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Prove or disprove that if a and b are rational numbers, then a^b is also rational.

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 $\dfrac{5}{2}$:

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

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

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

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