Proof: For all real numbers $x$ and $y$, if $\lfloor 2x\rfloor = \lfloor 2y \rfloor$ then $\lfloor x\rfloor = \lfloor y \rfloor$

Question

Here is what I have so far:

The statement is true. Let $x$ and $y$ be real numbers and assume that $\lfloor 2x\rfloor = \lfloor 2y \rfloor=n$, where $n$ is an integer (we are trying to prove $\lfloor x\rfloor =\lfloor y\rfloor$). Then $n \le 2x < n + 1$ and $n \le 2y < n + 1$.

I'm not quite sure where to go from here. Any suggestions/guidance would be much appreciated!

Hint: consider the two cases "$n=2k$ is even" and "$n=2k+1$ is odd". — Misha Lavrov, Oct 30 '20 at 19:10

score 1 · Answer 1 · answered Oct 30 '20 at 19:10

Since you want to prove something about $x$ and $y$, the natural thing is to divide those inequalities by $2$: $\frac{n}2\le x,y<\frac{n+1}2$. Now consider two cases, $n$ even and $n$ odd.

This works quite easily, but it’s even easier to prove the contrapositive: show that if $\lfloor x\rfloor\ne\lfloor y\rfloor$, then $\lfloor 2x\rfloor\ne\lfloor 2y\rfloor$.

Proof: For all real numbers $x$ and $y$, if $\lfloor 2x\rfloor = \lfloor 2y \rfloor$ then $\lfloor x\rfloor = \lfloor y \rfloor$

1 Answers1