Show that $5-\sqrt{3}$ is irrational.

(N/A) Let us assume,to the contrary,that $5-\sqrt{3}$ is rational.
That is,we can find coprime integers $a$ and $b$ $(b \neq 0)$ such that $5-\sqrt{3} = \frac{a}{b}$.
Rearranging the equation,we get $5 - \frac{a}{b} = \sqrt{3}$,which simplifies to $\sqrt{3} = \frac{5b - a}{b}$.
Since $a$ and $b$ are integers,$\frac{5b - a}{b}$ is a rational number. This implies that $\sqrt{3}$ is rational.
However,this contradicts the established fact that $\sqrt{3}$ is an irrational number.
This contradiction has arisen because of our incorrect assumption that $5-\sqrt{3}$ is rational.
Therefore,we conclude that $5-\sqrt{3}$ is irrational.