Prove that $\displaystyle 2+\sqrt{3}$ is an irrational number, given that $\displaystyle \sqrt{3}$ is an irrational number.

Let us assume that \(\displaystyle 2+\sqrt{3}\) is rational Let \(\displaystyle 2+\sqrt{3}=\frac{\mathrm{p}}{\mathrm{q}} ; \mathrm{q} \neq 0\) and p, q are integers \[\Rightarrow \sqrt{3}=\frac{p-2 q}{q} \] p and q are integers, \(\displaystyle \therefore \mathrm{p}-2 \mathrm{q}\) is an integer \(\displaystyle \Rightarrow \frac{\mathrm{p}-2 \mathrm{q}}{\mathrm{q}}\) is a rational number \(\displaystyle \Rightarrow \sqrt{3}\) is a rational number which contradicts our assumption that \(\displaystyle \sqrt{3}\) is an irrational number. \(\displaystyle \Rightarrow 2+\sqrt{3}\) is an irrational number