Prove that root3×root5 is irrational

If possible, let {tex}\sqrt{3}{/tex} + {tex}\sqrt{5}{/tex} be a rational number equal to x. Then,
x = {tex}\sqrt{3}{/tex} + {tex}\sqrt{5}{/tex}
{tex}\Rightarrow{/tex} x² = ({tex}\sqrt{3}{/tex} + {tex}\sqrt{5}{/tex})² 
{tex}\Rightarrow{/tex} x² = ({tex}\sqrt{3}{/tex})² + ({tex}\sqrt{5}{/tex})² + 2 {tex}\times{/tex} {tex}\sqrt{3}{/tex} {tex}\times{/tex} {tex}\sqrt{5}{/tex}
= 3 + 5 + 2{tex}\sqrt{15}{/tex}
= 8 + 2{tex}\sqrt{15}{/tex}
{tex}\Rightarrow{/tex} x² - 8 = 2{tex}\sqrt{15}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{x^2 -8}{2}{/tex} = {tex}\sqrt{15}{/tex}
Now, x is rational
{tex}\Rightarrow{/tex} x² is rational
{tex}\Rightarrow{/tex} {tex}\frac{x^2 - 8}{2}{/tex} is rational
{tex}\Rightarrow{/tex} {tex}\sqrt{15}{/tex} is rational
But, {tex}\sqrt{15}{/tex} is irrational.
Thus, we arrive at a contradiction. So, our supposition that {tex}\sqrt{3}{/tex} + {tex}\sqrt{5}{/tex} is rational is wrong.
Hence, {tex}\sqrt{3}{/tex} + {tex}\sqrt{5}{/tex} is an irrational number.