Question

prove that 3+√2 is irrational number​

Answer 1

Here is your answer user.

Answer 2

$\bold \red \star \: \underline{Given:-}$

$\: \: \: \: \: \tt{Prove \: that \: \red{ 3 + \sqrt{2}} \: is \: } \: \: \\ \: \: \tt{ \: \: \: irrational \: no.}$

$\bold \red \star \: \underline{Solution:-}$

$\tt{ \: \: \: \: \: Let's \: assume \: \red{ 3+ \sqrt{2}} \: is \: a \:} \\ \tt { \: \: \: \: \: rational \: no. }$

$\tt{ \: \: \: \: \: 3 + \sqrt{2} = \frac{p}{q} } \\ \\ \tt{ \: \: \: \: \: \sqrt{2} = \frac{p - 3q}{q} }$

$\rule{190}2$

$\\ \tt{ \: \: \: \: \: Here, We \: can \: see \: that \: \sqrt{2} \: is } \\ \tt{ \: \: \: \: \: an \: irrational \: no \: but \: \frac{p - 3q}{q} } \\ \tt{ \: \: \: \: \: is \: a \: rational \: no.}$

$\tt{ \: \: \: \: \: So, Our \: assumption \: is \: wrong } \\ \tt{ \: \: \: \: \: \red{3+ \sqrt{2}} \: is \: an \: irrational \: no.}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ \red{H}ence \: \red{P}roved.}$