Question

prove that
$\sqrt{2}$
is an irrational number?​

Answer 1

Solution:

√2 is an irrational number because it is not an integer number. It is recurring and it is non-terminating number.

√2=41421356237...e.t.c

Answer 2

Answer:

$\large {\underline {\underline {\mathbf {\blue{Given}}}}}$

$\small \bf{\sqrt{2} }$

$\large {\underline {\underline {\mathbf {\green{To \: prove:}}}}}$

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

$\large {\underline {\underline {\mathbf {\red{Proof:}}}}}$

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

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

$\small \bf{ \sqrt{2} = \frac{p}{q} }$

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

$\large {\underline {\underline {\bf {\purple{Solving}}}}}$

$\small \bf{ \sqrt{2} = \frac{p}{q} }$

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

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

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

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

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

$\small \bf{⇒ p = 2m}$

$\small \bf{⇒ {p}^{2} = 4 {m}^{2}…..(2)}$

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

$\small \bf{2 {q}^{2} = 4 {m}^{2} }$

$\small \bf{⇒ {q}^{2} = 2 {m}^{2} }$

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

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

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.

$\large {\boxed {\boxed {\mathbf {\color{red}{ \sqrt{2} \: is \: an \: irrational \: number.}}}}}$