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.

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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?