Question

prove that root 7 +root 3 is irrational​

Answer 1

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Answer 2

${ \large{ \sf{ \underbrace{\underline{\bigstar \: Answer}}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\pink{\bigstar}★ \large\underline{\boxed{\bf\green{√3+√7 is\: irrational.}}}$

${ \large{ \sf{ \underbrace{\underline{\bigstar \:Step\:-by\:-step\: explanation:}}}}}$

$\underline{\boxed{\sf\purple{Let \:us\: assume \:that \:√3+√7\: is\: rational.}}}$

That is , we can find coprimes a and b (b≠0) such that

$\sqrt{3}+\sqrt{7}=\sqrt{a}{b}$

$\bold{Therefore,}$

$\sqrt{7}=\frac{a}{b}-\sqrt{3}$

$\color{mangeta}\boxed{\sf{Squaring \:on\: both\: sides ,\:we \:get}}$

$7=\frac{a^{2}}{b^{2}}+3-2\times \frac{a}{b}\times \sqrt{3}$

$\color{mangeta}\boxed{\sf{Rearranging\: the\: terms}}$

$\begin{gathered}2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}}{b^{2}}+3-7\\=\frac{a^{2}}{b^{2}}-4\end{gathered}$

$\implies 2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}$

$\begin{gathered}\implies \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}\times \frac{b}{2a}\\=\frac{a^{2}-4b^{2}}{2ab}\end{gathered}$

$\color{purple}\boxed{\sf{Since,\: a\: and\: b\: are\: integers}}$

$\frac{(a^{2}-4b^{2})}{2ab}$

$\textbf{is rational ,and so √3 also rational.}$

But this contradicts the fact that √3 is irrational.

This contradiction has arisen because of our incorrect assumption that √3+√7 is rational.

$\large\bf\color{red}{\underline{\underline{Hence,\: √3+\:√7 \:is\: irrational.}}}$

Thankyou :)