Prove that root3×root5 is irrational

Prove that root3×root5 is irrational
  • 1 answers

Sia 🤖 8 months, 2 weeks ago

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.

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