Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs/Proof 1

Proof
Let $\alpha \in \C$ be a root of $f$.

Then $f \left({\alpha}\right) = 0$ by definition.

Suppose $\alpha$ is wholly real.

Then by Complex Number equals Conjugate iff Wholly Real:
 * $\alpha = \overline \alpha$

and so $\overline \alpha$ is a root of $f$ a priori.

Now let $\alpha \in \C$ not be wholly real.

By definition of complex conjugate, we have that:
 * $\overline 0 = 0$

and so:
 * $f \left({\alpha}\right) = \overline{f \left({\alpha}\right)}$

From Conjugate of Polynomial is Polynomial of Conjugate:
 * $\overline{f \left({\alpha}\right)} = f \left({\overline \alpha}\right)$

from which it follows that:
 * $f \left({\overline \alpha}\right) = 0$

That is, $\overline \alpha$ is also a root of $f$.