The problem is to show that for all $\alpha \in \text{Bad}$, and for any $n \in \mathbb{Z} \setminus \{0\}$, then $n\alpha \in \text{Bad}$. ($\text{Bad}$ being the set of all badly approximable numbers).
By definition, $\alpha \in \text{Bad}$ means that $\exists c \in \mathbb{R^+}$ such that for all $p,q \in \mathbb{Z}$, $q \geq 1$, we have:
\begin{equation} \bigg\vert \alpha - \frac{p}{q} \bigg\vert > \frac{c}{q^2} \end{equation}
What I have started by doing is noting that for $n=1$, this problem is trivially true. From there, I have proceeded by induction by assuming that for any $n >1$, say $n=k$, the statement holds, i.e. $\exists c_k \in \mathbb{R}$ such that $\forall p,q \in \mathbb{Z}$ ($q \geq 1$) we have:
\begin{equation} \bigg\vert k\alpha - \frac{p}{q} \bigg\vert > \frac{c_k}{q^2} \end{equation}
Now to try and show that the statement holds true for $n = k+1$, by finding some $c_{k+1} \in \mathbb{R}$, I have computed:
\begin{equation} \bigg\vert (k+1)\alpha - \frac{p}{q} \bigg\vert > \frac{c_{k+1}}{q^2} \end{equation} \begin{equation} \bigg\vert k\alpha - \frac{p}{q} + \alpha \bigg\vert > \frac{c_{k+1}}{q^2} \end{equation} By the triangle inequality... \begin{equation} \bigg\vert k\alpha - \frac{p}{q}\bigg\vert + \vert\alpha\vert > \frac{c_{k+1}}{q^2} \end{equation} Then our assumption step gives us... \begin{equation} \frac{c_k}{q^2} + \vert\alpha\vert > \frac{c_{k+1}}{q^2} \end{equation} So... \begin{equation} c_k + q^2\vert\alpha\vert >c_{k+1} \end{equation} Since the expression on the LHS is real, is this enough to conclude the proof for $n>0$? Also, I'm not sure how to start the proof for $n<0$. Any help is always appreciated!
I assume that $c$ has to be positive.
It's actually pretty easy. Suppose $c$ is such number that $|\alpha - \frac p q|>\frac c {q^2}$ for any integers $p, q$ as specified. Take positive integer $k$. Then $$ |k\alpha - \frac{kp}q| > \frac{ck}{q^2} $$ for every integers $kp, q$ with new $c := ck$. Therefore $k\alpha$ is Bad. On the other hand, if k is negative, $$ |(-k)\alpha - \frac{-kp}{q}| = |k\alpha - \frac{kp}{q}| > \frac{-ck}{q^2} $$