Prove that is `sqrt2` irrational number.
Let us assume, to the contrary, that `sqrt2` is rational. So, we can find integers r and s (≠0) such that `sqrt2=r/s`. Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get `sqrt2=a/b` where a and b are coprime.
So, b`sqrt2` = a.
Squaring on both sides and rearranging, we get 2b2 = a2.
Therefore, 2 divides a2. Now, by Theorem it following that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that `sqrt2` is rational.
So, we conclude that `sqrt2` is irrational.
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