Prove that $\displaystyle \sqrt{5}$ is an irrational number.

Let \(\displaystyle \sqrt{\mathbf{5}}\) be a rational number. \[\therefore \sqrt{\mathbf{5}}=\frac{\mathbf{p}}{\mathbf{q}} \text {, where } \mathrm{q} \neq 0 \text { and let } \mathrm{p} \& \mathrm{q} \text { be co-primes. } \] \[5 q^{2}=p^{2} \Rightarrow p^{2} \text { is divisible by } 5 \Rightarrow p \text { is divisible by } 5 \] \[\Rightarrow \mathrm{p}=5 \mathrm{a} \text {, where 'a' is some integer } \] \[25 \mathrm{a}^{2}=5 \mathrm{q}^{2} \Rightarrow \mathrm{q}^{2}=5 \mathrm{a}^{2} \Rightarrow \mathrm{q}^{2} \text { is divisible by } 5 \Rightarrow \mathrm{q} \text { is divisible by } 5 \] \[\Rightarrow \mathrm{q}=5 \mathrm{~b} \text {, where ' } \mathrm{b} \text { ' is some integer } \]