Question

Prove that $\sqrt{7}$ is irrational

Answer 1

Let us assume that $\sqrt{7}$ is rational.
It can be written as $\sqrt{7}=\frac{a}{b}$,
where $a$ and $b$ are integers in the lowest terms and $b\ne 0$

If $a$ and $b$ are integers in the lowest terms, it means that do not share any common factors.
So, our equation can be written as
$7=\frac{a^2}{b^2}$ (by squaring both sides)
$\Rightarrow 7b^2=a^2$

Since, both sides are equal, if $7$ divides the L.H.S., then it must also divide R.H.S.
Therefore, $a^2$ is divisible by $7$ and hence $a$ is also divisible by$7$.
So, we can rewrite '$a$' as $7k$, where $k$ is some integer and $k\ne 0$.

$\Rightarrow 7b^2=(7k{)}^2$

$\Rightarrow 7b^2=49k^2$
$\Rightarrow b^2=7k^2$
Similarly, $b$ is also divisible by $7$ as $b^2$ is divisible by $7$.

Thus, $a$ and $b$ are divisible by $7$.

But, $a$ and $b$ are in the lowest terms and do not share any common factors.

Thus, we have a contradiction which means our assumption is wrong.
Therefore, $\sqrt{7}$ is irrational.