Question

Prove that $\sqrt{2}$ is irrational and hence prove that $\frac{5-3\sqrt{2}}{7}$ is irrational.

Answer 1

Let us assume that $\sqrt{2}$ is a rational. Then there exist co-prime positive integers $a$ and $b$ such that,

$\sqrt{2}=\frac{a}{b}$

$⟹a=b\sqrt{2}$

Squaring on both sides, we get

$a^2=2b^2$

Therefore, $a^2$ is divisible by $2$ and hence $a$ is also divisible by $2$

So, we can write $a=2p$, for some integer $p$

substituting for $a$, we get

$4p^2=2b^2⟹b^2=2p^2$

This means, $b^2$ is divisible by $2$ and so, $b$ is also divisible by $2$.

Therefore, $a$ and $b$ have at least one common factor, i.e, $2$

But, this contradicts the fact that $a$ and $b$ are co-primes.

Thus, our supposition is wrong.

Hence, $\sqrt{2}$ is an irrational.

As $\sqrt{2}$ is an irrational , $5-3\sqrt{2}$ is an irrational,

And $\frac{5-3\sqrt{2}}{7}$ is also an irrational.