Prove that is `sqrt3` irrational number.
Let us assume, to contrary, that is rational. That is, we can find integers a and b (≠0) such that `sqrt2=a/b`
Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b`sqrt3`= a
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that `sqrt3` is rational.
So, we conclude that `sqrt3` is irrational.
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