Question

$\sqrt{3}$ \: is a irrational number.
Prove it.​

Answer 1

${\fbox{\boxed {\huge{\rm{\red{Answer}}}}}}$

Step-by-step explanation:

$\: \Huge\bold\blue{Solution} \:$

$\rightarrow \: \mathtt{ \sqrt{3}} \: = \large\mathtt {\frac{p}{q} }$

$\rightarrow \: \mathtt{3} = \large\mathtt{ \frac{ {p}^{2} }{ {q}^{2} } } \: \rightarrow(square \: both)$

$\rightarrow \mathtt{ {3q}^{2} } = \mathtt{ {P}^{2} }$

$\rightarrow \large\mathtt{3 \: divides \: by \: {P}^{2} } \\ \: \: \: \: \: \: \large \mathtt{3 \: divides \: by \: { P}^{2} }$

$\rightarrow \mathtt{Let \: P = 3m} \\ \: \: \: \: \: \: \: \mathtt{ {3q}^{2} } = \mathtt{(3m) {}^{2} } \\ \: \: \: \: \: \: \: \: \mathtt{ {q}^{2}} = \mathtt{ \cancel\frac{9m {}^{2} } {3} }$

$\: \bold{From\: (2)\: and\: (3)\: 3\: is \: a\: comman\: factor\: of \:P \:and\: Q.} \:$

$\: \bold{So \sqrt{3} \: is \: a\: irrational \:number.} \:$

${\boxed{\huge{\red{\mathtt{BeBrainly}}}}}$