# Disprove the following statement: For all real numbers $x$ and $y$, if $x + \lfloor x \rfloor = y + \lfloor y \rfloor$ then $x = y$.

Disprove the following statement: For all real numbers $$x$$ and $$y$$, if $$x + \lfloor x \rfloor = y + \lfloor y \rfloor$$ then $$x = y$$.

Aka: Prove the negation: There are real numbers $$x$$ and $$y$$, that $$x + \lfloor x \rfloor = y + \lfloor y \rfloor$$ and $$x \neq y$$.

I have tried plugging in many real number combinations to disprove it, and have tried a few properties to try and prove this but I am completely stuck!

Any hints or guidance on how to approach this question would be helpful in the least.

The claim you're trying to disprove looks true to me. (The function $$x\mapsto x+\lfloor x\rfloor$$ is strictly increasing and therefore injective). So you shouldn't be able to disprove it.
Note that $$x \mapsto \lfloor x \rfloor$$ is increasing, hence $$x \mapsto x + \lfloor x \rfloor$$ is strictly increasing, and hence injective.
Let $$x = n + p$$ and $$y = m + q$$, where $$n,m \in \mathbb{Z}$$ and $$0 \leq p,q < 1$$ (Note that $$n = \lfloor x \rfloor$$ and, similarly, $$m = \lfloor y \rfloor$$). Then we have: $$x + \lfloor x \rfloor = y + \lfloor y \rfloor \;\Rightarrow\; 2n + p = 2m +q \;\Rightarrow\; 2(n - m) = q-p$$ The LHS is an even integer. However $$(q - p) \in (-1, 1)$$, since $$p,q \in [0,1)$$ and the only even integer in the interval $$(-1,1)$$ is $$0$$.
Hint: note that every real number can be written in a unique way as $$a + b$$ with $$a \in {\mathbb Z}$$ and $$b \in [0,1)$$.