Prove that root 2 is an irrational number

 Prove that root 2 is an irrational number

Solution:- Let us assume that on the contrary that √2 is a rational number. Then there exist positive integers ‘a’ and ‘ b’ such that

⇒  \sqrt{2}= \frac{a}{b}

Where a and b are coprime i.e. there HCF is 1

Squaring both side

(\sqrt{2})^2 = (\frac{a}{b})^2

2 = \frac{a^2}{b^2}

\Rightarrow 2b^2 = a^2

a^2 is divisible by 2 then a also divisible by 2 –(i)

Hence a = 2c \Rightarrow a^2 = 4c^2

\Rightarrow 2b^2=4c^2

\Rightarrow b^2 = 2c^2

b^2 is divisble 2 then b is also divisible by 2  —(ii)

From (i) and (ii), we obtain that 2 is a common factor of a and b. But this contradicts the fact that a and b have no common factor other than 1. This means that our assumption is wrong.

Hence, \sqrt{2} is an irrational number

Some other question:

Question: Prove that  √5 is an irrational number



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