So I'm trying to prove that for every real number $a \in \mathbb{R}$, the set $M = \{a,2a,\dots,(k-1)a\}$ contains at least one element that is within $\frac{1}{k}$ of an integer. (Note that $k \in \mathbb{N}$)
So far I found that every $a$ can be expressed as $a = n + r$ where $n \in \mathbb{N}$ and $r \in \mathbb{R}, r \in [0,1)$, and so we can restrict the bounds of $a$ to $[0,1)$. If as a result of multiplication or addition, $a \geq 1$, then we set $a := a-1$ to bring $a$ back into its respective interval.
Now my problem is that I don't know where to go from here. I thought of partitioning the interval $[0,1]$ into subintervals of $\left[0,\frac{1}{k}\right),\left(\frac{1}{k},\frac{2}{k}\right),\dots,\left(\frac{k-1}{k},1\right]$ but I didn't go anywhere with this.
Could I have some help with the proof?
You've partitioned the interval into $k$ subintervals, and you have $k-1$ consecutive multiples of $a$.
If one of the multiples has fractional part in the first or last interval, you're done.
If two of the multiples have fractional part in the same interval, you're also done, since their difference is also one of the $k-1$ multiples.
Now show that one of the above must happen. (Note that you need to modify your intervals so that they include all the points - at the moment since the middle intervals are open you miss out points $1/k$ etc.)