# Prove that squre root 3 is an irrational number

Solution: Let us assume on the contrary that $\\\sqrt{3}$ is a rational number . then there exist positive integer a and b such that

$\\\sqrt{3}=a/b ,$ Where a and b co-prime i.e their HCF is 1

Now,

$\\\sqrt{3}=a/b \\ \\\Rightarrow 3=a^2/b^2\\ \\\Rightarrow 3b^2=a^2\\\\\Rightarrow 3/a^2\Rightarrow \because 3/3b^2\\ \\\Rightarrow 3/a$

$\\ a=3c$ 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 ,$\\\sqrt{3}$ is an irrational number

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