$\mathbf{\text{Step-1:}}\mathbf{\text{Assume}}\sqrt{\mathbf{3}},+\sqrt{\mathbf{5}}\mathbf{\text{as}}\mathbf{\text{rational}}\mathbf{\text{number.}}$

               $,\Rightarrow \sqrt{3},+\sqrt{5},={\displaystyle \frac{a}{b}},\text{where a and b are co - prime numbers and }b\mathrm{\ne }0$

               $\text{on squaring both side we get,}$

               $\Rightarrow {\left(\sqrt{3},+\sqrt{5}\right)}^2={\left({\displaystyle \frac{a}{b}}\right)}^2$

               $\Rightarrow 3+5+2\sqrt{3}\sqrt{5},={\displaystyle \frac{a^2}{b^2}}$

               $\Rightarrow 8+2\sqrt{15},={\displaystyle \frac{a^2}{b^2}}$

               $\Rightarrow 2\sqrt{15},={\displaystyle \frac{a^2}{b^2}}-8$

               $\Rightarrow \sqrt{15},={\displaystyle \frac{a^2-8b^2}{2b^2}}$

               $\text{a and b are integer then }{\displaystyle \frac{a^2-8b^2}{2b^2}}\text{ is a rational number}$

               $\text{Then},\sqrt{15}\text{ is also a rational number}$

$,\mathbf{\text{Step-2:}}\mathbf{\text{Prove}}\mathbf{\text{the}}\mathbf{\text{given}}\mathbf{\text{satement.}}$

               $\text{This contradicts the fact that }\sqrt{1}5\text{ is a irrational number.}$

               $\text{This contradiction has arisen due to the wrong assumption,that }$ $\sqrt{3},+\sqrt{5}\text{ is a rational number.}$

 $\mathbf{\text{Hence,}}\sqrt{\mathbf{3}}\text{ + }\sqrt{\mathbf{5}}\mathbf{\text{is}}\mathbf{\text{an}}\mathbf{\text{irrational}}\mathbf{\text{number}}\text{.}$