The **article aims to prove or disprove** that if **two numbers** a and b are **rational**, then *a^b* is also **rational**.

**Rational numbers** can be expressed as **fractions**, **positive**, **negative**, and **zero**. It can be written as **p/q**, where **q** is **not equal to zero.**

The **word** *rational***comes from the word** ** ratio**, a

**comparison of two or more numbers or whole numbers**, and is known as a fraction. In simple terms, the

**average of two whole numbers**. For example:

**3/5**is a rational number. It means that the number

**3**is divided by another number

**5**.

**Finite and recurring numbers** are also rational numbers. **Numbers** like $1.333$,$1.4$ and $1.7$ are** rational numbers.** Numbers having perfect squares are also included in rational numbers. For example: $9$,$16$,$25$ are rational numbers. The **nominator and the denominator are integers**, where the **denominator is not equal to zero.**

**Numbers** that are **not** **rational are irrational numbers**. It is not possible to write irrational numbers in the form of fractions; their $\dfrac{p}{q}$ form doesn’t exist. **Irrational numbers** can be written in the form of decimals. These consist of numbers that are **non-terminating and non-recurring**. Numbers like $1.3245$,$9.7654$,$0.654$ are irrational numbers. **Irrational numbers include** such $\sqrt 7$, $\sqrt 5$,$\sqrt 7$.

**Properties of Rational and Irrational Numbers**

**(a):** If two numbers are rational, their **sum** is also a **rational number.**

**Example:** $\dfrac{1}{4}+\dfrac{3}{4}=1$

**(b):** If two numbers are rational, their **product** is also a** rational number.**

**Example:** $\dfrac{1}{4}\times\dfrac{3}{4}=\dfrac{3}{4}$

**(c):** If two numbers are irrational, their** sum** is not always an **irrational number.**

**Example:** $\sqrt{2}+\sqrt{2}=2\sqrt{2}$ is irrational.

$2+2\sqrt{5}+(-2\sqrt{5}) = 2 $ is rational.

**(d):** If two numbers are irrational, their **product** is not always an **irrational number**.

**Example:** $\sqrt{4}\times\sqrt{3}=\sqrt{12}$ is irrational.

$\sqrt{2}+\sqrt{2} = 2 $ is rational.

## Expert Answer

If $a$ and $b$ are both **rational numbers,** then** prove or disprove** that $a^{b}$ is also rational.

Let’s **assume** that $a=5$ and $b=3$

**Plug** the values of the $a$ and $b$ in the** statement.**

\[a^{b}=5^{3}=125\]

$125$ is a **rational number.**

So, the **statement is true.**

Let’s **suppose values** of the $a=3$ and $b=\dfrac{1}{2}$

**Plug** the values into the** statement.**

\[a^{b}=(3)^\dfrac{1}{2}\]

$\sqrt{3}$ is not a **rational number.**

So, the **statement is false.**

Therefore, $a^{b}$ can be **rational or irrational.**

## Numerical Result

If $a$ and $b$ are **rational,** then $a^{b}$ **can be irrational or rational.** So the **statement is false.**

## Example

**Prove or disprove that if two numbers $x$ and $y$ are rational numbers, then $x^{y}$ is also rational.**

**Solution**

If $x$ and $y$ show** two rational numbers,** then prove that $x^{y}$ is also **rational.**

Let’s **assume** that $x=4$ and $y=2$

**Plug** the values of the $x$ and $y$ in the statement

\[x^{y}=4^{2}=16\]

$16$ is a **rational number.**

So, the **statement is true.**

Let’s suppose values of the $x=7$ and $y=\dfrac{1}{2}$

**Plug** the values into the statement.

\[x^{y}=(7)^\dfrac{1}{2}\]

$\sqrt{7}$ is not a **rational number.**

So, the **statement is false.**

Therefore, $x^{y}$ can be **rational or irrational.**

If $x$ and $y$ are **rational,** then $x^{y}$ can be **irrational or rational.** So the **statement is false.**