Question

prove that root 2 and root 7 irrational number by mathematically mathematically

Answer 1

Hello Mate!

$let \: \sqrt{2} \: be \: rational \: no. \\ \sqrt{2} = \frac{p}{q} \\ 2 = \frac{ {p}^{2} }{ {q}^{2} } \\ 2 {q}^{2} = {p}^{2} \\here \: 2 \: is \: factor \: of \: q \\ p \\ let \: some \: natural \: no. \: be \: m \\ 2m = p \\ {2}^{2} {m}^{2} = {p}^{2} \\ 4 {m}^{2} = 2 {q}^{2} \\ 2 {m}^{2} = {q}^{2} \\ hre \: 2 \: is \: factor \: of \: q$

But it can not be possible. So contradiction says that
$\sqrt{2} \: is \: irrational$

Similarly in case of root 7

$\sqrt{7} = \frac{p}{q} \\ 7 = \frac{ {p}^{2} }{ {q}^{2} } \\ 7 {q}^{2} = {p}^{2} \\ 7 \: is \: factor \: of \: p \\ 7m = p \\ {7}^{2} {m}^{2} = {p}^{2} \\ 49 {m}^{2} = 7 {q}^{2} \\ 7 {m}^{2} = {q}^{2} \\ 7 \: is \: factor \: of \: q$

But contradiction says that root 7 is irrational no. , because it cannot have common factor.