Irrationality of the Square Root of Prime Numbers

The square root of any prime number is irrational.

Proof by Contradiction

Let $ p $ be a prime number.

Suppose, for the sake of contradiction, that $ \sqrt{p} $ is rational.

This means there exist two integers $ m $ and $ n $, with $ m, n \in \mathbb{Z} $ and $ n \neq 0,1 $, such that:

$$ \sqrt{p} = \frac{m}{n} $$

Squaring both sides gives:

$$ p = \frac{m^2}{n^2} $$

Rearranging, we get:

$$ p n^2 = m^2 $$

By assumption, $ m $ and $ n $ are coprime, meaning their greatest common divisor is:

$$ \gcd(m, n) = 1 $$

Since $ p $ divides $ m^2 $, and $ p $ is prime, it must also divide $ m $. So we can write:

$$ m = p k $$

for some integer $ k \neq 0 $. Substituting this into our equation:

$$ p n^2 = (p k)^2 $$

which simplifies to:

$$ n^2 = p k^2 $$

This shows that $ n^2 $ is also divisible by $ p $, which means $ n $ must be a multiple of $ p $ as well.

But this contradicts our earlier assumption that $ m $ and $ n $ are coprime.

Therefore, our initial assumption must be false, and we conclude that the square root of any prime number is irrational.