Question

Prove that $\surd 7$ is an irrational number.

Answer 1

Let us assume that $\surd 7$ is a rational number.

So it t can be expressed in the form $\frac{p}{q}$ where $p,q$ are co-prime integers and $q\ne 0$

$\surd 7=\frac{p}{q}$

Here $p$ and $q$ are coprime numbers and $q\ne 0$

$\surd 7=\frac{p}{q}$

On squaring both the side we get,

${\left(\surd 7\right)}^2={\left(\frac{p}{q}\right)}^2$

$\Rightarrow 7={\left(\frac{p}{q}\right)}^2$

$\Rightarrow 7=\frac{p^2}{q^2}$

$\Rightarrow 7q^2=p^2$……………………………..$\left(1\right)$

$\Rightarrow \frac{p^2}{7}=q^2$

So $7$ divides $p$ and $p\mathrm{and}q$ are multiple of $7$.

$\Rightarrow p=7m$

$\Rightarrow p²=49m²$ ………………………………..$\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, we get,

$7q²=49m²$

$\Rightarrow q²=7m²$

$\Rightarrow q²\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}7$

$\Rightarrow q\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}7$

So,$p,q$ have a common factor $7$. This contradicts our assumption that they are co-primes. Therefore, $\frac{p}{q}$is not a rational number

$\surd 7$ is an irrational number.

Hence, it is proved that $\surd 7$ is an irrational number.