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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?

Kookie
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1 Answers1

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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.)

Especially Lime
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    How about if $a$ is irrational? – Kookie May 30 '20 at 07:49
  • It doesn't matter, all that matters is which intervals the multiples are in. The only problem is if $a$ is a rational with denominator $k$ (in lowest terms) when it is actually on the boundary of all your intervals and the minimum distance to an integer is exactly $1/k$. – Especially Lime May 30 '20 at 07:53
  • Ok, so how about we change each subinterval so that it always excludes its upper bound and always includes its lower bound. Since we cannot have any multiple of $a$ be in the first and last intervals, we are now left with $k-2$ intervals. If we had $k-2$ of $k-1$ mutliples of $a$ separately sit in each interval, then the last multiple will sit in the same interval as another. Their difference will give a smaller multiple of $a$ which lies in the very first interval. Hence, there exists an element in set $M$ which is within $\frac{1}{k}$ of an integer. How does this sound? – Kookie May 30 '20 at 08:06
  • @Kookie yes, that sounds good to me. – Especially Lime May 30 '20 at 08:22
  • Thank you so much! – Kookie May 30 '20 at 08:22