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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} x

^{2}= ({tex}\sqrt{3}{/tex} + {tex}\sqrt{5}{/tex})^{2}{tex}\Rightarrow{/tex} x

^{2}= ({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} x

^{2}- 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} x

^{2}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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