The main objective of this question is to prove or disprove that $a^b$ is a rational number provided that $a$ and $b$ are rational numbers.

In arithmetic, a rational number is a number that can be expressed as the ratio $\dfrac{a}{b}$ of two integers such that $b \neq 0$. Apart from fractions, the set of rational numbers contains all integers, where each integer can be expressed as a ratio wherein the numerator is an integer and the denominator is one. Fractions are composed of whole numbers, whereas rational numbers are composed of integers in the numerator and denominator. A rational number is the same as a real number.

Integers such as $-1,3,5$, fractions having integers in numerator and denominator, the decimals which terminate, and the non-terminating decimals having repeating patterns after the decimal point e.g, $0.2222\cdots$, $0.1313\cdots$, etc, are a few common types of rational numbers. The rational numbers along with multiplication and addition form a field containing integers. Rational numbers are a subset of real numbers in mathematical analysis. By using Cauchy sequences, Dedekind cuts, or infinite decimals, the real numbers can be formed from rational numbers.

## Expert Answer

We will either prove or disprove the given statement with a counter-example.

Suppose that $a=-1$ and $b=\dfrac{1}{2}$.

Now $a^b=(-1)^{\frac{1}{2}}$

$=\sqrt{-1}$

which is an imaginary number and not a rational number.

So, by counter-example, it is concluded that if $a$ and $b$ are rational numbers, then $a^b$ may not be a rational number.

## Example 1

Find $5$ rational numbers between $3$ and $4$.

### Solution

Since $3$ and $4$ are integers and both the integers can be expressed as:

$\dfrac{30}{10}$ and $\dfrac{40}{10}$

So, the rational numbers between $\dfrac{30}{10}$ and $\dfrac{40}{10}$ can be listed as:

$\dfrac{31}{10},\dfrac{32}{10},\dfrac{33}{10},\dfrac{34}{10},\dfrac{35}{10},\dfrac{36}{10},\dfrac{37}{10},\dfrac{38}{10},\dfrac{39}{10}$

So, $5$ rational numbers between $3$ and $4$ are:

$\dfrac{31}{10},\dfrac{32}{10},\dfrac{33}{10},\dfrac{34}{10},\dfrac{35}{10}$.

## Example 2

Find the sum, difference, product and the quotient of two rational numbers, $\dfrac{1}{2}$ and $\dfrac{3}{7}$.

### Solution

Sum:

$\dfrac{1}{2}+\dfrac{3}{7}$

$=\dfrac{7+3(2)}{14}$

$=\dfrac{7+6}{14}$

$=\dfrac{13}{14}$

Difference:

$\dfrac{1}{2}-\dfrac{3}{7}$

$=\dfrac{7-3(2)}{14}$

$=\dfrac{7-6}{14}$

$=\dfrac{1}{14}$

Product:

$\dfrac{1}{2}\times\dfrac{3}{7}$

$=\dfrac{1\times 3}{2\times 7}$

$=\dfrac{3}{14}$

Quotient:

$\dfrac{1}{2}\div\dfrac{3}{7}$

$=\dfrac{1}{2}\times\dfrac{7}{3}$

$=\dfrac{1\times 7}{2\times 3}$

$=\dfrac{7}{6}$

## Example 3

Let $-\dfrac{2}{3}$ be the product of two rational numbers. If one of the numbers is $\dfrac{2}{5}$, find the other number.

### Solution

Let $\dfrac{a}{b}$, $b\neq 0$, be the required rational number. Then according to the given statement:

$\dfrac{2}{5}\times \dfrac{a}{b}=-\dfrac{2}{3}$

Multiply both sides by $\dfrac{5}{2}$:

$\dfrac{5}{2}\times \dfrac{2}{5}\times \dfrac{a}{b}=-\dfrac{2}{3}\times \dfrac{5}{2}$

$\dfrac{a}{b}=-\dfrac{2}{3}\times \dfrac{5}{2}$

$\dfrac{a}{b}=-\dfrac{5}{3}$