Question

Prove that $4-\sqrt{3}$ is an irrational number.

Answer 1

Let us assume the contrary, that $4-\sqrt{3}$ is rational.

That is, we can find co-prime a and b $(b\ne 0)$ such that $4-\sqrt{3}=\frac{a}{b}$.

Therefore $4-\frac{a}{b}=\sqrt{3}.$

Rearranging this equation, we get $\sqrt{3}=4-\frac{a}{b}$

=$\frac{4b-a}{b}$

Since a and b are integers, we get $\frac{4b-a}{b}$ is rational, and so $\sqrt{3}$ is rational.

But this contradicts the fact that $\sqrt{3}$ is irrational,

This contradicts the fact that $\sqrt{3}$ is irrational.

This contradiction has arisen because of our incorrect assumption that 4-$\sqrt{3}$ is rational.

so, we conclude that 4$-\sqrt{3}$ is irrational.