Chapter 1 Rational Numbers

To find the sum, we can rearrange the terms using the commutative and associative properties of addition to group fractions with related denominators.

The expression is: $\frac{3}{7} + \left( \frac{-6}{11} \right) + \left( \frac{-8}{21} \right) + \left( \frac{5}{22} \right)$

Rearrange the terms:

$\left( \frac{3}{7} + \frac{-8}{21} \right) + \left( \frac{-6}{11} + \frac{5}{22} \right)$

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First, calculate the sum of the first group: $\left( \frac{3}{7} + \frac{-8}{21} \right)$

The Least Common Multiple (LCM) of the denominators 7 and 21 is 21.

Convert $\frac{3}{7}$ to an equivalent fraction with denominator 21:

$\frac{3}{7} = \frac{3 \times 3}{7 \times 3} = \frac{9}{21}$

Now add the fractions:

$\frac{9}{21} + \frac{-8}{21} = \frac{9 + (-8)}{21} = \frac{9 - 8}{21} = \frac{1}{21}$

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Next, calculate the sum of the second group: $\left( \frac{-6}{11} + \frac{5}{22} \right)$

The Least Common Multiple (LCM) of the denominators 11 and 22 is 22.

Convert $\frac{-6}{11}$ to an equivalent fraction with denominator 22:

$\frac{-6}{11} = \frac{-6 \times 2}{11 \times 2} = \frac{-12}{22}$

Now add the fractions:

$\frac{-12}{22} + \frac{5}{22} = \frac{-12 + 5}{22} = \frac{-7}{22}$

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Finally, add the results of the two groups:

$\frac{1}{21} + \frac{-7}{22}$

The Least Common Multiple (LCM) of the denominators 21 and 22 is needed.

Prime factors: $21 = 3 \times 7$ and $22 = 2 \times 11$.

LCM(21, 22) = $2 \times 3 \times 7 \times 11 = 462$.

Convert the fractions to equivalent fractions with denominator 462:

$\frac{1}{21} = \frac{1 \times 22}{21 \times 22} = \frac{22}{462}$

$\frac{-7}{22} = \frac{-7 \times 21}{22 \times 21} = \frac{-147}{462}$

Now add the equivalent fractions:

$\frac{22}{462} + \frac{-147}{462} = \frac{22 + (-147)}{462} = \frac{22 - 147}{462} = \frac{-125}{462}$

The fraction $\frac{-125}{462}$ cannot be simplified further as 125 ($5^3$) and 462 ($2 \times 3 \times 7 \times 11$) have no common factors other than 1.

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Final Answer:

The sum is $\mathbf{\frac{-125}{462}}$.

We need to calculate the product of the given rational numbers:

$\frac{-4}{5} \times \frac{3}{7} \times \frac{15}{16} \times \frac{-14}{9}$

We can rearrange the terms using the commutative and associative properties of multiplication to simplify the calculation.

$\left( \frac{-4}{5} \times \frac{15}{16} \right) \times \left( \frac{3}{7} \times \frac{-14}{9} \right)$

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First, calculate the product of the first group: $\left( \frac{-4}{5} \times \frac{15}{16} \right)$

$\frac{-4 \times 15}{5 \times 16}$

We can simplify by cancelling common factors:

$\frac{\cancel{-4}^ {-1}}{\cancel{5}_1} \times \frac{\cancel{15}^3}{\cancel{16}_4} = \frac{-1 \times 3}{1 \times 4} = \frac{-3}{4}$

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Next, calculate the product of the second group: $\left( \frac{3}{7} \times \frac{-14}{9} \right)$

$\frac{3 \times (-14)}{7 \times 9}$

We can simplify by cancelling common factors:

$\frac{\cancel{3}^1}{\cancel{7}_1} \times \frac{\cancel{-14}^{-2}}{\cancel{9}_3} = \frac{1 \times (-2)}{1 \times 3} = \frac{-2}{3}$

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Finally, multiply the results of the two groups:

$\left( \frac{-3}{4} \right) \times \left( \frac{-2}{3} \right)$

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

The product of two negative numbers is positive:

$\frac{3 \times 2}{4 \times 3}$

We can simplify by cancelling common factors:

$\frac{\cancel{3}^1 \times \cancel{2}^1}{\cancel{4}_2 \times \cancel{3}_1} = \frac{1 \times 1}{2 \times 1} = \frac{1}{2}$

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Alternate Method: Multiply all numerators and denominators directly.

$\frac{-4 \times 3 \times 15 \times (-14)}{5 \times 7 \times 16 \times 9}$

Since there are two negative signs, the result will be positive.

$\frac{4 \times 3 \times 15 \times 14}{5 \times 7 \times 16 \times 9}$

Now, cancel common factors:

$\frac{\cancel{4}^1 \times \cancel{3}^1 \times \cancel{15}^{\cancel{3}^1 \times \cancel{5}^1} \times \cancel{14}^2}{\cancel{5}_1 \times \cancel{7}_1 \times \cancel{16}_{4} \times \cancel{9}_{\cancel{3}_1}}$

Let's simplify step-by-step:

$\frac{4 \times 3 \times 15 \times 14}{5 \times 7 \times 16 \times 9} = \frac{\cancel{4}^1 \times 3 \times 15 \times 14}{5 \times 7 \times \cancel{16}_4 \times 9} = \frac{1 \times 3 \times 15 \times 14}{5 \times 7 \times 4 \times 9}$

$ = \frac{1 \times \cancel{3}^1 \times 15 \times 14}{5 \times 7 \times 4 \times \cancel{9}_3} = \frac{1 \times 1 \times 15 \times 14}{5 \times 7 \times 4 \times 3} $

$ = \frac{1 \times 1 \times \cancel{15}^3 \times 14}{\cancel{5}_1 \times 7 \times 4 \times 3} = \frac{1 \times 1 \times 3 \times 14}{1 \times 7 \times 4 \times 3} $

$ = \frac{1 \times 1 \times \cancel{3}^1 \times 14}{1 \times 7 \times 4 \times \cancel{3}_1} = \frac{1 \times 1 \times 1 \times 14}{1 \times 7 \times 4 \times 1} $

$ = \frac{14}{7 \times 4} = \frac{\cancel{14}^2}{\cancel{7}_1 \times 4} = \frac{2}{4} = \frac{\cancel{2}^1}{\cancel{4}_2} = \frac{1}{2} $

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Final Answer:

The result of the multiplication is $\mathbf{\frac{1}{2}}$.

The additive inverse of a number $a$ is the number $-a$ such that $a + (-a) = 0$. For a rational number $\frac{p}{q}$, its additive inverse is $-\frac{p}{q}$.

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(i) $\frac{-7}{19}$

The given number is $\frac{-7}{19}$.

The additive inverse is $- \left( \frac{-7}{19} \right)$.

$- \left( \frac{-7}{19} \right) = \frac{7}{19}$.

So, the additive inverse of $\frac{-7}{19}$ is $\frac{7}{19}$.

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(ii) $\frac{21}{112}$

The given number is $\frac{21}{112}$.

The additive inverse is $- \left( \frac{21}{112} \right)$.

$- \left( \frac{21}{112} \right) = -\frac{21}{112}$.

So, the additive inverse of $\frac{21}{112}$ is $-\frac{21}{112}$.

(Note: While $\frac{21}{112}$ can be simplified to $\frac{3}{16}$, the additive inverse of the given form $\frac{21}{112}$ is $-\frac{21}{112}$.)

We need to verify the property that the negative of the negative of a number is the number itself, i.e., $-(-x) = x$.

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(i) For $x = \frac{13}{17}$

First, find $-x$:

$-x = -\left(\frac{13}{17}\right) = \frac{-13}{17}$

Next, find $-(-x)$:

$-(-x) = -\left(\frac{-13}{17}\right)$

The negative of a negative number is the positive number.

$-(-x) = \frac{13}{17}$

Now, compare $-(-x)$ with $x$.

We found $-(-x) = \frac{13}{17}$, and we are given $x = \frac{13}{17}$.

Since $\frac{13}{17} = \frac{13}{17}$, we have verified that $-(-x) = x$ for $x = \frac{13}{17}$.

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(ii) For $x = \frac{-21}{31}$

First, find $-x$:

$-x = -\left(\frac{-21}{31}\right)$

The negative of a negative number is the positive number.

$-x = \frac{21}{31}$

Next, find $-(-x)$:

$-(-x) = -\left(\frac{21}{31}\right)$

$-(-x) = \frac{-21}{31}$

Now, compare $-(-x)$ with $x$.

We found $-(-x) = \frac{-21}{31}$, and we are given $x = \frac{-21}{31}$.

Since $\frac{-21}{31} = \frac{-21}{31}$, we have verified that $-(-x) = x$ for $x = \frac{-21}{31}$.

We need to evaluate the expression:

$\frac{2}{5} \times \frac{-3}{7} - \frac{1}{14} - \frac{3}{7} \times \frac{3}{5}$

Notice that the terms $\frac{2}{5} \times \frac{-3}{7}$ and $\frac{3}{7} \times \frac{3}{5}$ involve multiplication with $\frac{3}{7}$ (or its negative). We can use the commutative property to rearrange the last term: $\frac{3}{7} \times \frac{3}{5} = \frac{3}{5} \times \frac{3}{7}$.

The expression becomes:

$\frac{2}{5} \times \left(\frac{-3}{7}\right) - \frac{1}{14} - \frac{3}{5} \times \frac{3}{7}$

We can rewrite $\frac{2}{5} \times \left(\frac{-3}{7}\right)$ as $-\frac{2}{5} \times \frac{3}{7}$.

$-\frac{2}{5} \times \frac{3}{7} - \frac{1}{14} - \frac{3}{5} \times \frac{3}{7}$

Now, we can use the distributive property by factoring out $-\frac{3}{7}$ from the first and third terms, or $\frac{3}{7}$ and managing the signs.

Let's group the terms involving multiplication:

$\left(-\frac{2}{5} \times \frac{3}{7}\right) - \left(\frac{3}{5} \times \frac{3}{7}\right) - \frac{1}{14}$

Factor out $\frac{3}{7}$ from the first two terms (or $-\frac{3}{7}$):

$\frac{3}{7} \times \left(-\frac{2}{5} - \frac{3}{5}\right) - \frac{1}{14}$

Calculate the sum inside the parenthesis:

$-\frac{2}{5} - \frac{3}{5} = \frac{-2 - 3}{5} = \frac{-5}{5} = -1$

Substitute this back into the expression:

$\frac{3}{7} \times (-1) - \frac{1}{14}$

$-\frac{3}{7} - \frac{1}{14}$

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Now, perform the subtraction. The Least Common Multiple (LCM) of 7 and 14 is 14.

Convert $-\frac{3}{7}$ to an equivalent fraction with denominator 14:

$-\frac{3}{7} = -\frac{3 \times 2}{7 \times 2} = -\frac{6}{14}$

Perform the subtraction:

$-\frac{6}{14} - \frac{1}{14} = \frac{-6 - 1}{14} = \frac{-7}{14}$

Simplify the result:

$\frac{-7}{14} = -\frac{\cancel{7}^1}{\cancel{14}_2} = -\frac{1}{2}$

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Final Answer:

The value of the expression is $\mathbf{-\frac{1}{2}}$.

(i) $-\frac{2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}$

We can rearrange the terms using the commutative property to group terms with the common factor $\frac{3}{5}$:

$-\frac{2}{3} \times \frac{3}{5} - \frac{3}{5} \times \frac{1}{6} + \frac{5}{2}$

Now, use the distributive property by factoring out $-\frac{3}{5}$ from the first two terms:

$-\frac{3}{5} \times \left( \frac{2}{3} + \frac{1}{6} \right) + \frac{5}{2}$

First, calculate the sum inside the parenthesis. The LCM of 3 and 6 is 6.

$\frac{2}{3} + \frac{1}{6} = \frac{2 \times 2}{3 \times 2} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{4+1}{6} = \frac{5}{6}$

Substitute this back into the expression:

$-\frac{3}{5} \times \frac{5}{6} + \frac{5}{2}$

Perform the multiplication. Cancel common factors:

$-\frac{\cancel{3}^1}{\cancel{5}_1} \times \frac{\cancel{5}^1}{\cancel{6}_2} = -\frac{1 \times 1}{1 \times 2} = -\frac{1}{2}$

Now, perform the addition:

$-\frac{1}{2} + \frac{5}{2} = \frac{-1 + 5}{2} = \frac{4}{2}$

Simplify the result:

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

Therefore, $-\frac{2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6} = \mathbf{2}$.

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(ii) $\frac{2}{5} \times \left( -\frac{3}{7} \right) - \frac{1}{6} \times \frac{3}{2} + \frac{1}{14} \times \frac{2}{5}$

We can rearrange the terms using the commutative property to group terms with the common factor $\frac{2}{5}$:

$\frac{2}{5} \times \left( -\frac{3}{7} \right) + \frac{1}{14} \times \frac{2}{5} - \frac{1}{6} \times \frac{3}{2}$

Use the distributive property by factoring out $\frac{2}{5}$ from the first two terms:

$\frac{2}{5} \times \left( -\frac{3}{7} + \frac{1}{14} \right) - \frac{1}{6} \times \frac{3}{2}$

Calculate the sum inside the first parenthesis. The LCM of 7 and 14 is 14.

$-\frac{3}{7} + \frac{1}{14} = -\frac{3 \times 2}{7 \times 2} + \frac{1}{14} = -\frac{6}{14} + \frac{1}{14} = \frac{-6 + 1}{14} = -\frac{5}{14}$

Calculate the product in the last term:

$\frac{1}{6} \times \frac{3}{2} = \frac{1 \times 3}{6 \times 2} = \frac{3}{12} = \frac{1}{4}$

Substitute these back into the expression:

$\frac{2}{5} \times \left( -\frac{5}{14} \right) - \frac{1}{4}$

Perform the multiplication. Cancel common factors:

$\frac{\cancel{2}^1}{\cancel{5}_1} \times \frac{-\cancel{5}^{-1}}{\cancel{14}_7} = \frac{1 \times (-1)}{1 \times 7} = -\frac{1}{7}$

Now, perform the subtraction:

$-\frac{1}{7} - \frac{1}{4}$

The LCM of 7 and 4 is 28.

$-\frac{1 \times 4}{7 \times 4} - \frac{1 \times 7}{4 \times 7} = -\frac{4}{28} - \frac{7}{28} = \frac{-4 - 7}{28} = \frac{-11}{28}$

Therefore, $\frac{2}{5} \times \left( -\frac{3}{7} \right) - \frac{1}{6} \times \frac{3}{2} + \frac{1}{14} \times \frac{2}{5} = \mathbf{-\frac{11}{28}}$.

The additive inverse of a number $x$ is the number $-x$ such that $x + (-x) = 0$. For a rational number, the additive inverse is obtained by changing its sign.

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(i) $\frac{2}{8}$

The additive inverse of $\frac{2}{8}$ is $-\left(\frac{2}{8}\right) = \mathbf{-\frac{2}{8}}$ (or $\mathbf{-\frac{1}{4}}$ after simplification).

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(ii) $\frac{-5}{9}$

The additive inverse of $\frac{-5}{9}$ is $-\left(\frac{-5}{9}\right) = \mathbf{\frac{5}{9}}$.

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(iii) $\frac{-6}{-5}$

First, simplify the given number: $\frac{-6}{-5} = \frac{6}{5}$.

The additive inverse of $\frac{6}{5}$ is $-\left(\frac{6}{5}\right) = \mathbf{-\frac{6}{5}}$.

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(iv) $\frac{2}{-9}$

First, write the number in standard form: $\frac{2}{-9} = -\frac{2}{9}$.

The additive inverse of $-\frac{2}{9}$ is $-\left(-\frac{2}{9}\right) = \mathbf{\frac{2}{9}}$.

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(v) $\frac{19}{-6}$

First, write the number in standard form: $\frac{19}{-6} = -\frac{19}{6}$.

The additive inverse of $-\frac{19}{6}$ is $-\left(-\frac{19}{6}\right) = \mathbf{\frac{19}{6}}$.

We need to verify the identity $-(-x) = x$ for the given values of $x$. This identity states that the additive inverse of the additive inverse of a number is the number itself.

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(i) For $x = \frac{11}{15}$

We are given $x = \frac{11}{15}$.

The additive inverse of $x$ is $-x$.

$-x = -\left(\frac{11}{15}\right) = -\frac{11}{15}$.

Now, we find the additive inverse of $-x$, which is $-(-x)$.

$-(-x) = -\left(-\frac{11}{15}\right)$.

The negative of a negative number is the positive number.

$-(-x) = \frac{11}{15}$.

Comparing $-(-x)$ with the original value of $x$:

We found $-(-x) = \frac{11}{15}$, and we were given $x = \frac{11}{15}$.

Since $\frac{11}{15} = \frac{11}{15}$, the identity $-(-x) = x$ is verified for $x = \frac{11}{15}$.

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(ii) For $x = -\frac{13}{17}$

We are given $x = -\frac{13}{17}$.

The additive inverse of $x$ is $-x$.

$-x = -\left(-\frac{13}{17}\right)$.

The negative of a negative number is the positive number.

$-x = \frac{13}{17}$.

Now, we find the additive inverse of $-x$, which is $-(-x)$.

$-(-x) = -\left(\frac{13}{17}\right)$.

$-(-x) = -\frac{13}{17}$.

Comparing $-(-x)$ with the original value of $x$:

We found $-(-x) = -\frac{13}{17}$, and we were given $x = -\frac{13}{17}$.

Since $-\frac{13}{17} = -\frac{13}{17}$, the identity $-(-x) = x$ is verified for $x = -\frac{13}{17}$.

The multiplicative inverse of a non-zero number $x$ is $\frac{1}{x}$ such that their product $x \times \frac{1}{x} = 1$. For a non-zero rational number $\frac{p}{q}$, the multiplicative inverse is $\frac{q}{p}$.

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(i) – 13

The number is -13, which can be written as $\frac{-13}{1}$.

The multiplicative inverse is $\frac{1}{-13}$.

Writing in standard form, the multiplicative inverse is $\mathbf{-\frac{1}{13}}$.

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(ii) $\frac{-13}{19}$

The number is $\frac{-13}{19}$.

The multiplicative inverse is $\frac{19}{-13}$.

Writing in standard form, the multiplicative inverse is $\mathbf{-\frac{19}{13}}$.

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(iii) $\frac{1}{5}$

The number is $\frac{1}{5}$.

The multiplicative inverse is $\frac{5}{1}$.

The multiplicative inverse is 5.

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(iv) $\frac{-5}{8} \times \frac{-3}{7}$

First, calculate the product:

$\frac{-5}{8} \times \frac{-3}{7} = \frac{(-5) \times (-3)}{8 \times 7} = \frac{15}{56}$.

Now, find the multiplicative inverse of $\frac{15}{56}$.

The multiplicative inverse is $\mathbf{\frac{56}{15}}$.

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(v) – 1 × $\frac{-2}{5}$

First, calculate the product:

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

Now, find the multiplicative inverse of $\frac{2}{5}$.

The multiplicative inverse is $\mathbf{\frac{5}{2}}$.

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(vi) – 1

The number is -1, which can be written as $\frac{-1}{1}$.

The multiplicative inverse is $\frac{1}{-1}$.

Writing in standard form, the multiplicative inverse is $\mathbf{-1}$.

(i) $\frac{-4}{5} \times 1 = 1 \times \frac{-4}{5} = -\frac{4}{5}$

The property shown is that multiplying any rational number by 1 results in the same rational number. The number 1 is the multiplicative identity for rational numbers.

The property used is the Multiplicative Identity.

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(ii) $-\frac{13}{17} \times \frac{-2}{7} = \frac{-2}{7} \times \frac{-13}{17}$

The property shown is that the order in which two rational numbers are multiplied does not change the result ($a \times b = b \times a$).

The property used is the Commutative Property of Multiplication.

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(iii) $\frac{-19}{29} \times \frac{29}{-19} = 1$

The property shown is that the product of a non-zero rational number and its reciprocal (or multiplicative inverse) is 1 ($a \times \frac{1}{a} = 1$). Here, $\frac{29}{-19}$ is the multiplicative inverse of $\frac{-19}{29}$.

The property used is the Multiplicative Inverse Property.

To Find:

The result of multiplying $\frac{6}{13}$ by the reciprocal of $\frac{-7}{16}$.

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Solution:

First, find the reciprocal of $\frac{-7}{16}$.

The reciprocal of a non-zero rational number $\frac{a}{b}$ is $\frac{b}{a}$.

So, the reciprocal of $\frac{-7}{16}$ is $\frac{16}{-7}$.

We can write this in standard form as $-\frac{16}{7}$.

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Now, multiply $\frac{6}{13}$ by the reciprocal we found ($-\frac{16}{7}$):

$\frac{6}{13} \times \left(-\frac{16}{7}\right)$

Multiply the numerators and the denominators:

$\frac{6 \times (-16)}{13 \times 7}$

$\frac{-96}{91}$

The resulting fraction $\frac{-96}{91}$ cannot be simplified further, as 96 ($2^5 \times 3$) and 91 ($7 \times 13$) have no common factors other than 1.

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Final Answer:

The result is $\mathbf{-\frac{96}{91}}$.

We are asked to identify the property that allows changing the computation from:

$\frac{1}{3} \times \left( 6 \times \frac{4}{3} \right)$

to:

$\left( \frac{1}{3} \times 6 \right) \times \frac{4}{3}$

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Observe the change: The numbers involved in the multiplication are $\frac{1}{3}$, 6, and $\frac{4}{3}$. The order of the numbers remains the same, but the way they are grouped using parentheses has changed.

In the first expression, $6$ and $\frac{4}{3}$ are grouped together first.

In the second expression, $\frac{1}{3}$ and $6$ are grouped together first.

The property that states that the way in which factors are grouped in a multiplication problem does not change the product is called the Associative Property of Multiplication.

This property states that for any rational numbers $a$, $b$, and $c$:

$a \times (b \times c) = (a \times b) \times c$

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Therefore, the property that allows this computation is the Associative Property of Multiplication.

Two numbers are multiplicative inverses of each other if their product is equal to 1.

We need to check if the product of $\frac{8}{9}$ and $-1\frac{1}{8}$ is equal to 1.

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First, convert the mixed fraction $-1\frac{1}{8}$ into an improper fraction:

$-1\frac{1}{8} = -\left(1 + \frac{1}{8}\right) = -\left(\frac{8}{8} + \frac{1}{8}\right) = -\left(\frac{8+1}{8}\right) = -\frac{9}{8}$

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Now, multiply $\frac{8}{9}$ by $-\frac{9}{8}$:

$\frac{8}{9} \times \left(-\frac{9}{8}\right) = \frac{8 \times (-9)}{9 \times 8}$

$\frac{8 \times (-9)}{9 \times 8} = \frac{-72}{72}$

Simplify the result:

$\frac{-72}{72} = -1$

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The product of $\frac{8}{9}$ and $-1\frac{1}{8}$ is -1.

For two numbers to be multiplicative inverses, their product must be 1.

Since the product is -1 (which is not equal to 1), $\frac{8}{9}$ is not the multiplicative inverse of $-1\frac{1}{8}$.

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Final Answer:

No, $\frac{8}{9}$ is not the multiplicative inverse of $-1\frac{1}{8}$.

Reason: Because the product of $\frac{8}{9}$ and $-1\frac{1}{8}$ is -1, not 1.

Two numbers are multiplicative inverses of each other if their product equals 1.

We need to check if the product of 0.3 and $3\frac{1}{3}$ is equal to 1.

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First, let's convert both numbers into fraction form.

The decimal 0.3 can be written as a fraction:

$0.3 = \frac{3}{10}$

The mixed fraction $3\frac{1}{3}$ can be converted into an improper fraction:

$3\frac{1}{3} = \frac{(3 \times 3) + 1}{3} = \frac{9 + 1}{3} = \frac{10}{3}$

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Now, multiply the two numbers in their fraction forms:

Product = $0.3 \times 3\frac{1}{3}$

Product = $\frac{3}{10} \times \frac{10}{3}$

Product = $\frac{3 \times 10}{10 \times 3}$

Product = $\frac{30}{30}$

Product = 1

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Since the product of 0.3 and $3\frac{1}{3}$ is 1, they are multiplicative inverses of each other.

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Final Answer:

Yes, 0.3 is the multiplicative inverse of $3\frac{1}{3}$.

Reason: Because their product is 1 $\left( 0.3 \times 3\frac{1}{3} = \frac{3}{10} \times \frac{10}{3} = 1 \right)$.

(i) The rational number that does not have a reciprocal.

The reciprocal of a non-zero rational number $x$ is $\frac{1}{x}$. If $x = 0$, then the reciprocal would be $\frac{1}{0}$, which is undefined.

Therefore, the rational number that does not have a reciprocal is 0.

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(ii) The rational numbers that are equal to their reciprocals.

We are looking for rational numbers $x$ such that $x = \frac{1}{x}$.

If $x = 1$, its reciprocal is $\frac{1}{1} = 1$. So, $x$ is equal to its reciprocal.

If $x = -1$, its reciprocal is $\frac{1}{-1} = -1$. So, $x$ is equal to its reciprocal.

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

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(iii) The rational number that is equal to its negative.

We are looking for a rational number $x$ such that $x = -x$.

Adding $x$ to both sides, we get $x + x = -x + x$, which means $2x = 0$.

Dividing by 2, we get $x = 0$.

Checking: If $x=0$, then $-x = -0 = 0$. So $x=-x$ holds true.

Therefore, the rational number that is equal to its negative is 0.

(i) Zero has no reciprocal.

Elaboration:

The reciprocal of a number 'a' is a number 'b' such that their product is 1, i.e., $a \times b = 1$. The reciprocal of 'a' is also written as $\frac{1}{a}$.

If we try to find the reciprocal of zero, we would need to find a number 'b' such that $0 \times b = 1$.

However, the product of zero and any number is always zero. It is impossible to get 1. Furthermore, the expression for the reciprocal, $\frac{1}{0}$, is mathematically undefined. Therefore, zero has no reciprocal.

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(ii) The numbers 1 and -1 are their own reciprocals.

Elaboration:

A number is its own reciprocal if the number is equal to its reciprocal. Let the number be $x$.

This condition can be written as an equation: $x = \frac{1}{x}$.

To solve for $x$, we can multiply both sides by $x$ (assuming $x \neq 0$):

$x \times x = 1$

$x^2 = 1$

Taking the square root of both sides gives two possible solutions:

$x = \sqrt{1}$ or $x = -\sqrt{1}$

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

We can verify this: The reciprocal of 1 is $\frac{1}{1} = 1$. The reciprocal of -1 is $\frac{1}{-1} = -1$. Thus, 1 and -1 are their own reciprocals.

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(iii) The reciprocal of – 5 is $-\frac{1}{5}$.

Elaboration:

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

For the number -5, its reciprocal is $\frac{1}{-5}$, which is commonly written as $-\frac{1}{5}$.

We can check this by multiplying the number and its reciprocal: $(-5) \times (-\frac{1}{5}) = \frac{5}{5} = 1$. Since the product is 1, the answer is correct.

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(iv) Reciprocal of $\frac{1}{x}$, where x ≠ 0 is $x$.

Elaboration:

To find the reciprocal of a number, we divide 1 by that number. In this case, the number is $\frac{1}{x}$.

Reciprocal = $\frac{1}{(\frac{1}{x})}$

This is equivalent to $1 \div \frac{1}{x}$, which is calculated by multiplying 1 by the reciprocal of the divisor:

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

The condition $x \neq 0$ is important because if $x$ were 0, the original expression $\frac{1}{x}$ would be undefined.

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(v) The product of two rational numbers is always a rational number.

Elaboration:

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

Let's take two arbitrary rational numbers, say $\frac{a}{b}$ and $\frac{c}{d}$ (where a, b, c, d are integers, and $b \neq 0$, $d \neq 0$).

Their product is: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$.

The new numerator, $a \times c$, is a product of two integers, which is always an integer.

The new denominator, $b \times d$, is a product of two non-zero integers, which is always a non-zero integer.

Since the result is in the form of an integer divided by a non-zero integer, the product is always a rational number. This is known as the closure property of multiplication for rational numbers.

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(vi) The reciprocal of a positive rational number is positive.

Elaboration:

A positive rational number is a number greater than 0. It can be written as $\frac{p}{q}$ where the integers $p$ and $q$ have the same sign (both positive or both negative).

Let's take a positive rational number, $N = \frac{p}{q}$.

The reciprocal of $N$ is $\frac{1}{N} = \frac{1}{(p/q)} = \frac{q}{p}$.

Case 1: If $p$ and $q$ are both positive, then in the reciprocal $\frac{q}{p}$, both numerator and denominator are positive. Hence, the reciprocal is positive.

Case 2: If $p$ and $q$ are both negative, then in the reciprocal $\frac{q}{p}$, both numerator and denominator are negative. The ratio of two negative numbers is positive. Hence, the reciprocal is positive.

In both cases, the reciprocal of a positive rational number is always positive.

We need to find three rational numbers that are greater than -2 and less than 0.

First, we can write -2 and 0 as rational numbers with a common denominator. Let's choose a denominator like 10.

$-2 = \frac{-2}{1} = \frac{-2 \times 10}{1 \times 10} = \frac{-20}{10}$

$0 = \frac{0}{1} = \frac{0 \times 10}{1 \times 10} = \frac{0}{10}$

Now, we need to find rational numbers between $\frac{-20}{10}$ and $\frac{0}{10}$. We can choose any integer numerator between -20 and 0.

Possible integers are -19, -18, -17, ..., -1.

We can form rational numbers using these integers as numerators and 10 as the denominator.

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Three possible rational numbers between -2 and 0 are:

1. $\frac{-1}{10}$

2. $\frac{-5}{10}$ (which simplifies to $\mathbf{-\frac{1}{2}}$)

3. $\frac{-15}{10}$ (which simplifies to $\mathbf{-\frac{3}{2}}$)

Other examples could be $\frac{-19}{10}$, $\frac{-11}{10}$, $\frac{-7}{10}$, etc.

Alternatively, we can directly think of simple fractions or decimals between -2 and 0:

• $-1$ (which is $\frac{-1}{1}$)
• $-0.5$ (which is $\mathbf{-\frac{1}{2}}$)
• $-1.5$ (which is $\mathbf{-\frac{3}{2}}$)

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Let's list three such rational numbers:

$\mathbf{-\frac{1}{2}}$, $\mathbf{-1}$, $\mathbf{-\frac{3}{2}}$

To find rational numbers between two given rational numbers, we first need to express them with a common denominator.

The given rational numbers are $\frac{-5}{6}$ and $\frac{5}{8}$.

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Step 1: Find a common denominator.

We find the Least Common Multiple (LCM) of the denominators 6 and 8.

Prime factors of 6 are $2 \times 3$.

Prime factors of 8 are $2 \times 2 \times 2 = 2^3$.

LCM(6, 8) = $2^3 \times 3 = 8 \times 3 = 24$.

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Step 2: Convert the given rational numbers to equivalent fractions with the common denominator (24).

$\frac{-5}{6} = \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24}$

$\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}$

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Step 3: Find integers between the numerators.

We need to find rational numbers between $\frac{-20}{24}$ and $\frac{15}{24}$.

The integers between the numerators -20 and 15 are: -19, -18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14.

There are many integers between -20 and 15, so we can easily choose ten of them.

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Step 4: Write ten rational numbers.

We can form rational numbers by using any ten of these integers as numerators and 24 as the denominator.

Ten rational numbers between $\frac{-5}{6}$ (or $\frac{-20}{24}$) and $\frac{5}{8}$ (or $\frac{15}{24}$) are:

$\mathbf{\frac{-19}{24}, \frac{-18}{24} \left(or -\frac{3}{4}\right), \frac{-17}{24}, \frac{-16}{24} \left(or -\frac{2}{3}\right), \frac{-15}{24} \left(or -\frac{5}{8}\right), \frac{-1}{24}, \frac{0}{24} (or 0),} \ $$ \mathbf{\frac{1}{24}, \frac{2}{24} \left(or \frac{1}{12}\right), \frac{10}{24} \left(or \frac{5}{12}\right)}$

(Note: Many other sets of ten rational numbers are possible.)

We need to find one rational number that lies between $\frac{1}{4}$ and $\frac{1}{2}$.

One way to find a rational number between two given rational numbers $a$ and $b$ is to calculate their mean (average), which is given by $\frac{a+b}{2}$.

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Step 1: Find the sum of the two numbers.

Let $a = \frac{1}{4}$ and $b = \frac{1}{2}$.

Sum = $a + b = \frac{1}{4} + \frac{1}{2}$

To add these fractions, we need a common denominator. The Least Common Multiple (LCM) of 4 and 2 is 4.

Convert $\frac{1}{2}$ to an equivalent fraction with denominator 4:

$\frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4}$

Now add the fractions:

Sum = $\frac{1}{4} + \frac{2}{4} = \frac{1+2}{4} = \frac{3}{4}$

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Step 2: Calculate the mean (average).

Mean = $\frac{\text{Sum}}{2} = \frac{3/4}{2}$

Dividing by 2 is the same as multiplying by $\frac{1}{2}$.

Mean = $\frac{3}{4} \times \frac{1}{2}$

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

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The rational number $\frac{3}{8}$ lies between $\frac{1}{4}$ and $\frac{1}{2}$.

We can check this: $\frac{1}{4} = \frac{2}{8}$ and $\frac{1}{2} = \frac{4}{8}$. Clearly, $\frac{2}{8} < \frac{3}{8} < \frac{4}{8}$.

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Final Answer:

A rational number between $\frac{1}{4}$ and $\frac{1}{2}$ is $\mathbf{\frac{3}{8}}$.

(Note: There are infinitely many rational numbers between any two distinct rational numbers. $\frac{3}{8}$ is just one example, obtained by finding the average.)

We need to find three rational numbers that lie between $\frac{1}{4}$ and $\frac{1}{2}$.

Method 1: Using Equivalent Fractions with a Larger Common Denominator

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Step 1: Find a common denominator.

The Least Common Multiple (LCM) of the denominators 4 and 2 is 4.

Convert the fractions to equivalent fractions with denominator 4:

$\frac{1}{4} = \frac{1}{4}$

$\frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4}$

We are looking for numbers between $\frac{1}{4}$ and $\frac{2}{4}$. Since there are no integers between the numerators 1 and 2, we need to use a larger common denominator.

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Step 2: Find a larger common denominator to create more space between numerators.

Let's multiply the numerator and denominator of both equivalent fractions ($\frac{1}{4}$ and $\frac{2}{4}$) by 4 (we need at least 3+1=4 intervals).

$\frac{1}{4} = \frac{1 \times 4}{4 \times 4} = \frac{4}{16}$

$\frac{2}{4} = \frac{2 \times 4}{4 \times 4} = \frac{8}{16}$

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Step 3: Identify integers between the new numerators.

Now we need to find rational numbers between $\frac{4}{16}$ and $\frac{8}{16}$.

The integers between the numerators 4 and 8 are 5, 6, and 7.

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Step 4: Write the rational numbers.

Using the integers 5, 6, and 7 as numerators and 16 as the denominator, we get three rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$:

$\mathbf{\frac{5}{16}}$, $\mathbf{\frac{6}{16}}$ (which simplifies to $\frac{3}{8}$), $\mathbf{\frac{7}{16}}$

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Method 2: Using the Mean (Average) Method

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Step 1: Find the mean of the two given numbers.

Mean = $\frac{\frac{1}{4} + \frac{1}{2}}{2} = \frac{\frac{1}{4} + \frac{2}{4}}{2} = \frac{\frac{3}{4}}{2} = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$.

So, $\frac{3}{8}$ is one rational number between $\frac{1}{4}$ and $\frac{1}{2}$.

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Step 2: Find the mean of the first number and the mean found in Step 1.

Mean = $\frac{\frac{1}{4} + \frac{3}{8}}{2} = \frac{\frac{2}{8} + \frac{3}{8}}{2} = \frac{\frac{5}{8}}{2} = \frac{5}{8} \times \frac{1}{2} = \frac{5}{16}$.

So, $\frac{5}{16}$ is another rational number between $\frac{1}{4}$ and $\frac{1}{2}$ (specifically between $\frac{1}{4}$ and $\frac{3}{8}$).

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Step 3: Find the mean of the second number and the mean found in Step 1.

Mean = $\frac{\frac{3}{8} + \frac{1}{2}}{2} = \frac{\frac{3}{8} + \frac{4}{8}}{2} = \frac{\frac{7}{8}}{2} = \frac{7}{8} \times \frac{1}{2} = \frac{7}{16}$.

So, $\frac{7}{16}$ is a third rational number between $\frac{1}{4}$ and $\frac{1}{2}$ (specifically between $\frac{3}{8}$ and $\frac{1}{2}$).

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Final Answer:

Three rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$ are $\frac{5}{16}$, $\frac{3}{8}$, and $\frac{7}{16}$.

(i) To represent $\frac{7}{4}$ on the number line.

The given number is the rational number $\frac{7}{4}$.

This is an improper fraction. To understand its position on the number line, we can convert it into a mixed fraction:

$\frac{7}{4} = 1\frac{3}{4}$

From the mixed fraction, we can see that the number $\frac{7}{4}$ lies between the integers 1 and 2.

Steps for representation:

1. Draw a number line and mark integers on it.

2. The denominator of the fraction is 4. This tells us to divide the segment between each consecutive integer (e.g., 0 and 1, 1 and 2) into 4 equal parts.

3. The numerator is 7. Starting from 0, we count 7 of these smaller parts to the right (since the number is positive).

4. The point reached after 7 steps is the representation of $\frac{7}{4}$ on the number line. This point is also the third mark after the integer 1.

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(ii) To represent $\frac{-5}{6}$ on the number line.

The given number is the rational number $\frac{-5}{6}$.

This is a proper fraction and it is negative. This means its value lies between 0 and -1.

Steps for representation:

1. Draw a number line and mark integers on it.

2. Since the number is negative, it will be on the left side of 0.

3. The denominator of the fraction is 6. This tells us to divide the segment between 0 and -1 into 6 equal parts.

4. The numerator is -5. Starting from 0, we count 5 of these smaller parts to the left (due to the negative sign).

5. The point reached after 5 steps to the left of 0 is the representation of $\frac{-5}{6}$ on the number line.

To represent $\frac{-2}{11}$, $\frac{-5}{11}$, and $\frac{-9}{11}$ on the number line.

We observe that all the given rational numbers, $\frac{-2}{11}$, $\frac{-5}{11}$, and $\frac{-9}{11}$, have the same denominator, which is 11.

Also, they are all proper fractions (the absolute value of the numerator is less than the denominator) and are negative. This indicates that all three numbers lie on the number line between the integers 0 and -1.

Steps for representation:

1. Draw a number line and mark the integers 0 and -1.

2. Since the denominator is 11, we will divide the segment between 0 and -1 into 11 equal parts.

3. Starting from 0 and moving towards -1, these parts will represent $\frac{-1}{11}$, $\frac{-2}{11}$, $\frac{-3}{11}$, and so on, up to $\frac{-10}{11}$. The point -1 is equivalent to $\frac{-11}{11}$.

4. Now, we locate the given numbers on these divisions:

- To represent $\frac{-2}{11}$, we count 2 parts to the left of 0.

- To represent $\frac{-5}{11}$, we count 5 parts to the left of 0.

- To represent $\frac{-9}{11}$, we count 9 parts to the left of 0.

The representation of all three numbers on the same number line is shown below.

We need to list five rational numbers that are strictly less than 2.

A rational number is any number that can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. This includes integers, terminating decimals, and repeating decimals.

There are infinitely many rational numbers smaller than 2. We can choose any five that satisfy the condition.

Some examples include:

1. 1 (Since $1 < 2$ and 1 can be written as $\frac{1}{1}$)

2. 0 (Since $0 < 2$ and 0 can be written as $\frac{0}{1}$)

3. -1 (Since $-1 < 2$ and -1 can be written as $\frac{-1}{1}$)

4. $\mathbf{\frac{1}{2}}$ (Since $\frac{1}{2} = 0.5 < 2$)

5. $\mathbf{-\frac{3}{4}}$ (Since $-\frac{3}{4} = -0.75 < 2$)

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Therefore, five rational numbers smaller than 2 are 1, 0, -1, $\frac{1}{2}$, and $-\frac{3}{4}$.

(Other possible answers include $\frac{3}{2}$, $\frac{7}{4}$, $-5$, $\frac{-10}{3}$, etc.)

To find ten rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$, we first convert them to equivalent fractions with a common denominator.

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Step 1: Find a common denominator.

The denominators are 5 and 2. The Least Common Multiple (LCM) of 5 and 2 is 10.

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Step 2: Convert the fractions to equivalent fractions with the LCM as the denominator.

$\frac{-2}{5} = \frac{-2 \times 2}{5 \times 2} = \frac{-4}{10}$

$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$

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Step 3: Check if there are enough integers between the numerators.

We need to find numbers between $\frac{-4}{10}$ and $\frac{5}{10}$. The integers between -4 and 5 are -3, -2, -1, 0, 1, 2, 3, 4. There are only 8 integers, which is not enough to find 10 rational numbers.

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Step 4: Increase the common denominator to create more intermediate numbers.

Let's multiply the numerator and denominator of our equivalent fractions ($\frac{-4}{10}$ and $\frac{5}{10}$) by 2. (We can multiply by any integer greater than 1; 2 is simple).

$\frac{-4}{10} = \frac{-4 \times 2}{10 \times 2} = \frac{-8}{20}$

$\frac{5}{10} = \frac{5 \times 2}{10 \times 2} = \frac{10}{20}$

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Step 5: Identify integers between the new numerators.

Now we need to find rational numbers between $\frac{-8}{20}$ and $\frac{10}{20}$.

The integers between the numerators -8 and 10 are: -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

There are 17 integers, which is more than enough to choose 10.

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Step 6: Write ten rational numbers.

We can choose any ten integers between -8 and 10 and use them as numerators with the denominator 20.

Ten rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$ are:

$\mathbf{\frac{-7}{20}, \frac{-6}{20}, \frac{-5}{20}, \frac{-4}{20}, \frac{-3}{20}, \frac{-2}{20}, \frac{-1}{20}, \frac{0}{20}, \frac{1}{20}, \frac{2}{20}}$

(These can be simplified, for example, $\frac{-6}{20} = -\frac{3}{10}$, $\frac{-5}{20} = -\frac{1}{4}$, $\frac{0}{20}=0$, $\frac{2}{20}=\frac{1}{10}$, but the unsimplified forms are also correct rational numbers between the given limits.)

(Note: Many other sets of ten rational numbers are possible.)

To find rational numbers between two given rational numbers, we first find equivalent fractions with a common denominator. If needed, we find a larger common denominator to ensure there are enough integers between the numerators.

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(i) Between $\frac{2}{3}$ and $\frac{4}{5}$

1. Find the LCM of the denominators 3 and 5, which is 15.

2. Convert to equivalent fractions with denominator 15:

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

$\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}$

3. The only integer between 10 and 12 is 11. We need five numbers. So, we need a larger common denominator. Let's multiply the numerator and denominator by 4.

$\frac{10}{15} = \frac{10 \times 4}{15 \times 4} = \frac{40}{60}$

$\frac{12}{15} = \frac{12 \times 4}{15 \times 4} = \frac{48}{60}$

4. Integers between 40 and 48 are 41, 42, 43, 44, 45, 46, 47.

5. Five rational numbers between $\frac{40}{60}$ and $\frac{48}{60}$ (i.e., between $\frac{2}{3}$ and $\frac{4}{5}$) are:

$\frac{41}{60}, \frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60}$

(These can be simplified, e.g., $\frac{42}{60} = \frac{7}{10}$, $\frac{44}{60} = \frac{11}{15}$, $\frac{45}{60} = \frac{3}{4}$)

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(ii) Between $\frac{-3}{2}$ and $\frac{5}{3}$

1. Find the LCM of the denominators 2 and 3, which is 6.

2. Convert to equivalent fractions with denominator 6:

$\frac{-3}{2} = \frac{-3 \times 3}{2 \times 3} = \frac{-9}{6}$

$\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}$

3. Integers between -9 and 10 include -8, -7, ..., 0, ..., 8, 9. There are plenty of integers.

4. Five rational numbers between $\frac{-9}{6}$ and $\frac{10}{6}$ (i.e., between $\frac{-3}{2}$ and $\frac{5}{3}$) are:

$\frac{-8}{6}, \frac{-7}{6}, 0, \frac{1}{6}, \frac{2}{6}$

(Other examples: $\frac{-1}{6}, \frac{5}{6}, \frac{9}{6}$. Simplified examples: $-\frac{4}{3}, -\frac{7}{6}, 0, \frac{1}{6}, \frac{1}{3}$)

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(iii) Between $\frac{1}{4}$ and $\frac{1}{2}$

1. Find the LCM of the denominators 4 and 2, which is 4.

2. Convert to equivalent fractions with denominator 4:

$\frac{1}{4} = \frac{1}{4}$

$\frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4}$

3. There are no integers between 1 and 2. We need a larger common denominator. Let's multiply the numerator and denominator by 6 (since we need 5 numbers, we need $5+1=6$ spaces).

$\frac{1}{4} = \frac{1 \times 6}{4 \times 6} = \frac{6}{24}$

$\frac{2}{4} = \frac{2 \times 6}{4 \times 6} = \frac{12}{24}$

4. Integers between 6 and 12 are 7, 8, 9, 10, 11.

5. Five rational numbers between $\frac{6}{24}$ and $\frac{12}{24}$ (i.e., between $\frac{1}{4}$ and $\frac{1}{2}$) are:

$\frac{7}{24}, \frac{8}{24}, \frac{9}{24}, \frac{10}{24}, \frac{11}{24}$

(These can be simplified, e.g., $\frac{8}{24} = \frac{1}{3}$, $\frac{9}{24} = \frac{3}{8}$, $\frac{10}{24} = \frac{5}{12}$)

We need to list five rational numbers that are strictly greater than -2.

A rational number is any number that can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Numbers greater than -2 lie to the right of -2 on the number line.

There are infinitely many rational numbers greater than -2. We can choose any five.

Here are five examples:

1. -1 (Since $-1 > -2$ and -1 can be written as $\frac{-1}{1}$)

2. 0 (Since $0 > -2$ and 0 can be written as $\frac{0}{1}$)

3. $\mathbf{-\frac{1}{2}}$ (Since $-\frac{1}{2} = -0.5 > -2$)

4. $\mathbf{\frac{1}{4}}$ (Since $\frac{1}{4} = 0.25 > -2$)

5. 1 (Since $1 > -2$ and 1 can be written as $\frac{1}{1}$)

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Therefore, five rational numbers greater than -2 are -1, 0, $-\frac{1}{2}$, $\frac{1}{4}$, and 1.

(Other possible answers include $1.5$, $\frac{5}{2}$, $-1.9$, $\frac{-3}{2}$, $10$, etc.)

To find ten rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$, we will first convert them into equivalent rational numbers with a common denominator.

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Step 1: Find a common denominator.

The denominators are 5 and 4. The Least Common Multiple (LCM) of 5 and 4 is 20.

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Step 2: Convert the fractions to equivalent fractions with the LCM as the denominator.

$\frac{3}{5} = \frac{3 \times 4}{5 \times 4} = \frac{12}{20}$

$\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$

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Step 3: Check if there are enough integers between the numerators.

We need to find rational numbers between $\frac{12}{20}$ and $\frac{15}{20}$. The integers between the numerators 12 and 15 are 13 and 14. This only allows us to find two rational numbers ($\frac{13}{20}, \frac{14}{20}$). We need ten rational numbers.

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Step 4: Increase the common denominator to create more intermediate numbers.

We need to multiply the numerator and denominator of our equivalent fractions ($\frac{12}{20}$ and $\frac{15}{20}$) by a suitable integer to widen the gap between the numerators. Since we need 10 numbers, we need at least $10+1 = 11$ integer steps between the numerators. The current gap is $15-12=3$. Let's multiply by 4 (since $3 \times 4 = 12$, giving 11 integers in between).

$\frac{12}{20} = \frac{12 \times 4}{20 \times 4} = \frac{48}{80}$

$\frac{15}{20} = \frac{15 \times 4}{20 \times 4} = \frac{60}{80}$

Let's try multiplying by a larger number, say 8, to be safe and get more options:

$\frac{12}{20} = \frac{12 \times 8}{20 \times 8} = \frac{96}{160}$

$\frac{15}{20} = \frac{15 \times 8}{20 \times 8} = \frac{120}{160}$

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Step 5: Identify integers between the new numerators.

Now we need to find rational numbers between $\frac{96}{160}$ and $\frac{120}{160}$.

The integers between the numerators 96 and 120 are: 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119.

There are 23 integers, which is sufficient to choose 10.

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Step 6: Write ten rational numbers.

We can select any ten integers from the list above and use them as numerators with the denominator 160.

Ten rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$ are:

$\mathbf{\frac{97}{160}, \frac{98}{160}, \frac{99}{160}, \frac{100}{160}, \frac{101}{160}, \frac{102}{160}, \frac{103}{160}, \frac{104}{160}, \frac{105}{160}, \frac{106}{160}}$

(These can be simplified where possible, e.g., $\frac{98}{160} = \frac{49}{80}$, $\frac{100}{160} = \frac{5}{8}$, but the unsimplified forms are correct.)