Question

show that root p is irrational number ​

Answer 1

if a digit is not written in the form of p/q

where p is a rational number. and it is not divisible by q than it is an irrational number.

Answer 2

$Let \: us \: assume \: to \: the \: contrary \: that \: \sqrt{p} \: is \: rational. \\ Now, \: \sqrt{p} = \frac{a}{b} \: (where, \: a \: and \: b \: are \: coprime) \\ = > \: \: b \sqrt{p} = a$

$Squaring \: both \: sides, \\ = > {b}^{2} p = {a}^{2} ...............(i)\\ p \: divides \: {a}^{2} \: so \: p \: divides \: a$

$Let \: a = pc \: for \: some \: value \: c.$

$Substituting \: the \: value \: of \: a \: in \: equation \: (i) \\ = > {b}^{2} p = {(pc)}^{2} \\ = > {b}^{2} p = {p}^{2} {c}^{2} \\ = > \: \: {b}^{2} = {c}^{2} p \\ \\ p \: divides \: {b}^{2} \: so \: p \: divides \: b$

$So, \: a \: and \: b \: have \: at \: least \: p \: as \: their \: common \: factor \: \\ but \: this \: contradicts \: the \: fact \: that \: a \: and \: b \: have \: no \: \\ common \: factor \: other \: than \: 1.$

$This \: contradiction \: has \: arisen \: due \: to \: our \: incorrect \\ \: assumption \: that \: \sqrt{p} \: is \: irrational.$

$Hence \: we \: conclude \: that \: \sqrt{p} \: is \: irrational.$

:D