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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.

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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.

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Note that $x \mapsto \lfloor x \rfloor$ is increasing, hence $x \mapsto x + \lfloor x \rfloor$ is strictly increasing, and hence injective.

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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 $0$.

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Hint: note that every real number can be written in a unique way as $a + b$ with $a \in {\mathbb Z}$ and $b \in [0,1)$.

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