Question

prove that √3 is irrational by using contradiction method​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm \: Let \: assume \: that \: \sqrt{3} \: is \: not \: irrational.$

So,

$\rm\implies \:\sqrt{3} \: is \: rational.$

So, by definition of rational numbers,

$\rm \: Let \: \sqrt{3} = \dfrac{a}{b}$

where a and b are integers such that HCF (a, b) is 1 and b is non - zero.

$\rm\implies \:a = \sqrt{3}b$

$\rm\implies \: {a}^{2} = {3b}^{2}$

$\rm\implies \: {a}^{2} \: is \: divisible \: by \: 3$

$\rm\implies \: a \: is \: divisible \: by \: 3 - - - (1)$

$\rm \: Let \: assume \: that \: a = 3m$

$\rm\implies \: {a}^{2} = {9m}^{2}$

$\rm\implies \: {3b}^{2} = {9m}^{2}$

$\rm\implies \: {b}^{2} = {3m}^{2}$

$\rm\implies \: {b}^{2} \: is \: divisible \: by \: 3$

$\rm\implies \: b \: is \: divisible \: by \: 3 - - - (2)$

From equation (1) and (2), we concluded that

$\rm\implies \:a\:and \: b \: both \: are \: divisible \: by \: 3$

which is contradiction to the fact that HCF (a, b,) = 1.

Hence, our assumption is wrong.

Thus,

$\rm\implies \:\sqrt{3} \: is \: irrational.$

$\rule{190pt}{2pt}$

Irrational number :- Irrational number are those numbers whose decimal representation is neither terminating nor repeating.

Rational number :- Rational number are those numbers whom decimal representation is either terminating or non terminating but repeating. They can be represented in the form of a/b where b is non zero.