How can one prove that the square root of a prime number (e.g. $ \sqrt 61$) is irrational.
First we need to prove that $61$ is prime. This can be done by simply showing that $$ 49 < 61 < 64$$ and so $$ 7 < \sqrt 61 < 8$$ then the only possible prime factors of $61$ are $2 , 3, 5, 7$. Then by contradictions and DIC we can show that $61$ is prime.
Now how can we show that $\sqrt 61$ is irrational?
Suppose $61=a^2/b^2$, where $a$ and $b$ are both in lowest terms. Then $$ 61 \cdot b^2 =a^2 $$ since $a^2$ is multiple of $61^2$, or not at all, this simply cannot happen in lhs --- it has odd number of factor of 61, if any.