Question

root 3 is irrational number​

Answer 1

Answer:

$assume \: that \: \sqrt{3 \:} is \: irrational \: \\ therefore \: there \: exist \: positive \: i nteger\: \\ \: a \: and \: b \: \: \sqrt{3} = a \div b \\ a \: and \: b \: have \: no \: common \: factor \: \\ 3 = a ^{2} \div b ^{2} \\ a ^{2} = 3b ^{2} \: a \: is \: a \: multiple \: of \: 3 \\ a = 3c \\ a ^{2} = 3c ^{2} \\ a ^{2} = 9c ^{2} \\ 3b ^{2} = 9c ^{2} \\ b ^{2} = 9c ^{2} \div 3 \\ b ^{2} = 3c ^{2} \: \: b \: is \: a \: multiple \: of \: 3 \\ we \: have \: already \: show \: that \\ \: a \: is \: a \: \: multiple \: of \: 3 \: \: and \: b \: \\ is \: a \: multiple \: of \: 3 \: \\ a \: and \: b \: have \: common \: factor \: 3 \\ which \: is \: a \: contradiction \: to \: an \: assumption \: \\ a \: and \: b \: have \: no \: common \: factor \\ therefore \: our \: assumption \: is \: wrong \\ therefore \sqrt{3 \:} is \: irrational$

PLZZ MARK IT AS BRAINLIEST

Answer 2

We're asked to prove that √3 is an irrational number.

Assume to reach the contradiction that √3 is a rational number. The assumption means that √3 is nothing, but can be written in fractional form. So let,

$\sf{\sqrt3=\dfrac {a}{b}}$

for $\sf{b\neq0}$ and $\sf{a,\ b\in\mathbb{Z}}$ are assumed to have no common factors except 1, i.e., $\sf{\gcd(a,\ b)=1.}$

So,

$\sf{3=\dfrac {a^2}{b^2}}\\\\\\\sf{a^2=3b^2}$

This means $\sf{a^2}$ is a multiple of 3, so is $\sf{a}$ (since $\sf{a\in\mathbb{Z}}$).

Let $\sf{a=3m,\ m\in\mathbb{Z}}$

Then,

$\sf{(3m)^2=3b^2}\\\\\\\sf{b^2=3m^2}$

This means $\sf{b^2}$ is a multiple of 3, so is $\sf{b}$ since $\sf{b\in\mathbb{Z}}$ which contradicts our earlier consideration that $\sf{a}$ and $\sf{b}$ have no common factors.

Hence our assumption is contradicted and proved that √3 is an irrational number.