prove that √3 is irrational no.
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Let us assume on the contrary that 3 is a rational number.
Then, there exist positive integers a and b such that
3=ba where, a and b, are co-prime i.e. their HCF is 1
Now,
3=ba
⇒3=b2a2
⇒3b2=a2
⇒3 divides a2[∵3 divides 3b2]
⇒3 divides a...(i)
⇒a=3c for some integer c
⇒a2=9c2
⇒3b2=9c2[∵a2=3b2]
⇒b2=3c2
⇒3 divides b2[∵3 divides 3c2]
⇒3 divides b...(
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