Prove that the following number is irrational: $2+\sqrt{3}$.

(N/A) Assume,to the contrary,that $2+\sqrt{3}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $2+\sqrt{3} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{3} = \frac{a}{b} - 2$.
This simplifies to $\sqrt{3} = \frac{a - 2b}{b}$.
Since $a$ and $b$ are integers,$\frac{a - 2b}{b}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
However,this contradicts the established fact that $\sqrt{3}$ is an irrational number.
Therefore,our initial assumption is false,and $2+\sqrt{3}$ must be an irrational number.