Question

Show that $5–\sqrt{3}$ is irrational.

Answer 1

Let us assume, to the contrary, that $5-\sqrt{3}$ is rational.
That is, we can find co prime a and b (b $\ne$ 0) such that $5-\sqrt{3}$ = $\frac{a}{b}$
Therefore, $5-\frac{a}{b}=\sqrt{3}$
Rearranging this equation, we get $\sqrt{3}=5-\frac{a}{b}=\frac{(5b-a)}{b}$
Since a and b are integers, we get $5-\frac{a}{b}$ is rational, and so $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $5-\sqrt{3}$ is rational.
So, we conclude that $5-\sqrt{3}$ is irrational.