Prove that is sqrt2 irrational number. - Mathematics

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Prove that is `sqrt2` irrational number.

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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.

Concept: Proofs of Irrationality
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