Question

Prove that $\frac{\sqrt{7}}{4}$ is an irrational number.

Answer 1

Let us assume that $\sqrt{7}$ is a rational number which can be expressed in the form of $\frac{p}{q}$, where $p$ and $q$ are integers, $q\ne 0$ and $p$ and $q$ are co prime that is $HCF(p,q)=1$.

We have,

$\sqrt{7}=\frac{p}{q}\Rightarrow \sqrt{7}q=p......\left(1\right)\Rightarrow 7q^2=p^2\left(squaringbothsides\right)$

$\Rightarrow p^2$ is divisible by $7$

$\Rightarrow p$ is divisible by $7$$......\left(2\right)$

Therefore, for an integer $r$,

$p=7r\Rightarrow \sqrt{7}q=7r\left(from\right(1\left)\right)\Rightarrow 7q^2=49r^2\left(squaringbothsides\right)\Rightarrow q^2=\frac{49}{7}{r}^2\Rightarrow q^2=7r^2$

$\Rightarrow q^2$ is divisible by $7$

$\Rightarrow q$ is divisible by $7$$......\left(3\right)$

From equations 2 and 3, we get that $7$ is the common factor of $p$ and $q$ which contradicts that $p$ and $q$ are co prime. This means that our assumption was wrong.

Thus $\sqrt{7}$ is an irrational number.

Now, since division of a rational number to an irrational number is an irrational number.

Hence $\frac{\sqrt{7}}{4}$ is an irrational number.