Question

Prove that √2 is irrational number.
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Answer 1

Answer:

${\Huge {\mathbf {\color{navy}{Answer}}}}$

${\underline {\overline {\mathbf {\green{Given}:}}}}$

${\sf{\sqrt{2} }}$

${\underline {\overline {\bf {\color{plum}{To \: prove: \: }}}}}$

${\sf{  \sqrt{2} \: is \: an \: irrational \: number.}}$

${\overline {\underline {\mathbf {\color{skyblue}{Proof:}}}}}$

${\small {\sf {Let \: us \: assume \: that \: \sqrt{2} \: is \: a \: rational \: number.}}}$

So it can be expressed in the form p/q where p, q are co-prime integers and q≠0

${\sf{ \sqrt{2} = \frac{p}{q} }}$

${\small {\sf{Here \: p \:and \: q \: are \: coprime \: numbers \: and \: q ≠ 0}}}$

${\underline {\overline {\bf {\color{orange}{Solving}}}}}$

${\sf{ \sqrt{2} = \frac{p}{q} }}$

${\sf{On \: squaring \: both \: the \: side \: we \: get,}}$

${\bf{=>2 = { (\frac{p}{q} })^{2}}}$

${\bf {=> 2 {q}^{2}  = {p}^{2} …………..(1)}}$

$\bf{ \frac{ {p}^{2} }{2} = {q}^{2} }$

${\sf{So \: 2 \: divides \: p \: and \: p \: is \: a \: multiple \: of \: 2.}}$

$\bf{⇒ p = 2m}$

$\bf{⇒ {p}^{2} = 4 {m}^{2} ………..(2)}$

${\sf{From \: equations \: (1) \: and \: (2), we \: get,}}$

${\sf{2 {q}^{2} = 4 {m}^{2} }}$

$\bf{⇒ {q}^{2} = 2 {m}^{2} }$

$\bf{⇒ {q}^{2} \: is \: a \: multiple \: of \: 2}$

$\bf{⇒ \: q \: is \: a \: multiple \: of \: 2}$

Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

$\bf {\pmb {\sqrt{2} \: is \: an \: irrational \: number}}$