascending order, we compare their numerators: $84, -75, -80, -30, -72$.

Arranging the numerators in ascending order:

$-80 < -75 < -72 < -30 < 84$

So, the rational numbers in ascending order are:

$\frac{-80}{120}, \frac{-75}{120}, \frac{-72}{120}, \frac{-30}{120}, \frac{84}{120}$.

Replacing these with the original rational numbers:

$\frac{-80}{120}$ corresponds to $\frac{-2}{3}$ (or $\frac{2}{-3}$).

$\frac{-75}{120}$ corresponds to $\frac{-5}{8}$ (or $\frac{5}{-8}$).

$\frac{-72}{120}$ corresponds to $\frac{-3}{5}$.

$\frac{-30}{120}$ corresponds to $\frac{-1}{4}$.

$\frac{84}{120}$ corresponds to $\frac{7}{10}$.

Therefore, the rational numbers in ascending order are:

$\frac{2}{-3}$, $\frac{5}{-8}$, $\frac{-3}{5}$, $\frac{-1}{4}$, $\frac{7}{10}$.

Solution:

To represent a rational number on a number line, we first determine whether it is positive or negative. Positive numbers are represented to the right of zero, and negative numbers to the left. We identify the two integers between which the rational number lies, then divide the segment between these integers into parts based on the denominator and mark the point corresponding to the numerator.

1. Represent $\frac{3}{8}$ on the number line.

$\frac{3}{8}$ is a positive rational number. The numerator 3 is less than the denominator 8, so the fraction is less than 1. It lies between 0 and 1.

We divide the segment between 0 and 1 into 8 equal parts. Each part represents $\frac{1}{8}$.

Starting from 0 and moving to the right, the first mark is $\frac{1}{8}$, the second is $\frac{2}{8}$, the third is $\frac{3}{8}$, and so on.

We mark the third point from 0 to the right. This point represents $\frac{3}{8}$.

2. Represent $\frac{-7}{3}$ on the number line.

$\frac{-7}{3}$ is a negative rational number. We can write it as a mixed number:

$\frac{-7}{3} = -\frac{7}{3} = -(2 + \frac{1}{3}) = -2\frac{1}{3}$

This means $\frac{-7}{3}$ lies between the integers $-2$ and $-3$.

We divide the segment between $-2$ and $-3$ into 3 equal parts. Each part represents $\frac{1}{3}$.

Starting from $-2$ and moving to the left, the first mark is $-2 - \frac{1}{3} = -2\frac{1}{3}$, which is $\frac{-7}{3}$. The second mark is $-2 - \frac{2}{3} = -2\frac{2}{3}$, which is $\frac{-8}{3}$, and the third mark is $-2 - \frac{3}{3} = -3$, which is $\frac{-9}{3}$.

We mark the first point from $-2$ to the left. This point represents $\frac{-7}{3}$.

3. Represent $\frac{22}{-6}$ on the number line.

First, simplify the rational number and make the denominator positive:

$\frac{22}{-6} = \frac{\cancel{22}^{11}}{\cancel{-6}_{-3}} = \frac{11}{-3}$

To make the denominator positive, multiply numerator and denominator by $-1$:

$\frac{11}{-3} = \frac{11 \times (-1)}{-3 \times (-1)} = \frac{-11}{3}$

Now, we need to represent $\frac{-11}{3}$ on the number line. This is a negative rational number.

We write it as a mixed number:

$\frac{-11}{3} = -\frac{11}{3} = -(3 + \frac{2}{3}) = -3\frac{2}{3}$

This means $\frac{-11}{3}$ (or $\frac{22}{-6}$) lies between the integers $-3$ and $-4$.

We divide the segment between $-3$ and $-4$ into 3 equal parts. Each part represents $\frac{1}{3}$.

Starting from $-3$ and moving to the left, the first mark is $-3 - \frac{1}{3} = -3\frac{1}{3}$, which is $\frac{-10}{3}$. The second mark is $-3 - \frac{2}{3} = -3\frac{2}{3}$, which is $\frac{-11}{3}$. The third mark is $-3 - \frac{3}{3} = -4$, which is $\frac{-12}{3}$.

We mark the second point from $-3$ to the left. This point represents $\frac{-11}{3}$ or $\frac{22}{-6}$.

Note: A visual representation of the number line with these points marked would be drawn showing 0, positive integers to the right, negative integers to the left, and the specific points $\frac{3}{8}$, $\frac{-7}{3}$, and $\frac{22}{-6}$ placed correctly between the integers.

Solution:

We are given the equation $\frac{-5}{7} = \frac{x}{28}$. This equation states that the two rational numbers are equivalent.

To find the value of the unknown variable $x$, we can use the property of equivalent rational numbers which states that if $\frac{a}{b} = \frac{c}{d}$, then $a \times d = b \times c$ (cross-multiplication).

Applying cross-multiplication to the given equation $\frac{-5}{7} = \frac{x}{28}$, we get:

$(-5) \times 28 = 7 \times x$

Now, we perform the multiplication on both sides:

Left side: $(-5) \times 28$

$\begin{array}{cc}& & & 2 & 8 \\ \times & & & & 5 \\ \hline & & 1 & 4 & 0 \\ \hline \end{array}$

Since one number is negative and the other is positive, the product is negative.

$(-5) \times 28 = -140$

Right side: $7 \times x$

So, the equation becomes:

$-140 = 7x$

To find the value of $x$, we need to isolate $x$. We do this by dividing both sides of the equation by 7:

$x = \frac{-140}{7}$

Now, perform the division:

$\frac{-140}{7} = -20$

Thus, the value of $x$ is $-20$.

Alternatively, we can see that the denominator on the right side (28) is obtained by multiplying the denominator on the left side (7) by 4 (since $7 \times 4 = 28$). For the fractions to be equivalent, the numerator on the right side must also be obtained by multiplying the numerator on the left side by the same factor, 4.

So, $x = (-5) \times 4 = -20$.

The final answer is $\boxed{-20}$.

Solution:

To find rational numbers equivalent to a given rational number, we multiply both the numerator and the denominator by the same non-zero integer.

(i) Three rational numbers equivalent to $\frac{-3}{4}$.

Multiply numerator and denominator by 2:

$\frac{-3}{4} = \frac{-3 \times 2}{4 \times 2} = \frac{-6}{8}$

Multiply numerator and denominator by 3:

$\frac{-3}{4} = \frac{-3 \times 3}{4 \times 3} = \frac{-9}{12}$

Multiply numerator and denominator by 4:

$\frac{-3}{4} = \frac{-3 \times 4}{4 \times 4} = \frac{-12}{16}$

Three rational numbers equivalent to $\frac{-3}{4}$ are $\frac{-6}{8}$, $\frac{-9}{12}$, and $\frac{-12}{16}$.

(ii) Three rational numbers equivalent to $\frac{7}{11}$.

Multiply numerator and denominator by 2:

$\frac{7}{11} = \frac{7 \times 2}{11 \times 2} = \frac{14}{22}$

Multiply numerator and denominator by 3:

$\frac{7}{11} = \frac{7 \times 3}{11 \times 3} = \frac{21}{33}$

Multiply numerator and denominator by 4:

$\frac{7}{11} = \frac{7 \times 4}{11 \times 4} = \frac{28}{44}$

Three rational numbers equivalent to $\frac{7}{11}$ are $\frac{14}{22}$, $\frac{21}{33}$, and $\frac{28}{44}$.

Solution:

To find the next rational numbers in the pattern, we observe how the numerator and the denominator change from one term to the next. The pattern is usually formed by multiplying the numerator and the denominator of the first term by consecutive integers.

(i) Pattern: $\frac{4}{-5}$ , $\frac{8}{-10}$ , $\frac{12}{-15}$ , $\frac{16}{-20}$ , ______, ______,______

Observe the numerators: 4, 8, 12, 16. These are obtained by multiplying the numerator of the first term (4) by 1, 2, 3, 4 respectively.

$4 \times 1 = 4$

$4 \times 2 = 8$

$4 \times 3 = 12$

$4 \times 4 = 16$

Observe the denominators: -5, -10, -15, -20. These are obtained by multiplying the denominator of the first term (-5) by 1, 2, 3, 4 respectively.

$-5 \times 1 = -5$

$-5 \times 2 = -10$

$-5 \times 3 = -15$

$-5 \times 4 = -20$

The pattern is formed by multiplying the numerator and the denominator of the first term $\frac{4}{-5}$ by consecutive integers starting from 1. The general form of the rational numbers in this pattern is $\frac{4 \times n}{-5 \times n}$, where $n$ is a positive integer.

The given terms correspond to $n = 1, 2, 3, 4$. We need the next three terms, which correspond to $n = 5, 6, 7$.

For $n=5$: $\frac{4 \times 5}{-5 \times 5} = \frac{20}{-25}$.

For $n=6$: $\frac{4 \times 6}{-5 \times 6} = \frac{24}{-30}$.

For $n=7$: $\frac{4 \times 7}{-5 \times 7} = \frac{28}{-35}$.

The next three rational numbers are $\frac{20}{-25}$, $\frac{24}{-30}$, and $\frac{28}{-35}$.

(ii) Pattern: $\frac{-8}{7}$ , $\frac{-16}{14}$ , $\frac{-24}{21}$ , $\frac{-32}{28}$ , ______, ______, ______

Observe the numerators: -8, -16, -24, -32. These are obtained by multiplying the numerator of the first term (-8) by 1, 2, 3, 4 respectively.

$-8 \times 1 = -8$

$-8 \times 2 = -16$

$-8 \times 3 = -24$

$-8 \times 4 = -32$

Observe the denominators: 7, 14, 21, 28. These are obtained by multiplying the denominator of the first term (7) by 1, 2, 3, 4 respectively.

$7 \times 1 = 7$

$7 \times 2 = 14$

$7 \times 3 = 21$

$7 \times 4 = 28$

The pattern is formed by multiplying the numerator and the denominator of the first term $\frac{-8}{7}$ by consecutive integers starting from 1. The general form of the rational numbers in this pattern is $\frac{-8 \times n}{7 \times n}$, where $n$ is a positive integer.

The given terms correspond to $n = 1, 2, 3, 4$. We need the next three terms, which correspond to $n = 5, 6, 7$.

For $n=5$: $\frac{-8 \times 5}{7 \times 5} = \frac{-40}{35}$.

For $n=6$: $\frac{-8 \times 6}{7 \times 6} = \frac{-48}{42}$.

For $n=7$: $\frac{-8 \times 7}{7 \times 7} = \frac{-56}{49}$.

The next three rational numbers are $\frac{-40}{35}$, $\frac{-48}{42}$, and $\frac{-56}{49}$.

Solution:

To find rational numbers between two given rational numbers, we first express them with a common denominator. The larger the common denominator, the more rational numbers we can find between them.

The given rational numbers are $\frac{5}{7}$ and $\frac{7}{8}$.

The denominators are 7 and 8. We find the Least Common Multiple (LCM) of 7 and 8.

Since 7 and 8 are coprime, LCM$(7, 8) = 7 \times 8 = 56$.

Express each rational number with a denominator of 56:

$\frac{5}{7} = \frac{5 \times 8}{7 \times 8} = \frac{40}{56}$

$\frac{7}{8} = \frac{7 \times 7}{8 \times 7} = \frac{49}{56}$

Now we need to find four rational numbers between $\frac{40}{56}$ and $\frac{49}{56}$.

We look for fractions with denominator 56 and numerators between 40 and 49.

The integers between 40 and 49 are 41, 42, 43, 44, 45, 46, 47, 48.

We can choose any four of these as numerators.

Let's choose 41, 42, 43, and 44.

The four rational numbers are $\frac{41}{56}$, $\frac{42}{56}$, $\frac{43}{56}$, and $\frac{44}{56}$.

These numbers are indeed between $\frac{40}{56}$ and $\frac{49}{56}$, since $40 < 41 < 42 < 43 < 44 < 49$.

Therefore, four rational numbers between $\frac{5}{7}$ and $\frac{7}{8}$ are $\frac{41}{56}$, $\frac{42}{56}$, $\frac{43}{56}$, and $\frac{44}{56}$.

Note: There are infinitely many rational numbers between any two distinct rational numbers. Other valid answers are possible, for example, using a larger common denominator or choosing different numerators between 41 and 48.

Solution:

To find the sum of rational numbers:

1. If the denominators are different, find the Least Common Multiple (LCM) of the denominators and express each rational number with the LCM as the denominator.

2. If the denominators are the same, add the numerators and keep the common denominator.

3. Simplify the result to its lowest form.

(i) Find the sum of $\frac{8}{13}$ and $\frac{3}{11}$.

The denominators are 13 and 11. These are prime numbers, so their LCM is their product:

LCM$(13, 11) = 13 \times 11 = 143$.

Express each rational number with a denominator of 143:

$\frac{8}{13} = \frac{8 \times 11}{13 \times 11} = \frac{88}{143}$

$\frac{3}{11} = \frac{3 \times 13}{11 \times 13} = \frac{39}{143}$

Now, add the rational numbers with the common denominator:

$\frac{88}{143} + \frac{39}{143} = \frac{88 + 39}{143}$

Add the numerators:

$88 + 39 = 127$

So, the sum is $\frac{127}{143}$.

The numbers 127 and 143 have no common factors other than 1 (127 is a prime number, and $143 = 11 \times 13$), so the fraction is in its lowest form.

The sum of $\frac{8}{13}$ and $\frac{3}{11}$ is $\frac{127}{143}$.

(ii) Find the sum of $\frac{7}{3}$ and $\frac{-4}{3}$.

The denominators are the same (3).

Add the numerators and keep the common denominator:

$\frac{7}{3} + \frac{-4}{3} = \frac{7 + (-4)}{3}$

Add the numerators: $7 + (-4) = 7 - 4 = 3$.

So, the sum is $\frac{3}{3}$.

Simplify the fraction:

$\frac{3}{3} = 1$

The sum of $\frac{7}{3}$ and $\frac{-4}{3}$ is 1.

Solution:

(i) $\frac{29}{4}$ - $\frac{30}{7}$

To subtract rational numbers with different denominators, we first find the Least Common Multiple (LCM) of the denominators.

The denominators are 4 and 7. Since 4 and 7 are coprime, their LCM is their product.

LCM$(4, 7) = 4 \times 7 = 28$.

Now, we express each rational number with a denominator of 28.

For $\frac{29}{4}$: Multiply the numerator and denominator by $\frac{28}{4} = 7$.

$\frac{29}{4} = \frac{29 \times 7}{4 \times 7} = \frac{203}{28}$

For $\frac{30}{7}$: Multiply the numerator and denominator by $\frac{28}{7} = 4$.

$\frac{30}{7} = \frac{30 \times 4}{7 \times 4} = \frac{120}{28}$

Now, subtract the rational numbers with the common denominator:

$\frac{203}{28} - \frac{120}{28} = \frac{203 - 120}{28}$

Perform the subtraction in the numerator:

$\begin{array}{cc} & 2 & 0 & 3 \\ - & 1 & 2 & 0 \\ \hline & & 8 & 3 \\ \hline \end{array}$

So, $203 - 120 = 83$.

The result is $\frac{83}{28}$.

The fraction $\frac{83}{28}$ is in its lowest form because 83 is a prime number and 28 is not a multiple of 83.

(ii) $\frac{5}{13}$ − $\frac{-8}{26}$

First, rewrite the expression. Subtracting a negative number is the same as adding the corresponding positive number.

$\frac{5}{13} - \frac{-8}{26} = \frac{5}{13} + \frac{8}{26}$

To add rational numbers with different denominators, we find the LCM of the denominators.

The denominators are 13 and 26. Since $26 = 2 \times 13$, 26 is a multiple of 13.

LCM$(13, 26) = 26$.

Now, we express each rational number with a denominator of 26.

For $\frac{5}{13}$: Multiply the numerator and denominator by $\frac{26}{13} = 2$.

$\frac{5}{13} = \frac{5 \times 2}{13 \times 2} = \frac{10}{26}$

The second fraction $\frac{8}{26}$ already has the denominator 26.

Now, add the rational numbers with the common denominator:

$\frac{10}{26} + \frac{8}{26} = \frac{10 + 8}{26}$

Perform the addition in the numerator:

$\begin{array}{cc} & 1 & 0 \\ + & & 8 \\ \hline & 1 & 8 \\ \hline \end{array}$

So, $10 + 8 = 18$.

The result is $\frac{18}{26}$.

Finally, simplify the fraction to its lowest form. The GCD of 18 and 26 is 2.

$\frac{18}{26} = \frac{18 \div 2}{26 \div 2} = \frac{9}{13}$

The fraction $\frac{9}{13}$ is in its lowest form because the GCD of 9 and 13 is 1.

Solution:

To find the product of two rational numbers, we multiply the numerators together and multiply the denominators together. Then, we simplify the resulting fraction to its lowest form.

Product of $\frac{a}{b}$ and $\frac{c}{d}$ is $\frac{a \times c}{b \times d}$.

(i) Find the product of $\frac{-4}{5}$ and $\frac{-5}{12}$.

Product = $\frac{-4}{5} \times \frac{-5}{12}$

Multiply the numerators: $(-4) \times (-5) = 20$

Multiply the denominators: $5 \times 12 = 60$

The product is $\frac{20}{60}$.

Now, simplify the fraction $\frac{20}{60}$ to its lowest form by dividing the numerator and denominator by their greatest common divisor, which is 20.

$\frac{20}{60} = \frac{20 \div 20}{60 \div 20} = \frac{1}{3}$

Alternatively, using cancellation before multiplication:

$\frac{\cancel{-4}^{-1}}{\cancel{5}^{1}} \times \frac{\cancel{-5}^{-1}}{\cancel{12}^{3}} = \frac{(-1) \times (-1)}{1 \times 3} = \frac{1}{3}$

The product of $\frac{-4}{5}$ and $\frac{-5}{12}$ is $\frac{1}{3}$.

(ii) Find the product of $\frac{-22}{11}$ and $\frac{-21}{11}$.

Product = $\frac{-22}{11} \times \frac{-21}{11}$

Multiply the numerators: $(-22) \times (-21)$

$(-22) \times (-21) = 462$

Multiply the denominators: $11 \times 11 = 121$

The product is $\frac{462}{121}$.

Now, simplify the fraction $\frac{462}{121}$ to its lowest form. We notice that $121 = 11 \times 11$. Check if 462 is divisible by 11.

$462 \div 11 = 42$. So, $462 = 42 \times 11$.

The greatest common divisor of 462 and 121 is 11.

Divide the numerator and denominator by 11:

$\frac{462}{121} = \frac{462 \div 11}{121 \div 11} = \frac{42}{11}$

Alternatively, using cancellation before multiplication:

$\frac{\cancel{-22}^{-2}}{\cancel{11}^{1}} \times \frac{-21}{11} = \frac{-2 \times (-21)}{1 \times 11} = \frac{42}{11}$

The product of $\frac{-22}{11}$ and $\frac{-21}{11}$ is $\frac{42}{11}$.

Solution:

(i) Simplify $\frac{13}{11}$ × $\frac{-14}{5}$ + $\frac{13}{11}$ × $\frac{-7}{5}$ + $\frac{-13}{11}$ × $\frac{34}{5}$.

We can use the distributive property of multiplication over addition/subtraction: $a \times b + a \times c = a \times (b+c)$ and $a \times b - a \times c = a \times (b-c)$.

The expression is $\frac{13}{11} \times \frac{-14}{5} + \frac{13}{11} \times \frac{-7}{5} + \frac{-13}{11} \times \frac{34}{5}$.

Rewrite the third term so that the common factor $\frac{13}{11}$ is clear:

$\frac{-13}{11} \times \frac{34}{5} = \frac{13}{11} \times (\frac{-1}{1}) \times \frac{34}{5} = \frac{13}{11} \times \frac{-34}{5}$.

So the expression becomes:

$\frac{13}{11} \times \frac{-14}{5} + \frac{13}{11} \times \frac{-7}{5} + \frac{13}{11} \times \frac{-34}{5}$

Now, take the common factor $\frac{13}{11}$ out:

$\frac{13}{11} \times \left( \frac{-14}{5} + \frac{-7}{5} + \frac{-34}{5} \right)$

Add the rational numbers inside the parenthesis. They have a common denominator 5.

Sum of numerators: $-14 + (-7) + (-34) = -14 - 7 - 34$

$-14 - 7 = -21$

$-21 - 34 = -55$

So, the sum inside the parenthesis is $\frac{-55}{5}$.

Simplify the fraction $\frac{-55}{5}$:

$\frac{-55}{5} = -11$

Now, multiply the result by $\frac{13}{11}$:

$\frac{13}{11} \times (-11)$

$\frac{13}{\cancel{11}} \times \cancel{-11}^{-1}$

$= 13 \times (-1) = -13$

Thus, the simplified value of the expression is $-13$.

(ii) Simplify $\frac{6}{5}$ × $\frac{3}{7}$ - $\frac{1}{5}$ × $\frac{3}{7}$.

We can use the distributive property: $a \times b - c \times b = (a-c) \times b$.

In this expression, the common factor is $\frac{3}{7}$.

The expression can be written as:

$\left( \frac{6}{5} - \frac{1}{5} \right) \times \frac{3}{7}$

Perform the subtraction inside the parenthesis. The denominators are the same (5).

$\frac{6}{5} - \frac{1}{5} = \frac{6-1}{5} = \frac{5}{5}$

Simplify the result of the subtraction:

$\frac{5}{5} = 1$

Now, multiply the result by $\frac{3}{7}$:

$1 \times \frac{3}{7}$

Any number multiplied by 1 is the number itself.

$1 \times \frac{3}{7} = \frac{3}{7}$

Thus, the simplified value of the expression is $\frac{3}{7}$.

Solution:

To divide a rational number by another non-zero rational number, we multiply the first rational number by the reciprocal of the second rational number.

The reciprocal of a rational number $\frac{c}{d}$ is $\frac{d}{c}$, where $c \neq 0$.

(i) Simplify $\frac{3}{7}$ ÷ $\left( \frac{27}{-55} \right)$.

The first rational number is $\frac{3}{7}$.

The second rational number is $\frac{27}{-55}$. Its reciprocal is $\frac{-55}{27}$.

Now, multiply the first rational number by the reciprocal of the second:

$\frac{3}{7} \div \frac{27}{-55} = \frac{3}{7} \times \frac{-55}{27}$

Multiply the numerators and the denominators:

Numerator product: $3 \times (-55) = -165$

Denominator product: $7 \times 27 = 189$

The product is $\frac{-165}{189}$.

Now, simplify the fraction $\frac{-165}{189}$ to its lowest form. Find the Greatest Common Divisor (GCD) of 165 and 189.

Prime factorization of 165: $3 \times 5 \times 11$

Prime factorization of 189: $3 \times 3 \times 3 \times 7 = 3^3 \times 7$

The common factor is 3.

GCD$(165, 189) = 3$.

Divide the numerator and denominator by 3:

$\frac{-165}{189} = \frac{-165 \div 3}{189 \div 3} = \frac{-55}{63}$

The fraction $\frac{-55}{63}$ is in its lowest form because GCD($|-55|$, 63) = GCD(55, 63) = GCD($5 \times 11$, $3^2 \times 7$) = 1.

Alternatively, using cancellation before multiplication:

$\frac{\cancel{3}^{1}}{7} \times \frac{-55}{\cancel{27}^{9}} = \frac{1 \times (-55)}{7 \times 9} = \frac{-55}{63}$

The simplified value is $\frac{-55}{63}$.

(ii) Simplify 1 ÷ $\left( -\frac{1}{2} \right)$.

The first number is 1, which can be written as $\frac{1}{1}$.

The second rational number is $-\frac{1}{2}$. Its reciprocal is $\frac{2}{-1}$.

Note that $\frac{2}{-1} = -2$.

Now, multiply the first number by the reciprocal of the second:

$1 \div \left( -\frac{1}{2} \right) = 1 \times \left( \frac{2}{-1} \right) = 1 \times (-2)$

$1 \times (-2) = -2$

Thus, the simplified value is $-2$.

Solution:

To compare rational numbers, we can express them with a common denominator and then compare their numerators. For mixed numbers, it is often helpful to convert them to improper fractions or compare the integer and fractional parts separately.

(i) Compare $\frac{3}{4}$ and $\frac{7}{8}$.

The denominators are 4 and 8. The Least Common Multiple (LCM) of 4 and 8 is 8.

We express the first rational number, $\frac{3}{4}$, with a denominator of 8:

$\frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}$

The second rational number is $\frac{7}{8}$.

Now we compare $\frac{6}{8}$ and $\frac{7}{8}$. Since the denominators are the same and positive, we compare the numerators.

Comparing the numerators 6 and 7, we have $6 < 7$.

Therefore, $\frac{6}{8} < \frac{7}{8}$.

Replacing with the original numbers, we get $\frac{3}{4} < \frac{7}{8}$.

The greater rational number is $\frac{7}{8}$.

(ii) Compare $-3\frac{5}{7}$ and $3\frac{1}{9}$.

The first number is $-3\frac{5}{7}$, which is a negative mixed number. It can be written as $- (3 + \frac{5}{7})$. Since the integer part is -3 and there is a negative fractional part, this number is less than -3. It is located to the left of 0 on the number line.

The second number is $3\frac{1}{9}$, which is a positive mixed number. It can be written as $3 + \frac{1}{9}$. This number is greater than 3. It is located to the right of 0 on the number line.

A positive number is always greater than a negative number.

Since $3\frac{1}{9}$ is positive and $-3\frac{5}{7}$ is negative, we have $3\frac{1}{9} > -3\frac{5}{7}$.

The greater number is $3\frac{1}{9}$.

Conclusion:

(i) The greater number is $\frac{7}{8}$.

(ii) The greater number is $3\frac{1}{9}$.

Solution:

Let the rational number be $\frac{p}{q}$, where $p$ is the numerator and $q$ is the denominator.

We are given two conditions:

1. The numerator is less than $-7 \times 11$.

Calculate the value: $-7 \times 11 = -77$.

So, the numerator $p$ must be less than $-77$. This means $p < -77$.

2. The denominator is greater than $12 + 4$.

Calculate the value: $12 + 4 = 16$.

So, the denominator $q$ must be greater than 16. This means $q > 16$.

We need to find a rational number $\frac{p}{q}$ such that $p$ is an integer less than $-77$ and $q$ is a non-zero integer greater than 16.

Let's choose an integer for the numerator that is less than $-77$. For example, $p = -78$ satisfies $p < -77$ as $-78 < -77$.

Let's choose an integer for the denominator that is greater than 16. For example, $q = 17$ satisfies $q > 16$ as $17 > 16$. Also, $q$ must be non-zero.

Using these values for $p$ and $q$, we can form the rational number:

Rational Number = $\frac{p}{q} = \frac{-78}{17}$

We can verify the conditions:

Numerator is $-78$. Is $-78 < -77$? Yes.

Denominator is $17$. Is $17 > 16$? Yes.

The conditions are satisfied.

There are infinitely many possible answers. Any rational number $\frac{p}{q}$ where $p$ is an integer from the set $\{-78, -79, -80, \dots\}$ and $q$ is a non-zero integer from the set $\{17, 18, 19, \dots\}$ would be a valid answer.

For example, another possible rational number could be $\frac{-100}{20}$. Here, the numerator $-100$ is less than $-77$ (since $-100 < -77$), and the denominator $20$ is greater than $16$ (since $20 > 16$).

One possible rational number is $\frac{-78}{17}$.

Solution:

Given: $x = \frac{1}{10}$ and $y = \frac{-3}{8}$.

1. Evaluate x + y

$x + y = \frac{1}{10} + \frac{-3}{8}$

LCM of denominators 10 and 8 is 40.

Express fractions with denominator 40:

$\frac{1}{10} = \frac{1 \times 4}{10 \times 4} = \frac{4}{40}$

$\frac{-3}{8} = \frac{-3 \times 5}{8 \times 5} = \frac{-15}{40}$

Add the fractions:

$x + y = \frac{4}{40} + \frac{-15}{40} = \frac{4 + (-15)}{40} = \frac{4 - 15}{40}$

$x + y = \frac{-11}{40}$

2. Evaluate x – y

$x - y = \frac{1}{10} - \frac{-3}{8}$

Subtracting a negative is adding the positive:

$x - y = \frac{1}{10} + \frac{3}{8}$

LCM of denominators 10 and 8 is 40.

Express fractions with denominator 40:

$\frac{1}{10} = \frac{1 \times 4}{10 \times 4} = \frac{4}{40}$

$\frac{3}{8} = \frac{3 \times 5}{8 \times 5} = \frac{15}{40}$

Add the fractions:

$x - y = \frac{4}{40} + \frac{15}{40} = \frac{4 + 15}{40}$

$x - y = \frac{19}{40}$

3. Evaluate x × y

$x \times y = \frac{1}{10} \times \frac{-3}{8}$

Multiply numerators and denominators:

$x \times y = \frac{1 \times (-3)}{10 \times 8} = \frac{-3}{80}$

The fraction is in lowest form.

4. Evaluate x ÷ y

$x \div y = \frac{1}{10} \div \frac{-3}{8}$

Multiply by the reciprocal of the second fraction:

$x \div y = \frac{1}{10} \times \frac{8}{-3}$

Multiply numerators and denominators:

$x \div y = \frac{1 \times 8}{10 \times (-3)} = \frac{8}{-30}$

Make the denominator positive:

$\frac{8}{-30} = \frac{8 \times (-1)}{-30 \times (-1)} = \frac{-8}{30}$

Simplify the fraction by dividing numerator and denominator by their GCD, which is 2:

$\frac{-8}{30} = \frac{-8 \div 2}{30 \div 2} = \frac{-4}{15}$

Summary of results:

$x + y = \frac{-11}{40}$

$x - y = \frac{19}{40}$

$x \times y = \frac{-3}{80}$

$x \div y = \frac{-4}{15}$

Solution:

To find the reciprocal of an expression, we first evaluate the expression to get a single rational number and then find the reciprocal of that number. The reciprocal of a rational number $\frac{a}{b}$ is $\frac{b}{a}$, provided $a \neq 0$.

(i) Find the reciprocal of $\left( \frac{1}{2} × \frac{1}{4} \right)$ + $\left( \frac{1}{2} × 6 \right)$.

First, evaluate the expression:

$\left( \frac{1}{2} × \frac{1}{4} \right) = \frac{1 \times 1}{2 \times 4} = \frac{1}{8}$

$\left( \frac{1}{2} × 6 \right) = \frac{1 \times 6}{2} = \frac{6}{2} = 3$

Now, add the results:

$\frac{1}{8} + 3 = \frac{1}{8} + \frac{3}{1}$

Find the LCM of the denominators 8 and 1, which is 8.

$\frac{1}{8} + \frac{3 \times 8}{1 \times 8} = \frac{1}{8} + \frac{24}{8} = \frac{1 + 24}{8} = \frac{25}{8}$

The value of the expression is $\frac{25}{8}$.

The reciprocal of $\frac{25}{8}$ is $\frac{8}{25}$.

(ii) Find the reciprocal of $\frac{20}{50}$ × $\frac{4}{91}$.

First, evaluate the expression:

$\frac{20}{50} \times \frac{4}{91}$

Simplify $\frac{20}{50}$: $\frac{20}{50} = \frac{2}{5}$.

Now, multiply:

$\frac{2}{5} \times \frac{4}{91} = \frac{2 \times 4}{5 \times 91} = \frac{8}{455}$

The value of the expression is $\frac{8}{455}$.

The reciprocal of $\frac{8}{455}$ is $\frac{455}{8}$.

(iii) Find the reciprocal of $\frac{3}{13}$ ÷ $\frac{-4}{65}$.

First, evaluate the expression:

$\frac{3}{13} \div \frac{-4}{65}$

Divide by multiplying by the reciprocal of the second fraction ($\frac{-4}{65}$), which is $\frac{65}{-4}$.

$\frac{3}{13} \times \frac{65}{-4}$

Cancel the common factor 13 between 13 and 65 ($65 = 5 \times 13$).

$\frac{3}{\cancel{13}^1} \times \frac{\cancel{65}^5}{-4} = \frac{3 \times 5}{1 \times (-4)} = \frac{15}{-4}$

Make the denominator positive: $\frac{15}{-4} = \frac{-15}{4}$.

The value of the expression is $\frac{-15}{4}$.

The reciprocal of $\frac{-15}{4}$ is $\frac{4}{-15}$. Make the denominator positive: $\frac{4}{-15} = \frac{-4}{15}$.

(iv) Find the reciprocal of $\left( -5 × \frac{12}{15} \right)$ - $\left( -3 × \frac{2}{9} \right)$.

First, evaluate the expression:

Evaluate the first term: $-5 \times \frac{12}{15}$. Simplify $\frac{12}{15} = \frac{4}{5}$.

$-5 \times \frac{4}{5} = \cancel{-5}^{-1} \times \frac{4}{\cancel{5}^{1}} = -1 \times 4 = -4$.

Evaluate the second term: $-3 \times \frac{2}{9}$. Simplify $\frac{3}{9} = \frac{1}{3}$.

$-3 \times \frac{2}{9} = \cancel{-3}^{-1} \times \frac{2}{\cancel{9}^3} = -1 \times \frac{2}{3} = \frac{-2}{3}$.

Now, subtract the second result from the first:

$-4 - \left( \frac{-2}{3} \right) = -4 + \frac{2}{3}$

Write $-4$ as a fraction with denominator 3: $-4 = \frac{-4 \times 3}{1 \times 3} = \frac{-12}{3}$.

Add the fractions:

$\frac{-12}{3} + \frac{2}{3} = \frac{-12 + 2}{3} = \frac{-10}{3}$.

The value of the expression is $\frac{-10}{3}$.

The reciprocal of $\frac{-10}{3}$ is $\frac{3}{-10}$. Make the denominator positive: $\frac{3}{-10} = \frac{-3}{10}$.

Solution:

To complete the table, we need to find the sum of the rational number in each row header and the rational number in each column header. The value in each cell will be the sum of the corresponding row header and column header.

The row headers are $\frac{-1}{4}$, $\frac{2}{3}$, and $\frac{3}{5}$.

The column headers are $\frac{-2}{3}$, $\frac{4}{5}$, and $\frac{9}{-10}$. Note that $\frac{9}{-10} = \frac{-9}{10}$.

Let's calculate each sum:

Row 1 + Column 1: $\frac{-1}{4} + \frac{-2}{3}$

LCM(4, 3) = 12

$\frac{-1}{4} = \frac{-1 \times 3}{4 \times 3} = \frac{-3}{12}$

$\frac{-2}{3} = \frac{-2 \times 4}{3 \times 4} = \frac{-8}{12}$

Sum = $\frac{-3}{12} + \frac{-8}{12} = \frac{-3 - 8}{12} = \frac{-11}{12}$

Row 1 + Column 2: $\frac{-1}{4} + \frac{4}{5}$

LCM(4, 5) = 20

$\frac{-1}{4} = \frac{-1 \times 5}{4 \times 5} = \frac{-5}{20}$

$\frac{4}{5} = \frac{4 \times 4}{5 \times 4} = \frac{16}{20}$

Sum = $\frac{-5}{20} + \frac{16}{20} = \frac{-5 + 16}{20} = \frac{11}{20}$

Row 1 + Column 3: $\frac{-1}{4} + \frac{9}{-10} = \frac{-1}{4} + \frac{-9}{10}$

LCM(4, 10) = 20

$\frac{-1}{4} = \frac{-1 \times 5}{4 \times 5} = \frac{-5}{20}$

$\frac{-9}{10} = \frac{-9 \times 2}{10 \times 2} = \frac{-18}{20}$

Sum = $\frac{-5}{20} + \frac{-18}{20} = \frac{-5 - 18}{20} = \frac{-23}{20}$

Row 2 + Column 1: $\frac{2}{3} + \frac{-2}{3}$

Sum = $\frac{2 + (-2)}{3} = \frac{2 - 2}{3} = \frac{0}{3} = 0$

Row 2 + Column 2: $\frac{2}{3} + \frac{4}{5}$

LCM(3, 5) = 15

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

$\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}$

Sum = $\frac{10}{15} + \frac{12}{15} = \frac{10 + 12}{15} = \frac{22}{15}$

Row 2 + Column 3: $\frac{2}{3} + \frac{9}{-10} = \frac{2}{3} + \frac{-9}{10}$

LCM(3, 10) = 30

$\frac{2}{3} = \frac{2 \times 10}{3 \times 10} = \frac{20}{30}$

$\frac{-9}{10} = \frac{-9 \times 3}{10 \times 3} = \frac{-27}{30}$

Sum = $\frac{20}{30} + \frac{-27}{30} = \frac{20 - 27}{30} = \frac{-7}{30}$

Row 3 + Column 1: $\frac{3}{5} + \frac{-2}{3}$

LCM(5, 3) = 15

$\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}$

$\frac{-2}{3} = \frac{-2 \times 5}{3 \times 5} = \frac{-10}{15}$

Sum = $\frac{9}{15} + \frac{-10}{15} = \frac{9 - 10}{15} = \frac{-1}{15}$

Row 3 + Column 2: $\frac{3}{5} + \frac{4}{5}$

Denominators are the same.

Sum = $\frac{3 + 4}{5} = \frac{7}{5}$

Row 3 + Column 3: $\frac{3}{5} + \frac{9}{-10} = \frac{3}{5} + \frac{-9}{10}$

LCM(5, 10) = 10

$\frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10}$

Sum = $\frac{6}{10} + \frac{-9}{10} = \frac{6 - 9}{10} = \frac{-3}{10}$

Now we can fill the table with the calculated sums:

Solution:

We need to express each given number in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

(a) six-eighths

"Six-eighths" means $\frac{6}{8}$.

We can simplify this fraction by dividing the numerator and the denominator by their greatest common divisor, which is 2.

$\frac{6}{8} = \frac{6 \div 2}{8 \div 2} = \frac{3}{4}$

In the form $\frac{p}{q}$, six-eighths is $\frac{3}{4}$. Here, $p=3$ and $q=4$, which are integers and $q \neq 0$.

(b) three and half

"Three and half" is a mixed number $3\frac{1}{2}$.

To convert a mixed number $a\frac{b}{c}$ to an improper fraction, we use the formula $\frac{(a \times c) + b}{c}$.

$3\frac{1}{2} = \frac{(3 \times 2) + 1}{2} = \frac{6 + 1}{2} = \frac{7}{2}$

In the form $\frac{p}{q}$, three and half is $\frac{7}{2}$. Here, $p=7$ and $q=2$, which are integers and $q \neq 0$.

(c) opposite of 1

The opposite of a number is its additive inverse. The opposite of 1 is $-1$.

To express $-1$ in the form $\frac{p}{q}$, we can write it as a fraction with a denominator of 1.

$-1 = \frac{-1}{1}$

In the form $\frac{p}{q}$, the opposite of 1 is $\frac{-1}{1}$. Here, $p=-1$ and $q=1$, which are integers and $q \neq 0$. (Any other non-zero denominator would also work, e.g., $\frac{-2}{2}$, $\frac{-3}{3}$, etc., but $\frac{-1}{1}$ is the simplest form).

(d) one-fourth

"One-fourth" means 1 out of 4, which is $\frac{1}{4}$.

In the form $\frac{p}{q}$, one-fourth is $\frac{1}{4}$. Here, $p=1$ and $q=4$, which are integers and $q \neq 0$.

(e) zero

The number zero can be expressed as a fraction with a numerator of 0 and any non-zero integer as the denominator.

Zero $= \frac{0}{\text{any non-zero integer}}$

We can choose 1 as the denominator.

Zero $= \frac{0}{1}$

In the form $\frac{p}{q}$, zero is $\frac{0}{1}$. Here, $p=0$ and $q=1$, which are integers and $q \neq 0$.

(f) opposite of three-fifths

"Three-fifths" is the rational number $\frac{3}{5}$.

The opposite of $\frac{3}{5}$ is its additive inverse, which is $-\frac{3}{5}$.

$-\frac{3}{5} = \frac{-3}{5}$

In the form $\frac{p}{q}$, the opposite of three-fifths is $\frac{-3}{5}$. Here, $p=-3$ and $q=5$, which are integers and $q \neq 0$.

Solution:

Given that $p = m \times t$ and $q = n \times t$.

We need to express the rational number $\frac{p}{q}$ in terms of $m$, $n$, and $t$.

Substitute the expressions for $p$ and $q$ into the fraction $\frac{p}{q}$:

$\frac{p}{q} = \frac{m \times t}{n \times t}$

Assuming that $t \neq 0$, we can cancel the common factor $t$ from both the numerator and the denominator:

$\frac{m \times \cancel{t}}{\\n \times \cancel{t}} = \frac{m}{n}$

Therefore, $\frac{p}{q} = \frac{m}{n}$.

Completing the given expression:

If p = m × t and q = n × t, then $\frac{p}{q} = \frac{m}{n}$.

The completed expression is:

$\frac{p}{q} = \frac{m}{n}$

Solution:

Given that $\frac{p}{q}$ and $\frac{r}{s}$ are two rational numbers in standard form with different denominators. Since they are in standard form, their denominators $q$ and $s$ are positive integers.

To compare two rational numbers $\frac{p}{q}$ and $\frac{r}{s}$ with positive denominators, we can compare the cross-products $p \times s$ and $r \times q$.

Complete the following statements:

(a) $\frac{\text{____}}{\text{____}}$ < $\frac{\text{____}}{\text{____}}$, if $p \times s < r \times q$

This statement describes the condition for the first rational number being less than the second using cross-multiplication. The blanks should be filled with the given rational numbers.

The completed statement is: $\frac{p}{q}$ < $\frac{r}{s}$, if $p \times s < r \times q$.

(b) $\frac{p}{q} = \frac{r}{s}$, if ________ = _________

This statement describes the condition for the two rational numbers being equal. Based on cross-multiplication, $\frac{p}{q} = \frac{r}{s}$ if $p \times s = r \times q$. The blanks should be filled with the cross-products.

The completed statement is: $\frac{p}{q}$ = $\frac{r}{s}$, if $p \times s$ = $r \times q$.

(c) $\frac{\text{____}}{\text{____}}$ > $\frac{\text{____}}{\text{____}}$, if $p \times s > r \times q$

This statement describes the condition for the first rational number being greater than the second using cross-multiplication. The blanks should be filled with the given rational numbers.

The completed statement is: $\frac{p}{q}$ > $\frac{r}{s}$, if $p \times s > r \times q$.

Filling the blanks in the original format:

(a) $\frac{p}{q}$ < $\frac{r}{s}$ , if p × s < r × q

(b) $\frac{p}{q}$ = $\frac{r}{s}$ , if $p \times s$ = $r \times q$

(c) $\frac{p}{q}$ > $\frac{r}{s}$ , if p × s > r × q

Solution:

For each part, the first expression gives the value of the numerator, and the second expression gives the value of the denominator of a rational number. We need to evaluate these expressions and write the resulting fraction in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

(a) Numerator: $5 – 39$, Denominator: $54 – 6$

Calculate the numerator:

$5 - 39 = -34$

Calculate the denominator:

$54 - 6 = 48$

The rational number is $\frac{-34}{48}$.

To write this in the form $\frac{p}{q}$ where $p$ and $q$ are integers, we can simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2.

$\frac{-34}{48} = \frac{-34 \div 2}{48 \div 2} = \frac{-17}{24}$

In the form $\frac{p}{q}$, the rational number is $\frac{-17}{24}$. Here, $p=-17$ and $q=24$, which are integers and $q \neq 0$.

(b) Numerator: $(–4) \times 6$, Denominator: $8 \div 2$

Calculate the numerator:

$(-4) \times 6 = -24$

Calculate the denominator:

$8 \div 2 = 4$

The rational number is $\frac{-24}{4}$.

To write this in the form $\frac{p}{q}$ where $p$ and $q$ are integers, we simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

$\frac{-24}{4} = \frac{-24 \div 4}{4 \div 4} = \frac{-6}{1}$

In the form $\frac{p}{q}$, the rational number is $\frac{-6}{1}$. Here, $p=-6$ and $q=1$, which are integers and $q \neq 0$.

(c) Numerator: $35 \div (–7)$, Denominator: $35 –18$

Calculate the numerator:

$35 \div (-7) = -5$

Calculate the denominator:

$35 - 18 = 17$

The rational number is $\frac{-5}{17}$.

This fraction is already in its lowest form as the greatest common divisor of $|-5|=5$ and $17$ is 1. The numerator $p=-5$ and denominator $q=17$ are integers and $q \neq 0$.

(d) Numerator: $25 + 15$, Denominator: $81 \div 40$

Calculate the numerator:

$25 + 15 = 40$

Calculate the denominator:

$81 \div 40 = \frac{81}{40}$

The rational number has a numerator of 40 and a denominator of $\frac{81}{40}$. So the rational number is $\frac{40}{\frac{81}{40}}$.

To express this in the form $\frac{p}{q}$, we perform the division:

$\frac{40}{\frac{81}{40}} = 40 \div \frac{81}{40} = 40 \times \frac{40}{81}$

Multiply the numbers:

$40 \times 40 = 1600$

So, the rational number is $\frac{1600}{81}$.

This fraction is in the form $\frac{p}{q}$ where $p=1600$ and $q=81$, which are integers and $q \neq 0$. The greatest common divisor of 1600 ($2^6 \times 5^2$) and 81 ($3^4$) is 1, so the fraction is in its lowest form.

Solution:

A rational number is in standard form if its denominator is a positive integer and the only common divisor of the numerator and the denominator is 1 (their GCD is 1).

(a) Express 35% in standard form.

A percentage means 'out of 100'. So, 35% can be written as $\frac{35}{100}$.

Now, simplify the fraction $\frac{35}{100}$. Find the GCD of 35 and 100.

$35 = 5 \times 7$

$100 = 2^2 \times 5^2$

The common factor is 5. GCD$(35, 100) = 5$.

Divide the numerator and denominator by 5:

$\frac{35}{100} = \frac{35 \div 5}{100 \div 5} = \frac{7}{20}$

The denominator 20 is positive, and GCD(7, 20) = 1. So, the standard form is $\frac{7}{20}$.

(b) Express 1.2 in standard form.

The decimal number 1.2 can be written as a fraction $\frac{12}{10}$ because there is one digit after the decimal point.

Now, simplify the fraction $\frac{12}{10}$. Find the GCD of 12 and 10.

$12 = 2^2 \times 3$

$10 = 2 \times 5$

The common factor is 2. GCD$(12, 10) = 2$.

Divide the numerator and denominator by 2:

$\frac{12}{10} = \frac{12 \div 2}{10 \div 2} = \frac{6}{5}$

The denominator 5 is positive, and GCD(6, 5) = 1. So, the standard form is $\frac{6}{5}$.

(c) Express $-6\frac{3}{7}$ in standard form.

This is a negative mixed number. First, convert the mixed number to an improper fraction:

$-6\frac{3}{7} = -\left(6 + \frac{3}{7}\right) = -\left(\frac{6 \times 7 + 3}{7}\right) = -\left(\frac{42 + 3}{7}\right) = -\frac{45}{7}$

The rational number is $\frac{-45}{7}$.

The denominator 7 is positive. Find the GCD of $|-45|=45$ and 7.

$45 = 3^2 \times 5$

$7 = 7^1$

There are no common prime factors, so GCD$(45, 7) = 1$.

The fraction $\frac{-45}{7}$ is already in standard form.

(d) Express 240 ÷ (– 840) in standard form.

The expression means $\frac{240}{-840}$.

First, make the denominator positive by multiplying the numerator and denominator by $-1$:

$\frac{240}{-840} = \frac{240 \times (-1)}{-840 \times (-1)} = \frac{-240}{840}$

Now, simplify the fraction $\frac{-240}{840}$. Find the GCD of $|-240|=240$ and 840.

We can divide both by common factors:

$\frac{-240}{840} = \frac{-24 \times 10}{84 \times 10} = \frac{-24}{84}$ (Dividing by 10)

Now, find the GCD of 24 and 84.

$24 = 2^3 \times 3$

$84 = 2^2 \times 3 \times 7$

GCD$(24, 84) = 2^2 \times 3 = 4 \times 3 = 12$.

Divide the numerator and denominator by 12:

$\frac{-24}{84} = \frac{-24 \div 12}{84 \div 12} = \frac{-2}{7}$

The denominator 7 is positive, and GCD($|-2|$, 7) = GCD(2, 7) = 1. So, the standard form is $\frac{-2}{7}$.

(e) Express 115 ÷ 207 in standard form.

The expression means $\frac{115}{207}$.

The denominator 207 is positive. We need to simplify the fraction $\frac{115}{207}$ by finding the GCD of 115 and 207.

Prime factorization of 115: $115 = 5 \times 23$.

Prime factorization of 207:

207 is divisible by 3 (sum of digits $2+0+7=9$ is divisible by 3).

$207 = 3 \times 69 = 3 \times 3 \times 23 = 3^2 \times 23$.

The common prime factor is 23. GCD$(115, 207) = 23$.

Divide the numerator and denominator by 23:

$\frac{115}{207} = \frac{115 \div 23}{207 \div 23} = \frac{5}{9}$

The denominator 9 is positive, and GCD(5, 9) = 1. So, the standard form is $\frac{5}{9}$.

Solution:

To find the rational number exactly halfway between two rational numbers $a$ and $b$, we calculate their average, which is given by the formula $\frac{a+b}{2}$ or $\frac{1}{2} \times (a+b)$.

(a) Find the rational number halfway between $\frac{-1}{3}$ and $\frac{1}{3}$.

Let $a = \frac{-1}{3}$ and $b = \frac{1}{3}$.

Sum = $a + b = \frac{-1}{3} + \frac{1}{3}$

Since the denominators are the same, we add the numerators:

Sum = $\frac{-1 + 1}{3} = \frac{0}{3} = 0$

Halfway number = $\frac{1}{2} \times \text{Sum} = \frac{1}{2} \times 0 = 0$.

The rational number exactly halfway between $\frac{-1}{3}$ and $\frac{1}{3}$ is 0.

(b) Find the rational number halfway between $\frac{1}{6}$ and $\frac{1}{9}$.

Let $a = \frac{1}{6}$ and $b = \frac{1}{9}$.

Sum = $a + b = \frac{1}{6} + \frac{1}{9}$

Find the LCM of the denominators 6 and 9. LCM$(6, 9) = 18$.

Express each fraction with a denominator of 18:

$\frac{1}{6} = \frac{1 \times 3}{6 \times 3} = \frac{3}{18}$

$\frac{1}{9} = \frac{1 \times 2}{9 \times 2} = \frac{2}{18}$

Sum = $\frac{3}{18} + \frac{2}{18} = \frac{3 + 2}{18} = \frac{5}{18}$

Halfway number = $\frac{1}{2} \times \text{Sum} = \frac{1}{2} \times \frac{5}{18}$

Multiply the numerators and the denominators:

Halfway number = $\frac{1 \times 5}{2 \times 18} = \frac{5}{36}$

The rational number exactly halfway between $\frac{1}{6}$ and $\frac{1}{9}$ is $\frac{5}{36}$.

(c) Find the rational number halfway between $\frac{5}{-13}$ and $\frac{-7}{9}$.

First, rewrite the first rational number with a positive denominator: $\frac{5}{-13} = \frac{-5}{13}$.

Let $a = \frac{-5}{13}$ and $b = \frac{-7}{9}$.

Sum = $a + b = \frac{-5}{13} + \frac{-7}{9}$

Find the LCM of the denominators 13 and 9. LCM$(13, 9) = 117$.

Express each fraction with a denominator of 117:

$\frac{-5}{13} = \frac{-5 \times 9}{13 \times 9} = \frac{-45}{117}$

$\frac{-7}{9} = \frac{-7 \times 13}{9 \times 13} = \frac{-91}{117}$

Sum = $\frac{-45}{117} + \frac{-91}{117} = \frac{-45 + (-91)}{117} = \frac{-45 - 91}{117}$

Sum = $\frac{-136}{117}$

Halfway number = $\frac{1}{2} \times \text{Sum} = \frac{1}{2} \times \frac{-136}{117}$

Multiply the numerators and the denominators:

Halfway number = $\frac{1 \times (-136)}{2 \times 117} = \frac{-136}{234}$

Simplify the fraction $\frac{-136}{234}$ by dividing the numerator and denominator by their GCD. Both are divisible by 2:

$\frac{-136 \div 2}{234 \div 2} = \frac{-68}{117}$

The fraction $\frac{-68}{117}$ is in standard form.

The rational number exactly halfway between $\frac{5}{-13}$ and $\frac{-7}{9}$ is $\frac{-68}{117}$.

(d) Find the rational number halfway between $\frac{1}{15}$ and $\frac{1}{12}$.

Let $a = \frac{1}{15}$ and $b = \frac{1}{12}$.

Sum = $a + b = \frac{1}{15} + \frac{1}{12}$

Find the LCM of the denominators 15 and 12.

Prime factorization of 15: $3 \times 5$

Prime factorization of 12: $2^2 \times 3$

LCM$(15, 12) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$.

Express each fraction with a denominator of 60:

$\frac{1}{15} = \frac{1 \times 4}{15 \times 4} = \frac{4}{60}$

$\frac{1}{12} = \frac{1 \times 5}{12 \times 5} = \frac{5}{60}$

Sum = $\frac{4}{60} + \frac{5}{60} = \frac{4 + 5}{60} = \frac{9}{60}$

Halfway number = $\frac{1}{2} \times \text{Sum} = \frac{1}{2} \times \frac{9}{60}$

Multiply the numerators and the denominators:

Halfway number = $\frac{1 \times 9}{2 \times 60} = \frac{9}{120}$

Simplify the fraction $\frac{9}{120}$ by dividing the numerator and denominator by their GCD, which is 3.

$\frac{9 \div 3}{120 \div 3} = \frac{3}{40}$

The fraction $\frac{3}{40}$ is in standard form.

The rational number exactly halfway between $\frac{1}{15}$ and $\frac{1}{12}$ is $\frac{3}{40}$.

Solution:

Given the rational numbers $x = \frac{-4}{9}$, $y = \frac{5}{12}$, and $z = \frac{7}{18}$.

To perform addition and subtraction, we find a common denominator for 9, 12, and 18. The LCM of 9, 12, and 18 is 36.

We can rewrite the rational numbers with a denominator of 36:

$x = \frac{-4}{9} = \frac{-4 \times 4}{9 \times 4} = \frac{-16}{36}$

$y = \frac{5}{12} = \frac{5 \times 3}{12 \times 3} = \frac{15}{36}$

$z = \frac{7}{18} = \frac{7 \times 2}{18 \times 2} = \frac{14}{36}$

(a) The rational number which when added to x gives y.

Let the required rational number be $A$. According to the question, $x + A = y$.

To find $A$, we rearrange the equation: $A = y - x$.

$A = \frac{15}{36} - \left( \frac{-16}{36} \right)$

$A = \frac{15}{36} + \frac{16}{36}$

$A = \frac{15 + 16}{36}$

$A = \frac{31}{36}$

(b) The rational number which subtracted from y gives z.

Let the required rational number be $B$. According to the question, $y - B = z$.

To find $B$, we rearrange the equation: $B = y - z$.

$B = \frac{15}{36} - \frac{14}{36}$

$B = \frac{15 - 14}{36}$

$B = \frac{1}{36}$

(c) The rational number which when added to z gives us x.

Let the required rational number be $C$. According to the question, $z + C = x$.

To find $C$, we rearrange the equation: $C = x - z$.

$C = \frac{-16}{36} - \frac{14}{36}$

$C = \frac{-16 - 14}{36}$

$C = \frac{-30}{36}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 6.

$C = \frac{-30 \div 6}{36 \div 6} = \frac{-5}{6}$

(d) The rational number which when multiplied by y to get x.

Let the required rational number be $D$. According to the question, $y \times D = x$.

To find $D$, we rearrange the equation: $D = x \div y$.

$D = \frac{-4}{9} \div \frac{5}{12}$

Divide by multiplying by the reciprocal of $\frac{5}{12}$, which is $\frac{12}{5}$.

$D = \frac{-4}{9} \times \frac{12}{5}$

Multiply the numerators and denominators:

$D = \frac{-4 \times 12}{9 \times 5} = \frac{-48}{45}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 3.

$D = \frac{-48 \div 3}{45 \div 3} = \frac{-16}{15}$

(e) The reciprocal of x + y.

First, calculate $x + y$:

$x + y = \frac{-16}{36} + \frac{15}{36} = \frac{-16 + 15}{36} = \frac{-1}{36}$

The reciprocal of a rational number $\frac{a}{b}$ is $\frac{b}{a}$ (for $a \neq 0$).

The reciprocal of $\frac{-1}{36}$ is $\frac{36}{-1} = -36$.

(f) The sum of reciprocals of x and y.

Reciprocal of $x = \frac{-4}{9}$ is $\frac{9}{-4} = \frac{-9}{4}$.

Reciprocal of $y = \frac{5}{12}$ is $\frac{12}{5}$.

Now, find the sum of these reciprocals: $\frac{-9}{4} + \frac{12}{5}$.

Find the LCM of the denominators 4 and 5, which is 20.

$\frac{-9}{4} = \frac{-9 \times 5}{4 \times 5} = \frac{-45}{20}$

$\frac{12}{5} = \frac{12 \times 4}{5 \times 4} = \frac{48}{20}$

Sum = $\frac{-45}{20} + \frac{48}{20} = \frac{-45 + 48}{20} = \frac{3}{20}$

(g) Evaluate (x ÷ y) × z.

First, calculate $x \div y$. From part (d), $x \div y = \frac{-16}{15}$.

Now, multiply this result by $z = \frac{7}{18}$.

$\left( \frac{-16}{15} \right) \times \frac{7}{18}$

Multiply the numerators and the denominators:

$\frac{-16 \times 7}{15 \times 18} = \frac{-112}{270}$

Simplify the fraction by dividing the numerator and denominator by their GCD. Both are divisible by 2.

$\frac{-112 \div 2}{270 \div 2} = \frac{-56}{135}$

(h) Evaluate (x – y) + z.

First, calculate $x – y$. Use the fractions with common denominator 36.

$x - y = \frac{-16}{36} - \frac{15}{36} = \frac{-16 - 15}{36} = \frac{-31}{36}$

Now, add $z = \frac{14}{36}$ to this result.

$\left( \frac{-31}{36} \right) + \frac{14}{36} = \frac{-31 + 14}{36} = \frac{-17}{36}$

(i) Evaluate x + (y + z).

First, calculate $y + z$. Use the fractions with common denominator 36.

$y + z = \frac{15}{36} + \frac{14}{36} = \frac{15 + 14}{36} = \frac{29}{36}$

Now, add $x = \frac{-16}{36}$ to this result.

$x + (y + z) = \frac{-16}{36} + \left( \frac{29}{36} \right) = \frac{-16 + 29}{36} = \frac{13}{36}$

(j) Evaluate x ÷ (y ÷ z).

First, calculate $y \div z$.

$y \div z = \frac{5}{12} \div \frac{7}{18}$

Multiply by the reciprocal of $\frac{7}{18}$, which is $\frac{18}{7}$.

$y \div z = \frac{5}{12} \times \frac{18}{7}$

Multiply the numerators and denominators:

$y \div z = \frac{5 \times 18}{12 \times 7} = \frac{90}{84}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 6.

$\frac{90 \div 6}{84 \div 6} = \frac{15}{14}$

Now, calculate $x \div (y \div z) = \frac{-4}{9} \div \frac{15}{14}$.

Multiply by the reciprocal of $\frac{15}{14}$, which is $\frac{14}{15}$.

$\frac{-4}{9} \times \frac{14}{15}$

Multiply the numerators and the denominators:

$\frac{-4 \times 14}{9 \times 15} = \frac{-56}{135}$

(k) Evaluate x – (y + z).

First, calculate $y + z$. From part (i), $y + z = \frac{29}{36}$.

Now, calculate $x - (y + z) = \frac{-16}{36} - \frac{29}{36}$.

$x - (y + z) = \frac{-16 - 29}{36}$

$x - (y + z) = \frac{-45}{36}$

Simplify the fraction by dividing the numerator and denominator by their GCD, which is 9.

$\frac{-45 \div 9}{36 \div 9} = \frac{-5}{4}$

Solution:

The given rational number is $\frac{-1}{2}$. In decimal form, this is $-0.5$.

Natural numbers are positive integers starting from 1: $\{1, 2, 3, \dots\}$.

We need to find the natural number that is nearest to $-0.5$.

Let's look at the distances from $-0.5$ to the first few natural numbers:

Distance to 1: $|1 - (-0.5)| = |1 + 0.5| = |1.5| = 1.5$

Distance to 2: $|2 - (-0.5)| = |2 + 0.5| = |2.5| = 2.5$

The distance to any other natural number (3, 4, ...) will be greater than 1.5.

Thus, the nearest natural number to $\frac{-1}{2}$ is 1.

Let the rational number that should be added to $\frac{-1}{2}$ to obtain 1 be $R$.

We can write the equation:

$\frac{-1}{2} + R = 1$

To find $R$, we subtract $\frac{-1}{2}$ from 1:

$R = 1 - \left( \frac{-1}{2} \right)$

Subtracting a negative number is the same as adding its positive counterpart:

$R = 1 + \frac{1}{2}$

Write 1 as a fraction with denominator 2:

$R = \frac{2}{2} + \frac{1}{2}$

Add the fractions:

$R = \frac{2 + 1}{2} = \frac{3}{2}$

So, $\frac{3}{2}$ should be added to $\frac{-1}{2}$ to obtain the nearest natural number, which is 1.

Check: $\frac{-1}{2} + \frac{3}{2} = \frac{-1 + 3}{2} = \frac{2}{2} = 1$.

The rational number is $\frac{3}{2}$.

Solution:

The given rational number is $\frac{-2}{3}$.

First, let's determine the nearest integer to $\frac{-2}{3}$.

In decimal form, $\frac{-2}{3} \approx -0.666...$

We need to find the integer that is closest to $-0.666...$ on the number line.

The integers around $-0.666...$ are $-1$ and $0$.

The distance between $-0.666...$ and $-1$ is $|-1 - (-0.666...)| = |-1 + 0.666...| = |-0.333...| = 0.333... = \frac{1}{3}$.

The distance between $-0.666...$ and $0$ is $|0 - (-0.666...)| = |0.666...| = 0.666... = \frac{2}{3}$.

Comparing the distances, $\frac{1}{3} < \frac{2}{3}$.

So, the nearest integer to $\frac{-2}{3}$ is $-1$.

Let the rational number that should be subtracted from $\frac{-2}{3}$ to obtain $-1$ be $S$.

According to the question, we have the equation:

To find $S$, we rearrange the equation:

$-S = -1 - \frac{-2}{3}$

$-S = -1 + \frac{2}{3}$

To add $-1$ and $\frac{2}{3}$, we express $-1$ as a fraction with denominator 3:

$-1 = \frac{-3}{3}$

So, $-S = \frac{-3}{3} + \frac{2}{3}$

$-S = \frac{-3 + 2}{3}$

$-S = \frac{-1}{3}$

Multiply both sides by $-1$ to solve for $S$:

$S = \frac{1}{3}$

Thus, $\frac{1}{3}$ should be subtracted from $\frac{-2}{3}$ to obtain the nearest integer, $-1$.

Check: $\frac{-2}{3} - \frac{1}{3} = \frac{-2 - 1}{3} = \frac{-3}{3} = -1$.

The rational number is $\frac{1}{3}$.

Solution:

The given rational number is $\frac{-5}{8}$.

First, we need to find the integer that is nearest to $\frac{-5}{8}$.

In decimal form, $\frac{-5}{8} = -0.625$.

Let's consider the integers closest to $-0.625$ on the number line, which are $-1$ and $0$.

The distance from $-0.625$ to $-1$ is $|-1 - (-0.625)| = |-1 + 0.625| = |-0.375| = 0.375$.

The distance from $-0.625$ to $0$ is $|0 - (-0.625)| = |0.625| = 0.625$.

Comparing the distances, $0.375 < 0.625$.

So, the nearest integer to $\frac{-5}{8}$ is $-1$.

Let the rational number that should be multiplied with $\frac{-5}{8}$ to obtain $-1$ be $M$.

According to the question, we have the equation:

To find $M$, we divide both sides of the equation by $\frac{-5}{8}$. Dividing by a rational number is the same as multiplying by its reciprocal.

The reciprocal of $\frac{-5}{8}$ is $\frac{8}{-5}$.

$M = -1 \div \left( \frac{-5}{8} \right)$

$M = -1 \times \left( \frac{8}{-5} \right)$

Now, perform the multiplication:

$M = \frac{-1 \times 8}{1 \times (-5)} = \frac{-8}{-5}$

Simplify the fraction by dividing the numerator and denominator by $-1$:

$M = \frac{-8 \div (-1)}{-5 \div (-1)} = \frac{8}{5}$

Thus, $\frac{8}{5}$ should be multiplied with $\frac{-5}{8}$ to obtain the nearest integer, $-1$.

Check: $\frac{-5}{8} \times \frac{8}{5} = \frac{-5 \times 8}{8 \times 5} = \frac{-40}{40} = -1$.

The rational number is $\frac{8}{5}$.

Solution:

First, we need to identify the greatest negative integer.

The set of integers is $\{ \dots, -3, -2, -1, 0, 1, 2, \dots \}$.

The set of negative integers is $\{ \dots, -3, -2, -1 \}$.

The greatest among the negative integers is $-1$.

Let the rational number that should be divided by $\frac{1}{2}$ to obtain $-1$ be $X$.

According to the question, we can write the equation:

To solve for $X$, we use the property of division of rational numbers. Dividing by a rational number is equivalent to multiplying by its reciprocal.

The reciprocal of $\frac{1}{2}$ is $\frac{2}{1} = 2$.

So, the equation becomes:

Now, to find $X$, we divide both sides of the equation by 2:

Thus, $\frac{-1}{2}$ should be divided by $\frac{1}{2}$ to obtain the greatest negative integer, which is $-1$.

Check: $\frac{-1}{2} \div \frac{1}{2} = \frac{-1}{2} \times \frac{2}{1} = \frac{-1 \times 2}{2 \times 1} = \frac{-2}{2} = -1$. The result is the greatest negative integer.

The rational number is $\frac{-1}{2}$.

Given:

Total length of the rope = $68$ m.

Length of one piece = $4\frac{1}{4}$ m.

To Find:

The number of such pieces.

Solution:

To find the number of pieces, we need to divide the total length of the rope by the length of one piece.

Number of pieces = Total length $\div$ Length of one piece

The length of one piece is given as a mixed number, $4\frac{1}{4}$ m. Convert this mixed number into an improper fraction:

$4\frac{1}{4} = \frac{(4 \times 4) + 1}{4} = \frac{16 + 1}{4} = \frac{17}{4}$ m.

Now, we perform the division:

Number of pieces $= 68 \div \frac{17}{4}$

To divide by a rational number, we multiply by its reciprocal. The reciprocal of $\frac{17}{4}$ is $\frac{4}{17}$.

Number of pieces $= 68 \times \frac{4}{17}$

We can write 68 as $\frac{68}{1}$.

Number of pieces $= \frac{68}{1} \times \frac{4}{17}$

Now, we can simplify by cancelling the common factor between the numerator of the first fraction (68) and the denominator of the second fraction (17).

Since $68 = 4 \times 17$, 17 is a factor of 68.

Number of pieces $= \frac{\cancel{68}^{4}}{1} \times \frac{4}{\cancel{17}^{1}}$

Now, multiply the remaining terms:

Number of pieces $= \frac{4}{1} \times \frac{4}{1} = \frac{4 \times 4}{1 \times 1} = \frac{16}{1} = 16$

Thus, 16 pieces of length $4\frac{1}{4}$ m can be cut from a rope 68 m long.

The final answer is $\boxed{16}$.

Given:

Total length of cloth = $27$ m.

Number of shirts prepared = $12$.

To Find:

Length of cloth required for each shirt.

Solution:

To find the length of cloth required for each shirt, we need to divide the total length of the cloth by the number of shirts.

Length of cloth per shirt = Total length of cloth $\div$ Number of shirts

Length of cloth per shirt = $27 \text{ m} \div 12$

We can express this division as a fraction:

Length of cloth per shirt = $\frac{27}{12} \text{ m}$

Now, we simplify the fraction $\frac{27}{12}$ to its lowest form. Find the Greatest Common Divisor (GCD) of 27 and 12. The GCD of 27 and 12 is 3.

Divide the numerator and the denominator by 3:

$\frac{27}{12} = \frac{27 \div 3}{12 \div 3} = \frac{9}{4}$

So, the length of cloth required for each shirt is $\frac{9}{4}$ m.

This can also be expressed as a mixed number:

$\frac{9}{4} = 2\frac{1}{4}$

Or as a decimal:

$\frac{9}{4} = 2.25$

The length of cloth required for each shirt is $\frac{9}{4}$ m (or $2\frac{1}{4}$ m or 2.25 m).

The final answer is $\boxed{\frac{9}{4} \text{ m}}$.

Solution:

To insert rational numbers between two given rational numbers, we can express the given numbers with a common denominator. By using a sufficiently large common denominator, we can find integers between the new numerators, which then form the numerators of the rational numbers lying between the given numbers.

(i) Insert 3 rational numbers between $\frac{-1}{2}$ and $\frac{1}{5}$.

The given rational numbers are $\frac{-1}{2}$ and $\frac{1}{5}$.

Find the Least Common Multiple (LCM) of the denominators 2 and 5. LCM$(2, 5) = 10$.

Express each rational number with a denominator of 10:

$\frac{-1}{2} = \frac{-1 \times 5}{2 \times 5} = \frac{-5}{10}$

$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}$

Now we need to find 3 rational numbers between $\frac{-5}{10}$ and $\frac{2}{10}$. We can choose fractions with denominator 10 and numerators that are integers between -5 and 2. The integers between -5 and 2 are -4, -3, -2, -1, 0, 1.

We need 3 such rational numbers. We can pick any three from the following list:

$\frac{-4}{10}, \frac{-3}{10}, \frac{-2}{10}, \frac{-1}{10}, \frac{0}{10}, \frac{1}{10}$.

Let's choose $\frac{-4}{10}$, $\frac{-3}{10}$, and $\frac{-2}{10}$.

Simplify these fractions:

$\frac{-4}{10} = \frac{-4 \div 2}{10 \div 2} = \frac{-2}{5}$

$\frac{-3}{10}$ (already in simplest form)

$\frac{-2}{10} = \frac{-2 \div 2}{10 \div 2} = \frac{-1}{5}$

Thus, three rational numbers between $\frac{-1}{2}$ and $\frac{1}{5}$ are $\frac{-2}{5}$, $\frac{-3}{10}$, and $\frac{-1}{5}$.

(ii) Insert 3 rational numbers between 0 and –10.

The given numbers are 0 and –10. We want to find 3 rational numbers between them.

Rewrite the numbers as rational numbers: $0 = \frac{0}{1}$ and $-10 = \frac{-10}{1}$.

We need to find 3 rational numbers between $\frac{-10}{1}$ and $\frac{0}{1}$. To do this, we can use a larger common denominator.

Let's use a denominator of 4 (we need 3 numbers, $3+1=4$).

$\frac{-10}{1} = \frac{-10 \times 4}{1 \times 4} = \frac{-40}{4}$

$\frac{0}{1} = \frac{0 \times 4}{1 \times 4} = \frac{0}{4}$

Now we need to find 3 rational numbers between $\frac{-40}{4}$ and $\frac{0}{4}$. We can choose fractions with denominator 4 and numerators that are integers between -40 and 0. The integers between -40 and 0 are -39, -38, -37, ..., -1.

We need 3 such rational numbers. Let's choose the first three integers from the list: -39, -38, -37.

The rational numbers are $\frac{-39}{4}$, $\frac{-38}{4}$, and $\frac{-37}{4}$.

Simplify these fractions if possible:

$\frac{-39}{4}$ (already in simplest form)

$\frac{-38}{4} = \frac{-38 \div 2}{4 \div 2} = \frac{-19}{2}$

$\frac{-37}{4}$ (already in simplest form)

Thus, three rational numbers between 0 and –10 are $\frac{-39}{4}$, $\frac{-19}{2}$, and $\frac{-37}{4}$.

Note: Many other sets of 3 rational numbers are possible between the given numbers.

Solution:

We will classify each given number and mark the appropriate categories with (√).

Recall the definitions of the number sets:

Natural Numbers: The set of positive integers $\{1, 2, 3, \dots\}$.

Whole Numbers: The set of non-negative integers $\{0, 1, 2, 3, \dots\}$.

Integers: The set of whole numbers and their additive inverses $\{\dots, -2, -1, 0, 1, 2, \dots\}$.

Fraction: A number that can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. In the context of distinguishing from integers, this often refers to rational numbers that are not integers.

Rational Number: Any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. All integers and fractions are rational numbers.

Let's analyse each number:

(a) – 114: This is a negative integer. It is not a natural number or a whole number. It is an integer and a rational number (as it can be written as $\frac{-114}{1}$). In the context of distinguishing from integers, it is not typically called a fraction.

(b) $\frac{19}{27}$: This is a positive rational number that is not an integer. It is in the form of a fraction. It is not a natural number, whole number, or integer.

(c) $\frac{623}{1}$: This number is equal to 623. This is a positive integer. It is a natural number, whole number, and integer. As an integer, it is also a rational number ($\frac{623}{1}$). It is written in fractional form, but its value is an integer, so in the context of distinguishing from integers, it is not typically marked as a fraction.

(d) $-19\frac{3}{4}$: This is a negative mixed number. Converting to an improper fraction gives $\frac{(-19 \times 4) + 3}{4} = \frac{-76 + 3}{4} = \frac{-73}{4}$ (or $-\frac{79}{4}$ - check calculation... $-19\frac{3}{4} = -(19 + 3/4) = -((19 \times 4 + 3)/4) = -(76+3)/4 = -79/4$). It is a negative rational number that is not an integer. It is in the form of a mixed fraction, so it is a fraction. It is not a natural number, whole number, or integer.

(e) $\frac{73}{71}$: This is a positive improper fraction. Its value is $1\frac{2}{71}$, which is not an integer. It is in the form of a fraction. It is not a natural number, whole number, or integer. It is a rational number.

(f) 0: This is the integer zero. It is a whole number and an integer. It is not a natural number. It is a rational number (as it can be written as $\frac{0}{1}$). In the context of distinguishing from integers, it is not typically called a fraction.

Based on this analysis, the completed table is:

Solution:

We are given two different numbers, $a$ and $b$, taken from the set of integers $\{1, 2, 3, \dots, 50\}$. We need to find the largest values of two expressions involving $a$ and $b$.

Part 1: Largest value of $\frac{a - b}{a + b}$.

The numbers $a$ and $b$ are positive integers, so their sum $a+b$ is always positive ($a+b \geq 1+2 = 3$).

To maximize the fraction $\frac{a - b}{a + b}$, we want the numerator $a-b$ to be as large as possible (positive) and the denominator $a+b$ to be as small as possible (positive).

For $a-b$ to be positive and large, $a$ must be greater than $b$, and $a$ should be large while $b$ should be small.

The largest possible value for $a$ is 50.

The smallest possible value for $b$ is 1.

Since $a$ and $b$ must be different, we can choose $a = 50$ and $b = 1$. Both are in the set $\{1, \dots, 50\}$ and $a \neq b$.

Let's calculate the value of the expression with these numbers:

Numerator $a - b = 50 - 1 = 49$

Denominator $a + b = 50 + 1 = 51$

Value of the expression = $\frac{a - b}{a + b} = \frac{49}{51}$.

Any other choice where $a > b$ would result in a smaller $a$ or a larger $b$, which would either decrease the numerator $a-b$ or increase the denominator $a+b$ (or both), leading to a smaller fractional value compared to $\frac{49}{51}$. For example, if $a=49, b=1$, the value is $\frac{49-1}{49+1} = \frac{48}{50} = \frac{24}{25}$. $\frac{49}{51} \approx 0.96$ while $\frac{24}{25} = 0.96$. Let's check carefully: $\frac{49}{51}$ vs $\frac{48}{50}$. $49 \times 50 = 2450$, $51 \times 48 = 2448$. Since $2450 > 2448$, $\frac{49}{51} > \frac{48}{50}$.

If $a < b$, the numerator $a-b$ will be negative, making the fraction negative. Any positive value is greater than any negative value. Thus, the maximum value occurs when $a-b$ is positive.

The largest value that $\frac{a - b}{a + b}$ can have is $\frac{49}{51}$.

Part 2: Largest value of $\frac{a + b}{a - b}$.

To maximize the fraction $\frac{a + b}{a - b}$, we want the numerator $a+b$ to be as large as possible and the denominator $a-b$ to be as small as possible (but not zero, and ideally positive for a large value). Note that $a \neq b$, so $a-b \neq 0$.

Case 1: $a > b$. The denominator $a-b$ is a positive integer. To make the denominator as small as possible, we choose the smallest possible positive integer value for $a-b$, which is 1.

If $a-b = 1$, then $a = b + 1$. To maximize the numerator $a+b$, we need to maximize $a$ and $b$. Since $a=b+1$ and $a, b \in \{1, \dots, 50\}$, the largest possible value for $b$ is 49 (when $a=50$). So, we choose $a=50$ and $b=49$. Both are in the set $\{1, \dots, 50\}$ and $a \neq b$.

Let's calculate the value of the expression with these numbers:

Numerator $a + b = 50 + 49 = 99$

Denominator $a - b = 50 - 49 = 1$

Value of the expression = $\frac{a + b}{a - b} = \frac{99}{1} = 99$.

Case 2: $a < b$. The denominator $a-b$ is a negative integer. Let $a-b = -k$, where $k$ is a positive integer. The expression becomes $\frac{a+b}{-k}$. This will result in a negative value.

To make the value largest (least negative, closest to 0), we would want the numerator $a+b$ to be small and the denominator $|a-b|$ to be large. For example, if $a=1, b=50$, $\frac{1+50}{1-50} = \frac{51}{-49}$.

Comparing the positive value 99 (from Case 1) with any negative value (from Case 2), the positive value 99 is always greater.

Thus, the largest value that $\frac{a + b}{a - b}$ can have is 99.

The largest value that $\frac{a - b}{a + b}$ can have is $\frac{49}{51}$.

The largest value that $\frac{a + b}{a - b}$ can have is 99.

Given:

Total number of students = 150.

Percentage of students studying English = 62%.

Percentage of students studying Maths = 68%.

All 150 students are studying English, Maths, or both.

To Find:

Number of students studying both English and Maths.

Solution:

Let E be the set of students studying English, and M be the set of students studying Maths.

We are given that every student studies at least one subject, which means the total number of students is equal to the number of students in the union of the sets E and M, i.e., $|\text{E} \cup \text{M}| = 150$.

In terms of percentages, this means that the percentage of students studying English or Maths or both is 100% of the total students.

We are given the percentage of students studying English, $|\text{E}|/\text{Total} = 62\%$, and the percentage of students studying Maths, $|\text{M}|/\text{Total} = 68\%$.

When we add the percentage of students studying English and the percentage of students studying Maths, the students who are studying both subjects are counted twice (once in English and once in Maths).

The sum of the given percentages is $62\% + 68\%$.

$62\% + 68\% = 130\%$.

This sum (130%) represents the percentage of students studying English plus the percentage of students studying Maths. It is greater than the total percentage of students (100%) because the group studying both is included in both percentages.

The excess percentage over 100% is the percentage of students studying both subjects.

Percentage of students studying both = (Percentage studying English) + (Percentage studying Maths) - (Percentage studying English or Maths or both)

Percentage of students studying both = $130\% - 100\% = 30\%$.

So, 30% of the total students are studying both English and Maths.

Now, we calculate the number of students who are studying both subjects.

Number of students studying both = $30\%$ of 150.

Number of students studying both = $\frac{30}{100} \times 150$

Number of students studying both = $\frac{30 \times 150}{100}$

Number of students studying both = $\frac{4500}{100}$

Number of students studying both = 45.

Alternatively, we can calculate the number of students for each subject first.

Number of students studying English = 62% of 150 = $\frac{62}{100} \times 150 = \frac{62 \times 3}{2} = 31 \times 3 = 93$.

Number of students studying Maths = 68% of 150 = $\frac{68}{100} \times 150 = \frac{68 \times 3}{2} = 34 \times 3 = 102$.

Total count when summing individuals studying English and Maths = $93 + 102 = 195$.

The total number of students is 150. The excess count (195 - 150) represents the students who were counted twice, i.e., those studying both subjects.

Number of students studying both = $195 - 150 = 45$.

The final answer is $\boxed{45}$.

Given:

A body floats with $\frac{2}{9}$ of its volume above the surface.

To Find:

The ratio of the body's submerged volume to its exposed volume.

Express this ratio as a rational number.

Solution:

Let the total volume of the body be $V$.

The volume of the body exposed above the surface (exposed volume) is given as $\frac{2}{9}$ of the total volume.

Exposed volume = $\frac{2}{9} \times V$

The volume of the body submerged below the surface (submerged volume) is the total volume minus the exposed volume.

Submerged volume = Total volume - Exposed volume

Submerged volume = $V - \frac{2}{9} V$

Submerged volume = $\left( 1 - \frac{2}{9} \right) V$

Submerged volume = $\left( \frac{9}{9} - \frac{2}{9} \right) V$

Submerged volume = $\frac{9 - 2}{9} V = \frac{7}{9} V$

We need to find the ratio of the submerged volume to the exposed volume.

Ratio = $\frac{\text{Submerged volume}}{\text{Exposed volume}}$

Ratio = $\frac{\frac{7}{9} V}{\frac{2}{9} V}$

Assuming $V \neq 0$, we can cancel $V$ from the numerator and denominator.

Ratio = $\frac{\frac{7}{9}}{\frac{2}{9}}$

To divide the fractions, we multiply the numerator by the reciprocal of the denominator.

Ratio = $\frac{7}{9} \div \frac{2}{9} = \frac{7}{9} \times \frac{9}{2}$

We can cancel the common factor 9 from the numerator and denominator.

Ratio = $\frac{7}{\cancel{9}} \times \frac{\cancel{9}}{2} = \frac{7}{2}$

The ratio of the submerged volume to its exposed volume is $7:2$.

Finally, we need to re-write this ratio as a rational number in the form $\frac{p}{q}$.

A ratio $a:b$ can be written as a rational number $\frac{a}{b}$.

The ratio $7:2$ as a rational number is $\frac{7}{2}$.

This rational number is in its standard form as the denominator (2) is positive and the greatest common divisor of the numerator (7) and the denominator (2) is 1.

The ratio of the body submerged volume to its exposed volume is $7:2$.

As a rational number, this is $\frac{7}{2}$.

To find the odd one out, we need to evaluate each given expression.

Evaluation of (a):

The expression is $\frac{4}{3} \times \frac{3}{4}$.

Product = $\frac{4}{3} \times \frac{3}{4} = \frac{\cancel{4}^1}{\cancel{3}^1} \times \frac{\cancel{3}^1}{\cancel{4}^1} = \frac{1}{1} \times \frac{1}{1} = 1$.

Evaluation of (b):

The expression is $\frac{-3}{2} \times \frac{-2}{3}$.

Product = $\frac{-3}{2} \times \frac{-2}{3} = \frac{\cancel{-3}^{-1}}{\cancel{2}^1} \times \frac{\cancel{-2}^{-1}}{\cancel{3}^1} = \frac{-1}{1} \times \frac{-1}{1} = (-1) \times (-1) = 1$.

Evaluation of (c):

The expression is $2 \times \frac{1}{2}$.

Product = $2 \times \frac{1}{2} = \frac{2}{1} \times \frac{1}{2} = \frac{\cancel{2}^1}{1} \times \frac{1}{\cancel{2}^1} = \frac{1}{1} \times \frac{1}{1} = 1$.

Evaluation of (d):

The expression is $\frac{-1}{3} \times \frac{3}{1}$.

Product = $\frac{-1}{3} \times \frac{3}{1} = \frac{-1}{\cancel{3}^1} \times \frac{\cancel{3}^1}{1} = \frac{-1}{1} \times \frac{1}{1} = -1$.

Comparing the results of the evaluations:

(a) Result = 1

(b) Result = 1

(c) Result = 1

(d) Result = -1

The results for options (a), (b), and (c) are all $1$, while the result for option (d) is $-1$.

The numbers in options (a), (b), and (c) are pairs of multiplicative inverses (reciprocals), because their product is $1$. The numbers in option (d) are not multiplicative inverses as their product is $-1$.

Therefore, the odd one out is (d) $\frac{-1}{3} \times \frac{3}{1}$.

To find the odd one out, we will simplify each fraction to its lowest terms.

(a) $\frac{4}{-9}$

This fraction is already in its lowest terms. The standard form is $\frac{-4}{9}$.

(b) $\frac{-16}{36}$

We can divide both the numerator and the denominator by their greatest common divisor, which is 4.

$\frac{-16}{36} = \frac{\cancel{-16}^{-4}}{\cancel{36}^9} = \frac{-4}{9}$.

(c) $\frac{-20}{-45}$

We can divide both the numerator and the denominator by their greatest common divisor, which is 5. Also, the division of a negative number by a negative number results in a positive number.

$\frac{-20}{-45} = \frac{\cancel{-20}^{-4}}{\cancel{-45}^{-9}} = \frac{-4}{-9} = \frac{4}{9}$.

(d) $\frac{28}{-63}$

We can divide both the numerator and the denominator by their greatest common divisor, which is 7.

$\frac{28}{-63} = \frac{\cancel{28}^{4}}{\cancel{-63}^{-9}} = \frac{4}{-9} = \frac{-4}{9}$.

Comparing the simplified forms:

(a) $\frac{-4}{9}$

(b) $\frac{-4}{9}$

(c) $\frac{4}{9}$

(d) $\frac{-4}{9}$

The simplified forms of options (a), (b), and (d) are $\frac{-4}{9}$, which is a negative rational number. The simplified form of option (c) is $\frac{4}{9}$, which is a positive rational number.

Thus, option (c) is different from the others.

The odd one out is (c) $\frac{-20}{-45}$ because it represents a positive rational number $\frac{4}{9}$, while the other options represent the same negative rational number $\frac{-4}{9}$.