Content of Scalar Multiple

Theorem
Let $f$ be a polynomial with rational coefficients.

Let $\operatorname{cont}\left({f}\right)$ be the content of $f$.

Let $q \in \Q$ be a rational number.

Then $\operatorname{cont}\left({qf}\right) = q\operatorname{cont}\left({f}\right)$.

Proof
Let $q = a/b$ with $a,b \in \Z$.

Let $n \in \Z$ such that $nf \in \Z[X]$.

Then we have $bn\left({qf}\right) = anf \in \Z[X]$.

Now by the definition of content, and using that $a \in \Z$, we have
 * $(1) \quad \operatorname{cont}\left({bnqf}\right) = a \operatorname{cont}\left({nf}\right)$

And by definition
 * $\displaystyle\operatorname{cont}\left({qf}\right) = \frac1{bn}\operatorname{cont}\left({bnqf}\right)$

Therefore, using $(1)$ and the definition of $\operatorname{cont}\left({f}\right)$, we find that:
 * $\displaystyle\operatorname{cont}\left({qf}\right) = \frac a b \frac1n\operatorname{cont}\left({nf}\right) = q \operatorname{cont}\left({f}\right)$