Example to disprove the statement: for all real numbers $x$ and $y$, if $x + \lfloor x \rfloor = y + \lfloor y \rfloor$ then $x = y$ I am supposed to disprove this statement but I can't find an example.
Could anybody give me a hint/guide?

For all real numbers $x$ and $y$, if $x + \lfloor x \rfloor = y + \lfloor y \rfloor$ then $x = y$.

I can't find any example that disproves this yet so I am becoming to believe this is true..
Any insight would be appreciated.. :)
 A: Suppose that $x<y$, then if $0<y-x<1$, we have that $\lfloor x\rfloor = \lfloor y \rfloor$, or $\lfloor x \rfloor < \lfloor y \rfloor$, in either case, since $x<y$, we have $x + \lfloor x \rfloor < y +\lfloor y \rfloor$. If on the other hand $y-x>1$, then the result is clear since we also have $\lfloor x \rfloor <\lfloor y \rfloor$. Therefore the function is strictly increasing. And by the hint in the comments, it is also injective.
A: Let $\lfloor x \rfloor = n \in \mathbb Z$
Then $n \le x < n+1$.  Let $\alpha = x - n$.  Then $0 \le \alpha < 1$ and $x = n + \alpha$.
Likewise 
Let $\lfloor y \rfloor = j \in \mathbb Z$ Then $j \le y < j+1$.  Let $\beta = y - j$.  Then $0 \le \beta < 1$ and $y = j +\beta$.
Suppose $x + \lfloor x \rfloor = n + \alpha + n = 2n + \alpha = \lfloor y \rfloor = j + \beta + j = 2j + \beta$.
So $2n + \alpha = 2j + \beta$ and then $\alpha - \beta = 2j - 2n$.
$2j - 2n\in \mathbb Z$.  
And because $0 \le \alpha < 1$ we know $-beta \le \alpha - \beta < 1 - \beta$.
And because $0 \le \alpha < 1$ we know $-1 < -beta \le \alpha -\beta$.  And we know that $1 -\beta \ge 1 - beta > \alpha - \beta$.
So $-1 < \alpha -\beta < 1$.  But $\alpha - \beta = 2j -2n$ is an integer.  The only integer between $-1$ and $1$ is $0$.  So $\alpha - \beta = 0$.
So $\alpha = \beta$.
And $2j - 2n = 0$ so $j =n$.  So $x = n +\alpha = j + \beta = y$.
and $0 \le \beta$ then $-1 < \alpha - \beta < 1$
A: Let $\{x\}$ be the fractional part of $x$.  We will use the following results.
Lemma: Let $x$ be real and $n$ be an integer.  Then $\lfloor x+n \rfloor=\lfloor x \rfloor+n$, and $\{x+n\}=\{x\}$.
Lemma: If $\lfloor x \rfloor=\lfloor y \rfloor$ and $\{x\}=\{y\}$, then $x=y$.
Now, assume $x+\lfloor x \rfloor=y+\lfloor y \rfloor$.  By the lemma, $\{x+\lfloor x \rfloor\}=\{x\}$, and so $\{x\}=\{y\}$.  Similarly, taking the least integer, since $\lfloor x+\lfloor x \rfloor\rfloor=2\lfloor x \rfloor$, we get $2 \lfloor x \rfloor = 2 \lfloor y \rfloor$.  By the second lemma, $x=y$.
A: We know x=[x]+{x},where []-floor function,{}-fractional part.
The given equation thus becomes,
2[x]-{x}=2[y]-{y}
[x]-[y]=$\frac{\{x\}-\{y\}}{2}$
For any real number 0<{x}<1 and hence -1<{x}-{y}<1 
-1/2<$\frac{\{x\}-\{y\}}{2}$<1/2
Since [x]-[y] is always an integer the only value possible in this range is 0.
[x]-[y]=0 and {x}-{y}=0
Therefore x=y.
