The problem I'm having with this proof is that I'm not sure if my proof actually proves the theorem correct or if I'm using circular reasoning.
Theorem:
Prove that the square root of any irrational number is irrational.
Proof:
=> Suppose not. The square root of any irrational number is rational.
=> Let $m$ be some irrational number. It follows that $\sqrt{m}$ is rational.
=> By definition of a rational number, there are two positive integers $p$ and $q$ such that $\sqrt{m} = \dfrac{q}{p}$
=> $m = \dfrac{q^{2}}{p^{2}}$
=> $q^{2}$ and $p^{2}$ are integers, and by definition of a rational number, $\dfrac{q^{2}}{p^{2}}$ is rational
=> $m$ is irrational and is equal to the rational number $\dfrac{q^{2}}{p^{2}}$. This is a contradiction.
=> Thus, the square root of any irrational number is irrational.
I've seen this proof also done with the addition of these steps:
$m = \dfrac{q^{2}}{p^{2}}$
$m \times p^{2} = q^{2}$
Because a an irrational number times a rational number is irrational, we have an irrational number equaling a rational number which is a contradiction.
My question is: Is this step really necessary?