Question

6. Prove that √3 is irrational.​

Answer 1

$\huge \huge\bold\color{blue}{\underbrace{\overbrace{\mathfrak{\color{red}{✍Answer⭐}}}}}$

Assume that √3 is rational number.

Then, there exist a coprime number a/b such that ( b ≠ 0 ).

$\boxed{\therefore \sqrt{3} = \frac{a}{b} }$

Let a & b have a common factor other than 1. Then we can divide them by common factor, assuming that a and b are coprime,

$\boxed{a = \sqrt{3} b}$

Squaring on both side.

$\boxed{ {a}^{2} = 3 {b}^{2} }$

Therefore a² is divisible by 5 and it can be said that a is divisible by 3.

• Let a = 5k ,where k is an integer.

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$\boxed{ {(3k)}^{2} = 3{b}^{2} }$This means that b² is divisible by 3 and hence b is divisible by 3.

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$\boxed{ {b}^{2} = 3 {k}^{2} }$This implies that a & b have 5 as a common factor.

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And this is a contradiction to that fact that a and b are coprime.

$\color{orange}{\mathtt{ Hence \: \: \sqrt{5} \: \: is \: \: irrational. }}$