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Prove that is `sqrt2` irrational number.

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#### Solution 1

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 2b^{2} = a^{2}.

Therefore, 2 divides a^{2}. Now, by Theorem it following that 2 divides a.

So, we can write a = 2c for some integer c.

Substituting for a, we get 2b^{2} = 4c^{2}, that is, b^{2} = 2c^{2}.

This means that 2 divides b^{2}, 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.

#### Solution 2

Let `sqrt(2)` be rational.

∴ `sqrt(2) = "p"/"q"` where p and q are co-prime integers and, q ≠ 0

Implies that `sqrt(2"q") = "p"`

2q^{2 }- p^{2} ...(i)

⇒ 2 divides p^{2}

⇒ 2 divides p ...(A)

Let p = 2C for some integer c

p^{2} = 4c^{2}

⇒ 2q^{2} = 4C^{2}

⇒ q^{2 }= 2c^{2}

⇒ 2 divides q^{2}

⇒ 2 divides q ...(B)

From (A) and (B), we get

∴ 2 is common factor of both p and q. But this contradicts the fact that p and q have no common factor other than 1.

∴ Our supposition is wrong Hence, `sqrt(2)` is an irrational number.

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