Prove √3 irrational number
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Let us assume on the contrary that root 3 is a rational number.
Then, there exist positive integers a and b such that
root 3= a/b
3=a^2/b^2
3b^2=a^2
3 divides a^2 [∵3 divides 3b^2]
3 divides a
a=3c for some integer c
a^2=9c^2
⇒3 divides b^2 [∵3 divides 3c^2 ]
⇒3 divides b...(ii)
From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.
Hence, root 3 is an irrational number.
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