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Prove that squre root 3 is an irrational number

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Solution: Let us assume on the contrary that \sqrt3 is a rational number . then there exist positive integer a and b such that

    \sqrt3=a/b , Where a and b co-prime i.e their HCF is 1

Now,

\sqrt3=a/b \ \Rightarrow 3=a^2/b^2\ \Rightarrow 3b^2=a^2\\Rightarrow 3/a^2Rightarrow ecause 3/3b^2\ \Rightarrow 3/a

\ a=3c for some integer c

\ Rightarrow a^2=9c^2\ \Rightarrow 3b^2=9c^2\ \Rightarrow b^2=3c^2\ \Rightarrow 3/b^2\ \Rightarrow 3/b

Hence ,\sqrt3 is an irrational number

Posted by

Deependra Verma

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