Comparison and Ordering of Numbers
Comparison Methods for Different Number Types
Comparing numbers is a fundamental skill in mathematics, allowing us to determine the relative size or value of two or more numbers. Comparison involves stating whether one number is greater than, less than, or equal to another. Ordering involves arranging a set of numbers based on their values, typically in ascending (smallest to largest) or descending (largest to smallest) order.
The standard mathematical symbols used for comparison are:
- $a > b$: $a$ is greater than $b$.
- $a < b$: $a$ is less than $b$.
- $a = b$: $a$ is equal to $b$.
- $a \ge b$: $a$ is greater than or equal to $b$.
- $a \le b$: $a$ is less than or equal to $b$.
The method used for comparing and ordering numbers often depends on the type of numbers involved, such as integers, fractions, or decimals.
Comparing Integers ($\mathbb{Z}$)
Integers ($\mathbb{Z} = \{..., -3, -2, -1, 0, 1, 2, 3, ...\}$) are discrete points on the number line. Their order is clearly defined by their position relative to the origin (0) and to each other.
The most intuitive way to compare integers is to visualize them on the number line:
For any two distinct integers, the integer that appears further to the right on the standard horizontal number line is the greater number.
- Comparing positive integers: A positive integer with a larger value is greater. E.g., $7 > 3$, $100 > 99$.
- Comparing positive and negative integers: Any positive integer is always greater than any negative integer. E.g., $5 > -100$, $1 > -1$.
- Comparing zero with other integers: Zero is greater than any negative integer and less than any positive integer. E.g., $0 > -50$, $0 < 20$.
- Comparing two negative integers: For two negative integers, the one with the smaller absolute value is greater. This is because it is closer to zero on the number line, and numbers increase as you move right towards zero from the negative side. E.g., compare $-2$ and $-5$. The absolute value of $-2$ is $|-2|=2$. The absolute value of $-5$ is $|-5|=5$. Since $2 < 5$, $-2$ is closer to 0 than $-5$. On the number line, $-2$ is to the right of $-5$. Therefore, $-2 > -5$. Similarly, $-1 > -10$.
Consider the integers $-5, -2, 0, 3$. Their positions on the number line clearly show their order:

(Note: The image shows a number line with marks at -5, -2, 0, and 3. The visual arrangement confirms their order.)
Based on their positions from left to right, we can order them:
$-5 < -2 < 0 < 3$
Example 1. Arrange the integers $-15, 7, 0, -3, 10$ in ascending order.
Answer:
We need to arrange the integers $-15, 7, 0, -3, 10$ from smallest to largest.
The negative integers are $-15$ and $-3$. Comparing their absolute values, $|-15|=15$ and $|-3|=3$. Since $3 < 15$, $-3$ is closer to zero than $-15$. Therefore, $-15 < -3$.
Zero is greater than any negative integer.
The positive integers are $7$ and $10$. Comparing them, $7 < 10$.
Combining these comparisons, the ascending order is:
$-15 < -3 < 0 < 7 < 10$
The integers in ascending order are -15, -3, 0, 7, 10.
Comparing Fractions ($\mathbb{Q}$ in fractional form)
Comparing fractions involves determining which fraction represents a larger or smaller part of a whole. This can be done effectively by expressing the fractions with respect to a common reference.
Method 1: Finding a Common Denominator
This is a reliable method, especially for comparing more than two fractions. To compare fractions like $\frac{a}{b}$ and $\frac{c}{d}$ (where $b \neq 0, d \neq 0$), we convert them into equivalent fractions that share the same denominator.
- Find the Least Common Multiple (LCM) of the denominators: The LCM of the denominators $b$ and $d$ is the smallest positive integer that is a multiple of both $b$ and $d$. This LCM will serve as the common denominator. For multiple fractions, find the LCM of all denominators.
- Convert each fraction to an equivalent fraction with the common denominator: For each fraction, multiply both its numerator and its denominator by the factor that makes the denominator equal to the LCM.
For a fraction $\frac{a}{b}$, the multiplying factor is $\frac{\text{LCM}(b,d)}{b}$. The equivalent fraction is $\frac{a \times \left(\frac{\text{LCM}(b,d)}{b}\right)}{b \times \left(\frac{\text{LCM}(b,d)}{b}\right)}$.
For a fraction $\frac{c}{d}$, the multiplying factor is $\frac{\text{LCM}(b,d)}{d}$. The equivalent fraction is $\frac{c \times \left(\frac{\text{LCM}(b,d)}{d}\right)}{d \times \left(\frac{\text{LCM}(b,d)}{d}\right)}$.
- Compare the numerators: Once all fractions have the same denominator, their magnitudes are directly proportional to their numerators. The fraction with the larger numerator is the greater fraction.
If you are comparing $\frac{N_1}{D}$ and $\frac{N_2}{D}$ (where $D > 0$), then:
- If $N_1 > N_2$, then $\frac{N_1}{D} > \frac{N_2}{D}$.
- If $N_1 < N_2$, then $\frac{N_1}{D} < \frac{N_2}{D}$.
- If $N_1 = N_2$, then $\frac{N_1}{D} = \frac{N_2}{D}$.
If the common denominator $D$ is negative, the comparison of numerators is reversed, but for simplicity, we usually work with positive common denominators.
Example 1. Arrange the fractions $\frac{2}{3}, \frac{5}{6}, \frac{3}{4}$ in ascending order.
Answer:
The denominators of the given fractions are $3, 6,$ and $4$.
Find the LCM of 3, 6, and 4:
$\begin{array}{c|cc} 2 & 3 \;, & 6 \;, & 4 \\ \hline 3 & 3 \; , & 3 \; , & 2 \\ \hline & 1 \; , & 1 \; , & 2 \end{array}$LCM$(3, 6, 4) = 2 \times 3 \times 1 \times 1 \times 2 = 12$. The common denominator is 12.
Convert each fraction to an equivalent fraction with a denominator of 12:
$$ \frac{2}{3} = \frac{2 \times (12 \div 3)}{3 \times (12 \div 3)} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} $$ $$ \frac{5}{6} = \frac{5 \times (12 \div 6)}{6 \times (12 \div 6)} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12} $$ $$ \frac{3}{4} = \frac{3 \times (12 \div 4)}{4 \times (12 \div 4)} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} $$Now compare the numerators of the equivalent fractions: $8, 10,$ and $9$.
Comparing these integers, we have $8 < 9 < 10$.
Therefore, comparing the fractions:
$\frac{8}{12} < \frac{9}{12} < \frac{10}{12}$
Substitute back the original fractions:
$\frac{2}{3} < \frac{3}{4} < \frac{5}{6}$
The fractions in ascending order are $\mathbf{\frac{2}{3}, \frac{3}{4}, \frac{5}{6}}$.
Method 2: Cross-Multiplication
This method is particularly useful for quickly comparing just two fractions, say $\frac{a}{b}$ and $\frac{c}{d}$, where the denominators $b$ and $d$ are positive ($b > 0, d > 0$).
- Calculate the two cross-products: $a \times d$ and $c \times b$.
- Compare the two products:
- If $a \times d > c \times b$, then $\frac{a}{b} > \frac{c}{d}$.
- If $a \times d < c \times b$, then $\frac{a}{b} < \frac{c}{d}$.
- If $a \times d = c \times b$, then $\frac{a}{b} = \frac{c}{d}$.
Derivation of the method: To compare $\frac{a}{b}$ and $\frac{c}{d}$, we can use the common denominator method. The common denominator is $bd$. The equivalent fractions are $\frac{ad}{bd}$ and $\frac{cb}{bd}$. Since $b>0$ and $d>0$, $bd > 0$. Comparing $\frac{ad}{bd}$ and $\frac{cb}{bd}$ is equivalent to comparing their numerators, $ad$ and $cb$. Thus, comparing $\frac{a}{b}$ and $\frac{c}{d}$ is equivalent to comparing $ad$ and $cb$.
Example 2. Compare $\frac{7}{9}$ and $\frac{5}{7}$ using cross-multiplication.
Answer:
We are comparing $\frac{7}{9}$ and $\frac{5}{7}$. The denominators $9$ and $7$ are positive.
Calculate the cross-products:
$$ 7 \times 7 = 49 $$ $$ 5 \times 9 = 45 $$Compare the two products: $49$ and $45$.
$$ 49 > 45 $$Since the first cross-product ($7 \times 7 = 49$) is greater than the second cross-product ($5 \times 9 = 45$), the first fraction is greater than the second fraction.
Therefore, $\mathbf{\frac{7}{9} > \frac{5}{7}}$.
Special Cases for Comparing Fractions
- Fractions with the Same Denominator: If two fractions have the same positive denominator, the fraction with the larger numerator is the greater fraction. E.g., $\frac{3}{5} > \frac{2}{5}$ because $3 > 2$. If the denominator is negative, the comparison of numerators is reversed.
- Fractions with the Same Numerator: If two positive fractions have the same positive numerator, the fraction with the smaller positive denominator is the greater fraction. E.g., $\frac{1}{2} > \frac{1}{3}$ because dividing a whole into 2 parts results in larger parts than dividing it into 3 parts. For negative fractions with the same numerator (and positive denominators), the one with the smaller denominator is less (e.g., $-\frac{1}{2} < -\frac{1}{3}$).
- Comparing Negative Fractions: To compare negative fractions, you can compare their absolute values. The negative fraction with the smaller absolute value is greater (because it is closer to $0$ on the number line). E.g., compare $-\frac{3}{4}$ and $-\frac{5}{6}$. We know $\frac{3}{4} < \frac{5}{6}$ (since $3 \times 6 = 18$ and $5 \times 4 = 20$, and $18 < 20$). Since $|\!-\!\frac{3}{4}| < |\!-\!\frac{5}{6}|$, we have $-\frac{3}{4} > -\frac{5}{6}$. Alternatively, you can use the common denominator method or cross-multiplication directly with signed numerators.
Comparing Decimal Numbers
Decimal numbers are compared by examining their digits from left to right, starting from the highest place value.
- Compare the whole number parts: Compare the integers to the left of the decimal point. The number with the larger whole number part is the greater number. (Handle negative numbers by comparing absolute values first, then reversing the comparison as done for integers).
- If whole number parts are equal: Move to the decimal part and compare the digits from left to right, position by position (Tenths, Hundredths, Thousandths, and so on).
- Identify the first differing digits: At the first place value where the digits of the two numbers are different, the number with the larger digit at that specific place is the greater number.
- Adding trailing zeros: It can be helpful to add trailing zeros to the decimal with fewer decimal places so that both numbers have the same number of digits after the decimal point. This does not change the value of the number (e.g., $2.5 = 2.50 = 2.500$), but it aligns the place values and can make the digit-by-digit comparison clearer.
Example 1. Compare $12.345$ and $12.35$.
Answer:
The numbers are $12.345$ and $12.35$. To make comparison easier, add a trailing zero to $12.35$ so it has the same number of decimal places as $12.345$. So, compare $12.345$ and $12.350$.
Align the decimal points and compare digits from left to right:
- Whole number part: $12 = 12$ (Equal)
- Tenths place (1st digit after decimal): $3 = 3$ (Equal)
- Hundredths place (2nd digit after decimal): $4 < 5$ (First difference found)
Since the digit in the hundredths place of the second number ($12.350$), which is $5$, is greater than the digit in the hundredths place of the first number ($12.345$), which is $4$, the second number is greater.
Therefore, $\mathbf{12.345 < 12.350}$ or $\mathbf{12.345 < 12.35}$.
Example 2. Compare $-0.8$ and $-0.75$.
Answer:
We are comparing two negative decimal numbers. A common way to compare negative numbers is to compare their absolute values and then reverse the inequality.
Absolute value of $-0.8$ is $|-0.8| = 0.8$.
Absolute value of $-0.75$ is $|-0.75| = 0.75$.
Now, compare the positive decimals $0.8$ and $0.75$. Add a trailing zero to $0.8$ to get $0.80$. Compare $0.80$ and $0.75$ digit by digit:
- Whole number part: $0 = 0$ (Equal)
- Tenths place: $8 > 7$ (First difference found)
Since $8 > 7$, we conclude that $0.80 > 0.75$, which means $|-0.8| > |-0.75|$.
For negative numbers, the number with the smaller absolute value is the greater number (it is closer to zero on the number line). Since $|-0.75| < |-0.8|$, we have $-0.75 > -0.8$.
Therefore, $\mathbf{-0.8 < -0.75}$.
Alternatively, you can directly compare $-0.8$ and $-0.75$ (or $-0.80$ and $-0.75$) by thinking about their positions on the number line. As you move from $-1$ towards $0$, you encounter $-0.80$ first, and then $-0.75$. Since $-0.75$ is to the right of $-0.80$, $-0.75$ is greater.
Comparing Different Types of Numbers
To compare numbers presented in different formats (e.g., a fraction and a decimal, a rational number and an irrational number), the most effective strategy is usually to convert them to a common format, typically decimal form.
- Convert all numbers to decimal form:
- If a number is a fraction $\frac{p}{q}$, perform the division $p \div q$ to get its decimal expansion.
- If a number is already a decimal, keep it as is.
- If a number is an irrational number (like $\sqrt{2}$ or $\pi$), use its known decimal approximation to a sufficient number of decimal places.
- Compare the decimal expansions: Once all numbers are in decimal form, compare them using the digit-by-digit method described for decimal numbers. For non-terminating decimals (repeating or non-repeating), compare as many digits as necessary until a difference is found or until you have enough confidence for the comparison required.
Example 1. Compare $\frac{3}{8}$ and $0.37$.
Answer:
Convert the fraction $\frac{3}{8}$ to a decimal by dividing 3 by 8:
$\frac{3}{8} = 3 \div 8 = 0.375$
Now compare the decimals $0.375$ and $0.37$. Add a trailing zero to $0.37$ to get $0.370$ to match decimal places.
Compare $0.375$ and $0.370$ digit by digit:
- Whole number part: $0 = 0$ (Equal)
- Tenths place: $3 = 3$ (Equal)
- Hundredths place: $7 = 7$ (Equal)
- Thousandths place: $5 > 0$ (First difference found)
Since the digit in the thousandths place of $0.375$ ($5$) is greater than the digit in the thousandths place of $0.370$ ($0$), $0.375$ is greater.
Therefore, $\mathbf{\frac{3}{8} > 0.37}$.
Example 2. Compare $\pi$ and $\frac{22}{7}$.
Answer:
$\pi$ is an irrational number. Its decimal expansion starts as $3.14159265...$
$\frac{22}{7}$ is a rational number (a common approximation for $\pi$). Let's convert it to a decimal by dividing 22 by 7:
$\frac{22}{7} = 3.\overline{142857} = 3.142857142857...$
Now compare $\pi$ and $\frac{22}{7}$ by comparing their decimal expansions digit by digit:
- $\pi \quad = 3.14159265...$
- $\frac{22}{7} = 3.14285714...$
Compare digits from left to right:
- Whole number part: $3 = 3$ (Equal)
- Tenths place: $1 = 1$ (Equal)
- Hundredths place: $4 = 4$ (Equal)
- Thousandths place: $1 < 2$ (First difference found)
Since the digit in the thousandths place of $\pi$ ($1$) is less than the digit in the thousandths place of $\frac{22}{7}$ ($2$), $\pi$ is less than $\frac{22}{7}$.
Therefore, $\mathbf{\pi < \frac{22}{7}}$.
(Note: $\frac{22}{7}$ is a close rational approximation of $\pi$, but $\pi$ is slightly smaller).
Summary of Comparison Methods
Number Type | Primary Comparison Method | Notes/Alternative Methods |
---|---|---|
Integers ($\mathbb{Z}$) | Position on the Number Line | Compare magnitude for positive numbers; Compare absolute values for negative numbers (smaller abs. value is greater); Any positive > 0 > any negative. |
Fractions ($\mathbb{Q}$) | Find a Common Denominator | Cross-Multiplication (for two fractions); Convert to decimals. |
Decimals ($\mathbb{Q}$ or $\mathbb{R}$) | Digit-by-digit comparison from left to right. | Align decimal points; Add trailing zeros; Compare whole parts first, then decimal parts. |
Mixed Types (e.g., Fraction vs. Decimal, Rational vs. Irrational) |
Convert all numbers to Decimal Form | Compare the decimal expansions digit by digit. For roots, sometimes squaring both numbers can help if applicable (e.g., compare $\sqrt{3}$ and $1.7$; compare $(\sqrt{3})^2=3$ and $(1.7)^2=2.89$). |
Mastering these comparison methods is essential for solving inequalities and ordering various sets of numbers.
Absolute Value of Integers and Rational Numbers
Definition and Meaning of Absolute Value
The absolute value of a number is its distance from zero on the number line. Distance is always a non-negative quantity, so the absolute value of a number is always a non-negative number.
The absolute value of a number $x$ is denoted by placing the number between two vertical bars: $|x|$.
The formal definition of the absolute value of a real number $x$ is:
$$ |x| = \begin{cases} x & , & \text{if } x \ge 0 \\ -x & , & \text{if } x < 0 \end{cases} $$
[Definition of Absolute Value]
This definition can be interpreted as follows:
- If the number $x$ is positive or zero ($x \ge 0$), its absolute value is the number itself. For example, $|5| = 5$ and $|0| = 0$.
- If the number $x$ is negative ($x < 0$), its absolute value is the opposite of the number, which is positive. For example, $|-5| = -(-5) = 5$.
Visually, on the number line, the absolute value represents the magnitude of the number without considering its direction (positive or negative). Both a number and its additive inverse (opposite) are the same distance from zero.

(Note: The image shows a number line with the origin 0. Points are marked at -4 and 4. Arrows or labels indicate that the distance from 0 to -4 is 4 units, and the distance from 0 to 4 is also 4 units. Thus, $|-4|=4$ and $|4|=4$.)
Absolute Value of Integers ($\mathbb{Z}$)
The concept of absolute value applies directly to integers. For any integer $n$, its absolute value $|n|$ is its non-negative value, representing its distance from 0 on the number line.
- For a positive integer $n$, $|n| = n$. Example: $|10| = 10$, $|7| = 7$.
- For the integer zero, $|0| = 0$.
- For a negative integer $-n$ (where $n$ is a positive integer), $|-n| = -(-n) = n$. Example: $|-10| = -(-10) = 10$, $|-7| = -(-7) = 7$.
In simpler terms, to find the absolute value of a non-zero integer, just remove its sign.
Example 1. Find the absolute value of $-25$ and $18$.
Answer:
For $-25$:
Using the definition, since $-25 < 0$, we take the opposite of $-25$.
$\quad |-25| = -(-25) = 25$
The distance of -25 from 0 on the number line is 25 units.
For $18$:
Using the definition, since $18 > 0$, the absolute value is the number itself.
$\quad |18| = 18$
The distance of 18 from 0 on the number line is 18 units.
Absolute Value of Rational Numbers ($\mathbb{Q}$)
The definition and concept of absolute value apply equally to all rational numbers, which include integers, fractions, and terminating or repeating decimals. The absolute value of a rational number $q$ is its non-negative value, representing its distance from 0 on the number line.
For any rational number $q$:
- If $q \ge 0$, then $|q| = q$.
- $|\frac{3}{4}| = \frac{3}{4}$
- $|5.12| = 5.12$
- $|0.\overline{7}| = 0.\overline{7}$
- If $q < 0$, then $|q| = -q$.
- $|-\frac{3}{4}| = -(-\frac{3}{4}) = \frac{3}{4}$
- $|-5.12| = -(-5.12) = 5.12$
- $|-0.\overline{7}| = -(-0.\overline{7}) = 0.\overline{7}$
Example 1. Find the absolute value of $-\frac{7}{8}$.
Answer:
The number is $-\frac{7}{8}$. Since $-\frac{7}{8} < 0$, its absolute value is the opposite of the number.
$\quad |-\frac{7}{8}| = -(-\frac{7}{8}) = \frac{7}{8}$
The distance of $-\frac{7}{8}$ from 0 on the number line is $\frac{7}{8}$ units.
Example 2. Find the absolute value of $3.05$.
Answer:
The number is $3.05$. Since $3.05 > 0$, its absolute value is the number itself.
$\quad |3.05| = 3.05$
The distance of $3.05$ from 0 on the number line is $3.05$ units.
Key Properties of Absolute Value
The absolute value function has several important properties that are true for all real numbers, and therefore specifically for all integers and rational numbers. Let $x$ and $y$ be any real numbers.
- Non-negativity: The absolute value of any number is always greater than or equal to zero.
$\quad |x| \ge 0$
[Absolute value is non-negative]
Also, the absolute value is zero if and only if the number itself is zero.
$\quad |x| = 0 \iff x = 0$
- Symmetry: A number and its additive inverse (opposite) have the same absolute value.
$\quad |-x| = |x|$
[Symmetry property]
Example: $|-15| = 15$ and $|15| = 15$. So, $|-15| = |15|$.
- Multiplicativity: The absolute value of a product of two numbers is equal to the product of their absolute values.
$\quad |x \times y| = |x| \times |y|$
[Multiplicativity property]
Example: Let $x = -4$ and $y = 5$.
$|x \times y| = |(-4) \times 5| = |-20| = 20$.
$|x| \times |y| = |-4| \times |5| = 4 \times 5 = 20$.
Since $20 = 20$, $|(-4) \times 5| = |-4| \times |5|$.
- Divisibility (for non-zero divisor): The absolute value of a quotient of two numbers is equal to the quotient of their absolute values, provided the divisor is not zero.
$\quad |\frac{x}{y}| = \frac{|x|}{|y|}$, provided $y \neq 0$
[Divisibility property]
Example: Let $x = 12$ and $y = -3$.
$|\frac{x}{y}| = |\frac{12}{-3}| = |-4| = 4$.
$\frac{|x|}{|y|} = \frac{|12|}{|-3|} = \frac{12}{3} = 4$.
Since $4 = 4$, $|\frac{12}{-3}| = \frac{|12|}{|-3|}$.
- Triangle Inequality: The absolute value of the sum of two numbers is less than or equal to the sum of their absolute values.
$\quad |x + y| \le |x| + |y|$
[Triangle Inequality]
Example: Let $x = 2$ and $y = 3$. $|2+3|=|5|=5$. $|2|+|3|=2+3=5$. $5 \le 5$.
Let $x = -2$ and $y = 3$. $|-2+3|=|1|=1$. $|-2|+|3|=2+3=5$. $1 \le 5$.
Let $x = -2$ and $y = -3$. $|-2+(-3)|=|-5|=5$. $|-2|+|-3|=2+3=5$. $5 \le 5$.
This property is fundamental in various areas of mathematics, especially in defining distances and convergence.
- Reverse Triangle Inequality: A related property states that the absolute value of the difference of two absolute values is less than or equal to the absolute value of their difference.
$\quad ||x| - |y|| \le |x - y|$
[Reverse Triangle Inequality]
Example: Let $x = 5$ and $y = 2$. $||5| - |2|| = |5 - 2| = |3| = 3$. $|5 - 2| = |3| = 3$. $3 \le 3$.
Let $x = 2$ and $y = 5$. $||2| - |5|| = |2 - 5| = |-3| = 3$. $|2 - 5| = |-3| = 3$. $3 \le 3$.
Let $x = -5$ and $y = 2$. $||-5| - |2|| = |5 - 2| = |3| = 3$. $|-5 - 2| = |-7| = 7$. $3 \le 7$.
Absolute value is used in various mathematical contexts, such as measuring errors (absolute error), defining distance between points on the number line ($|x-y|$ is the distance between $x$ and $y$), solving equations and inequalities involving absolute values, and in defining norms and metrics in higher mathematics.
Finding Numbers Between Two Given Numbers (Rational and Irrational)
The set of real numbers ($\mathbb{R}$) is continuous and dense, meaning there are no "gaps" on the number line. This density applies not only to the set of rational numbers ($\mathbb{Q}$) but also to the set of irrational numbers ($\mathbb{I}$). Furthermore, the real number line is "filled" in such a way that between any two distinct real numbers, no matter how close they are, there are infinitely many rational numbers and infinitely many irrational numbers.
Finding Rational Numbers Between Two Given Rational Numbers
Given two distinct rational numbers, say $r_1$ and $r_2$, with $r_1 < r_2$, we know that the set of rational numbers is dense. This means there are infinitely many rational numbers located between $r_1$ and $r_2$. Here are common methods to find such numbers:
Method 1: Using the Midpoint (Average)
The simplest rational number to find between two distinct rational numbers is their average (arithmetic mean). The average of $r_1$ and $r_2$ is $\frac{r_1 + r_2}{2}$.
As proven in the section on Rational Numbers, if $r_1$ and $r_2$ are rational, their sum $r_1+r_2$ is rational (due to closure under addition for rationals), and dividing a rational number by a non-zero rational number (2 in this case) results in a rational number (due to closure under division for rationals). Thus, $\frac{r_1+r_2}{2}$ is always rational.
Also, as proven in the section on Rational Numbers' density, if $r_1 < r_2$, then $r_1 < \frac{r_1 + r_2}{2} < r_2$. So, the average lies strictly between the two numbers.
Example 1. Find a rational number between $\frac{1}{4}$ and $\frac{1}{3}$.
Answer:
Let $r_1 = \frac{1}{4}$ and $r_2 = \frac{1}{3}$. We will find their average.
$\text{Rational number} = \frac{r_1 + r_2}{2} = \frac{\frac{1}{4} + \frac{1}{3}}{2}$
First, add the fractions in the numerator. Find a common denominator for 4 and 3, which is LCM$(4, 3) = 12$.
$\quad \frac{1}{4} + \frac{1}{3} = \frac{1 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4} = \frac{3}{12} + \frac{4}{12} = \frac{3+4}{12} = \frac{7}{12}$
Now, divide the sum by 2 (which is the same as multiplying by $\frac{1}{2}$):
$\quad \frac{\frac{7}{12}}{2} = \frac{7}{12} \times \frac{1}{2} = \frac{7 \times 1}{12 \times 2} = \frac{7}{24}$
The number $\frac{7}{24}$ is a rational number. To verify it is between $\frac{1}{4}$ and $\frac{1}{3}$, we can convert them to fractions with the common denominator 24:
$\quad \frac{1}{4} = \frac{1 \times 6}{4 \times 6} = \frac{6}{24}$
$\quad \frac{1}{3} = \frac{1 \times 8}{3 \times 8} = \frac{8}{24}$
Comparing the fractions $\frac{6}{24}, \frac{7}{24}, \frac{8}{24}$, we see that $\frac{6}{24} < \frac{7}{24} < \frac{8}{24}$. So, $\frac{1}{4} < \frac{7}{24} < \frac{1}{3}$.
Thus, $\mathbf{\frac{7}{24}}$ is a rational number between $\frac{1}{4}$ and $\frac{1}{3}$. To find more rational numbers, you can repeat the process (e.g., find the average of $\frac{1}{4}$ and $\frac{7}{24}$, etc.), generating infinitely many rational numbers between the original two.
Method 2: Using Equivalent Fractions with a Larger Denominator
This method is useful for finding a specific number of rational numbers between two given rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ (assume $b > 0, d > 0$ and $\frac{a}{b} < \frac{c}{d}$).
- Find a common denominator: Find the LCM of $b$ and $d$, let it be $D$. Convert $\frac{a}{b}$ to $\frac{a'}{D}$ and $\frac{c}{d}$ to $\frac{c'}{D}$.
- Create space between numerators: If there are already enough integers between $a'$ and $c'$ (i.e., $c' - a' > n$, where $n$ is the number of rationals you want to find), you can use $\frac{a'+1}{D}, \frac{a'+2}{D}, ..., \frac{a'+n}{D}$ as rational numbers between them.
- If there are not enough integers between $a'$ and $c'$ (i.e., $c' - a' \le n$), multiply the numerator and denominator of both equivalent fractions $\frac{a'}{D}$ and $\frac{c'}{D}$ by a suitable integer $k$. A simple choice for $k$ is $n+1$, or you can choose any integer large enough, e.g., 10, 100, etc., to ensure there are at least $n$ integers between the new numerators $a'k$ and $c'k$. The new equivalent fractions are $\frac{a'k}{Dk}$ and $\frac{c'k}{Dk}$.
- Choose intermediate numerators: The rational numbers between $\frac{a'k}{Dk}$ and $\frac{c'k}{Dk}$ are fractions with denominator $Dk$ and numerators that are integers strictly between $a'k$ and $c'k$. You can pick any $n$ such integers, say $N_1, N_2, ..., N_n$, such that $a'k < N_1 < N_2 < ... < N_n < c'k$. The corresponding rational numbers are $\frac{N_1}{Dk}, \frac{N_2}{Dk}, ..., \frac{N_n}{Dk}$.
Example 2. Find five rational numbers between $-\frac{2}{3}$ and $\frac{1}{4}$.
Answer:
We need to find 5 rational numbers between $-\frac{2}{3}$ and $\frac{1}{4}$.
Find the common denominator for the given fractions. Denominators are 3 and 4. LCM$(3, 4) = 12$.
Convert the fractions to equivalent fractions with denominator 12:
$-\frac{2}{3} = -\frac{2 \times 4}{3 \times 4} = -\frac{8}{12}$
$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$
We need 5 rational numbers between $-\frac{8}{12}$ and $\frac{3}{12}$. These are fractions with denominator 12 and numerators that are integers between -8 and 3. The integers between -8 and 3 are -7, -6, -5, -4, -3, -2, -1, 0, 1, 2. There are 10 such integers.
We can pick any five of these integers as numerators. For example, let's pick the first five:
The integers are $-7, -6, -5, -4, -3$.
The corresponding rational numbers with denominator 12 are:
$\quad \frac{-7}{12}, \frac{-6}{12}, \frac{-5}{12}, \frac{-4}{12}, \frac{-3}{12}$
These fractions can be simplified:
$\quad -\frac{7}{12}, -\frac{1}{2}, -\frac{5}{12}, -\frac{1}{3}, -\frac{1}{4}$
All these are rational numbers and lie between $-\frac{2}{3}$ (or $-\frac{8}{12}$) and $\frac{1}{4}$ (or $\frac{3}{12}$).
Another set of five rational numbers could be $\frac{-2}{12}, \frac{-1}{12}, \frac{0}{12}, \frac{1}{12}, \frac{2}{12}$, which simplify to $-\frac{1}{6}, -\frac{1}{12}, 0, \frac{1}{12}, \frac{1}{6}$. There are infinitely many possibilities.
Finding Irrational Numbers Between Two Given Numbers (Rational or Irrational)
The set of irrational numbers is also dense. Furthermore, between any two distinct real numbers (which could be two rationals, two irrationals, or one of each), there are infinitely many irrational numbers.
Between Two Rational Numbers ($r_1 < r_2$)
To find an irrational number between two rationals $r_1$ and $r_2$: One common strategy is to construct a non-terminating, non-repeating decimal whose value is strictly between $r_1$ and $r_2$. You can do this by starting with the decimal representation of $r_1$ and appending a non-repeating sequence of digits such that the resulting number remains less than $r_2$.
Example 1. Find an irrational number between $0.1$ and $0.2$.
Answer:
We need a non-terminating, non-repeating decimal that is greater than $0.1$ and less than $0.2$.
We can start with $0.1$ and append a non-repeating sequence of digits. For example, consider a sequence like $01001000100001...$ where the number of zeros between the ones increases by one each time. The number $0.101001000100001...$ is irrational because its decimal expansion is non-terminating and does not repeat.
Comparing this number with $0.1$ and $0.2$:
$0.1 = 0.100000000000...$
$0.2 = 0.200000000000...$
$0.101001000100001...$
Clearly, the digits of $0.1010010001...$ after the tenths place (0) are such that the number is greater than $0.1$ (since the first digit after $0.1$ is $0$, then $1$, which makes it greater than $0.100...$) and less than $0.2$ (since the first digit after $0.$ is $1$, which is less than $2$).
Thus, $0.1 < 0.101001000100001... < 0.2$.
So, $\mathbf{0.101001000100001...}$ is an irrational number between $0.1$ and $0.2$.
Another method: Since $0.1^2 = 0.01$ and $0.2^2 = 0.04$, any irrational number of the form $\sqrt{x}$ where $0.01 < x < 0.04$ will lie between $0.1$ and $0.2$. For example, $\sqrt{0.02} \approx 0.1414...$ or $\sqrt{0.03} \approx 0.1732...$. Since 0.02 and 0.03 are not perfect squares of rational numbers, their square roots are irrational. And $0.1 < \sqrt{0.02} < 0.2$ and $0.1 < \sqrt{0.03} < 0.2$.
Between Two Irrational Numbers ($i_1 < i_2$)
Between any two distinct irrational numbers, there are also infinitely many irrational numbers. Similar to the previous case, you can construct a non-terminating, non-repeating decimal that falls between their decimal expansions.
Example 2. Find an irrational number between $\sqrt{2}$ and $\sqrt{3}$.
Answer:
We know the approximate decimal expansions of $\sqrt{2}$ and $\sqrt{3}$:
$\sqrt{2} \approx 1.41421356...$
$\sqrt{3} \approx 1.73205080...$
We need an irrational number between $1.41421356...$ and $1.73205080...$. We can construct a non-terminating, non-repeating decimal that starts with digits that fall between the initial digits of $\sqrt{2}$ and $\sqrt{3}$. For example, we can start with $1.5$ and append a non-repeating pattern like $010010001...$
Consider the number $1.5010010001...$
This number is irrational because its decimal expansion is non-terminating and non-repeating. Comparing it with $\sqrt{2}$ and $\sqrt{3}$:
$1.41421356... < 1.5010010001... < 1.73205080...$
Thus, $\sqrt{2} < 1.5010010001... < \sqrt{3}$.
So, $\mathbf{1.5010010001...}$ is an irrational number between $\sqrt{2}$ and $\sqrt{3}$.
Another approach: Any irrational number of the form $\sqrt{x}$ where $(\sqrt{2})^2 < x < (\sqrt{3})^2$ and $x$ is not a perfect square of a rational number will lie between $\sqrt{2}$ and $\sqrt{3}$. That is, $2 < x < 3$. For example, $\sqrt{2.5}$ is irrational and $2 < 2.5 < 3$, so $\sqrt{2} < \sqrt{2.5} < \sqrt{3}$.
Between Two Real Numbers (Including Rational and Irrational)
Between any two distinct real numbers, there are infinitely many rational numbers and infinitely many irrational numbers. The methods used above can be adapted.
Example 3. Find a rational number between $\sqrt{3}$ and $1.75$.
Answer:
We know $\sqrt{3} \approx 1.73205080...$ and $1.75 = 1.75000000...$
We need a rational number (terminating or repeating decimal) between $1.73205080...$ and $1.75000000...$
We can choose a simple terminating decimal that falls in this range. For example, consider a decimal with a few digits that starts like the smaller number's expansion but becomes larger at some decimal place, while remaining smaller than the larger number.
Consider the number $1.733$. This is a terminating decimal, so it is rational.
Comparing $1.733$ with $\sqrt{3}$ and $1.75$:
$1.73205080... < 1.73300000... < 1.75000000...$
Thus, $\sqrt{3} < 1.733 < 1.75$.
So, $\mathbf{1.733}$ is a rational number between $\sqrt{3}$ and $1.75$.
Other rational numbers could be $1.74$, $1.745$, or even $1.\overline{74}$ (which is between $1.74$ and $1.75$).
Example 4. Find an irrational number between $0.5$ and $\sqrt{0.3}$.
Answer:
We know $0.5 = 0.500000...$
$\sqrt{0.3}$ is irrational (since 0.3 is not the perfect square of a rational number). $\sqrt{0.3} = \sqrt{\frac{3}{10}} = \frac{\sqrt{30}}{10}$. We know $\sqrt{30}$ is between $\sqrt{25}=5$ and $\sqrt{36}=6$, specifically $\sqrt{30} \approx 5.477...$.
So, $\sqrt{0.3} \approx \frac{5.477...}{10} = 0.5477...$
We need an irrational number between $0.500000...$ and $0.5477...$.
We can construct a non-terminating, non-repeating decimal that starts with $0.51$ (which is greater than 0.5 and less than 0.54...) and append a non-repeating pattern.
Consider the number $0.5121121112...$
This number is irrational. Comparing it with $0.5$ and $\sqrt{0.3}$:
$0.500000... < 0.5121121112... < 0.5477...$
Thus, $0.5 < 0.5121121112... < \sqrt{0.3}$.
So, $\mathbf{0.5121121112...}$ is an irrational number between $0.5$ and $\sqrt{0.3}$.
Another approach: Find an irrational number $i$ such that $0.5^2 < i^2 < (\sqrt{0.3})^2$ and $i$ is irrational. This means $0.25 < i^2 < 0.3$. For example, $i^2 = 0.26$. Then $i = \sqrt{0.26}$. Since 0.26 is not a perfect square of a rational, $\sqrt{0.26}$ is irrational. And $0.25 < 0.26 < 0.3$, so $\sqrt{0.25} < \sqrt{0.26} < \sqrt{0.3}$, which means $0.5 < \sqrt{0.26} < \sqrt{0.3}$.
The ability to find infinitely many rational and irrational numbers between any two distinct real numbers demonstrates the dense nature of both sets and the continuous nature of the real number line.
Comparing Very Large and Very Small Numbers
Very large and very small numbers are frequently encountered in various fields like science (e.g., distance to stars, size of atoms), engineering, economics (e.g., national debt, tiny tolerances in manufacturing), and technology. Writing these numbers in their full form (like $50,000,000,000$ or $0.0000000001$) is not only tedious but also makes it challenging to compare their magnitudes quickly and accurately.
Scientific notation (also known as standard form) is a widely accepted mathematical notation used to express such numbers concisely and to simplify calculations and comparisons.
Scientific Notation (Standard Form)
A number is written in scientific notation if it is expressed as the product of a number between $1$ and $10$ (including 1, but not 10) and an integer power of $10$.
The general form of a number in scientific notation is $a \times 10^n$, where:
- $a$ is the coefficient (or mantissa). It is a real number such that $1 \le |a| < 10$. This means the absolute value of $a$ is greater than or equal to 1 and strictly less than 10. The decimal point is placed after the first non-zero digit.
- $10^n$ is the power of 10.
- $n$ is the exponent. It is an integer (positive, negative, or zero) and indicates the order of magnitude of the number.
Converting a number to Scientific Notation:
To write a number in scientific notation, move the decimal point until there is only one non-zero digit to the left of the decimal point. The number of places the decimal point was moved determines the absolute value of the exponent $n$.
- If the original number is large (greater than 10) and the decimal point is moved to the left, the exponent $n$ is positive. The number of places moved is the value of $n$.
Example: Speed of light $\approx 300,000,000 \text{ m/s}$. Move decimal 8 places left: $3.0 \times 10^8$. Here $a=3.0$ (which is $1 \le 3.0 < 10$) and $n=8$.
Example: Population of India $\approx 1,400,000,000$. Move decimal 9 places left: $1.4 \times 10^9$. Here $a=1.4$ and $n=9$.
- If the original number is small (between 0 and 1) and the decimal point is moved to the right, the exponent $n$ is negative. The number of places moved is the absolute value of $n$.
Example: Mass of an electron $\approx 0.0000000000000000000000000009109 \text{ kg}$. Move decimal 31 places right: $9.109 \times 10^{-31}$. Here $a=9.109$ and $n=-31$.
Example: Diameter of a virus $\approx 0.000000025 \text{ m}$. Move decimal 8 places right: $2.5 \times 10^{-8}$. Here $a=2.5$ and $n=-8$.
- If the number is already between $1$ and $10$ (not including 10), or is $0$, the exponent $n$ is $0$.
Example: $7.8 = 7.8 \times 10^0$.
Example: $4.005 = 4.005 \times 10^0$.
Note: The number 0 is sometimes written as $0 \times 10^n$ for any $n$, but it doesn't fit the $1 \le |a| < 10$ rule. For non-zero numbers, the coefficient $a$ must be non-zero.
Comparing Numbers in Scientific Notation
Comparing two numbers written in scientific notation, $a \times 10^n$ and $b \times 10^m$ (where $1 \le |a| < 10$ and $1 \le |b| < 10$), is straightforward. The comparison is primarily determined by the exponents of $10$, and secondarily by the coefficients if the exponents are the same.
- Compare the Exponents ($n$ and $m$): This is the most important step. The number with the larger exponent is the greater number.
- If $n > m$, then $a \times 10^n$ is greater than $b \times 10^m$.
If $n > m$, then $a \times 10^n > b \times 10^m$.
Example: $5 \times 10^8$ vs $9 \times 10^6$. Exponent 8 is greater than exponent 6. So $5 \times 10^8 > 9 \times 10^6$.
Example: $2 \times 10^{-4}$ vs $8 \times 10^{-5}$. Exponent -4 is greater than exponent -5 (since -4 is to the right of -5 on the number line). So $2 \times 10^{-4} > 8 \times 10^{-5}$.
- If $n > m$, then $a \times 10^n$ is greater than $b \times 10^m$.
- Compare the Coefficients ($a$ and $b$) if Exponents are Equal ($n = m$): If the exponents are the same, the numbers have the same order of magnitude. The comparison then depends on the coefficients.
- If $n = m$, then compare the coefficients $a$ and $b$ using standard decimal comparison rules. The number with the larger coefficient is the greater number.
If $n = m$ and $a > b$, then $a \times 10^n > b \times 10^m$.
If $n = m$ and $a < b$, then $a \times 10^n < b \times 10^m$.
If $n = m$ and $a = b$, then $a \times 10^n = b \times 10^m$.
Example: $4.5 \times 10^7$ vs $3.2 \times 10^7$. Exponents are both 7. Compare 4.5 and 3.2. Since $4.5 > 3.2$, $4.5 \times 10^7 > 3.2 \times 10^7$.
Example: $6.1 \times 10^{-9}$ vs $8.0 \times 10^{-9}$. Exponents are both -9. Compare 6.1 and 8.0. Since $6.1 < 8.0$, $6.1 \times 10^{-9} < 8.0 \times 10^{-9}$.
- If $n = m$, then compare the coefficients $a$ and $b$ using standard decimal comparison rules. The number with the larger coefficient is the greater number.
Examples of Comparing Numbers in Scientific Notation
Example 1. Compare the estimated population of India ($1.4 \times 10^9$) and the estimated population of the United States ($3.3 \times 10^8$).
Answer:
We are comparing the numbers $1.4 \times 10^9$ and $3.3 \times 10^8$, which are already in scientific notation.
Compare the exponents of $10$ first:
The exponent for the population of India is 9.
The exponent for the population of the United States is 8.
Comparing the exponents: $9 > 8$.
Since the exponent 9 is greater than the exponent 8, the number with the exponent 9 is greater.
Therefore, $1.4 \times 10^9 > 3.3 \times 10^8$.
The estimated population of India is greater than the estimated population of the United States.
Example 2. Compare the diameter of a red blood cell ($7 \times 10^{-6}$ meters) and the diameter of a bacterium ($1 \times 10^{-6}$ meters).
Answer:
We are comparing $7 \times 10^{-6}$ and $1 \times 10^{-6}$. Both numbers are in scientific notation.
Compare the exponents of $10$ first:
The exponent for the red blood cell diameter is -6.
The exponent for the bacterium diameter is -6.
The exponents are equal ($-6 = -6$).
Since the exponents are equal, we compare the coefficients:
The coefficient for the red blood cell diameter is 7.
The coefficient for the bacterium diameter is 1.
Comparing the coefficients: $7 > 1$.
Since the coefficient 7 is greater than the coefficient 1, the number with the coefficient 7 is greater.
Therefore, $7 \times 10^{-6} > 1 \times 10^{-6}$.
The diameter of a red blood cell is greater than the diameter of a bacterium.
Example 3. Arrange the following numbers in ascending order: $3.5 \times 10^4, 1.2 \times 10^5, 8.9 \times 10^3, 4.1 \times 10^4$.
Answer:
The given numbers are $3.5 \times 10^4, 1.2 \times 10^5, 8.9 \times 10^3, 4.1 \times 10^4$. All are in scientific notation.
Identify the exponents of $10$ for each number:
- $3.5 \times 10^4$: exponent is 4
- $1.2 \times 10^5$: exponent is 5
- $8.9 \times 10^3$: exponent is 3
- $4.1 \times 10^4$: exponent is 4
To arrange the numbers in ascending order (smallest to largest), we first order them based on their exponents from smallest to largest:
The exponents are 3, 4, 4, 5. So the order of magnitude is $10^3 < 10^4 < 10^5$.
- The number with exponent 3 is $8.9 \times 10^3$. This will be the smallest.
- The numbers with exponent 4 are $3.5 \times 10^4$ and $4.1 \times 10^4$. We need to compare these two.
- The number with exponent 5 is $1.2 \times 10^5$. This will be the largest.
Compare the numbers with the same exponent ($10^4$). We compare their coefficients: $3.5$ and $4.1$.
Comparing $3.5$ and $4.1$, we find $3.5 < 4.1$.
So, $3.5 \times 10^4 < 4.1 \times 10^4$.
Now, combine the ordered numbers:
Smallest exponent: $8.9 \times 10^3$
Next exponent: $3.5 \times 10^4$ (because its coefficient is smaller)
Next number with same exponent: $4.1 \times 10^4$
Largest exponent: $1.2 \times 10^5$
The numbers in ascending order are:
$\mathbf{8.9 \times 10^3, \quad 3.5 \times 10^4, \quad 4.1 \times 10^4, \quad 1.2 \times 10^5}$.
Scientific notation simplifies working with numbers across a vast range of scales and makes comparing their magnitudes intuitive and efficient.