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How can we prove that $$\lfloor x + y \rfloor = \lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor$$ for all real $x$, where $ \lfloor x \rfloor$ denotes greatest integer less than or equal to $x$?

I was able to prove that $\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor $ or $\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + 1$ by using the property that $\lfloor x \rfloor = m$ means $m \le x \lt m + 1$ but cannot prove above one by any means. I have tried a lot, can any one please help and please give a simple proof?

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  • $\begingroup$ I know this page, actually I used this site to prove the formula in below mentioned way. $\endgroup$ – Matt Aug 16 '16 at 16:28
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    $\begingroup$ This is not correct as written. For example, if $x = y = 1$, then $[x+y] = 2$ but $[y+x-[x]] = 1$. $\endgroup$ – Omnomnomnom Aug 16 '16 at 16:33
  • $\begingroup$ You also have to add [x] = 1. Hence it becomes 2. $\endgroup$ – Matt Aug 16 '16 at 17:14
  • $\begingroup$ Sorry I skipped it. I have edited the question. $\endgroup$ – Matt Aug 16 '16 at 17:15
  • $\begingroup$ Let [x] be floor and {x} = x - [x] be fractional part and note 0<= {x} < 1. So 0 < = {y} + {x} < 2. So we have two cases i) 0 <= {y} + {x} < 1$ and ii)1 <= {y} + {x} < 2. Case i)[y + x] = [[y] + [x] + {x} + {y}] = [[y] + [x]]. As [y] and [x] are integers [[y] + [x]] = [y] + [x]. A {x}+{y} < 1, [y] = [[y] + {x} + {y}] = [y + x - [x]] and we are done. Case ii) [y + x] = [[y] + [x] + {x} + {y}] = [[y] + [x]] + 1 = [y] + [x] + 1 = [y + 1] + [x]=[[y] + {x} + {y}] + [x] = [y + x - [x]] + [x] and we are done. $\endgroup$ – fleablood Aug 16 '16 at 18:30
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If $n$ is an integer and $t$ any real number, then it is straightforward to show that$\lfloor t + n \rfloor = \lfloor t \rfloor + n.$ Therefore, since $\lfloor x \rfloor$ is an integer, \begin{align*} \quad &\lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor \\ = &\lfloor y + x \rfloor - \lfloor x \rfloor + \lfloor x \rfloor \\ = &\lfloor x + y \rfloor. \end{align*}

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    $\begingroup$ That was simple and helpful. Thanks. $\endgroup$ – Matt Aug 16 '16 at 18:28
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If you introduced the notation {$x$} = $x - [x]$ and note $0 \le \{x\} < 1$ this becomes easy.

$0 \le \{x\} + \{y\} < 2$.

Case 1: $0 \le \{x\} + \{y\} < 1$.

Note for integers $n, m$ and $0 \le r < 1$ that: $[n + m + r] = n+m$ and $[n + m] = n + m$. As $[x]$ and $[y]$ are integers:

Then $[x + y] = [[x]+ [y] + \{x\} + \{y\}] = [[x] + [y]] = [x] + [y]$.

Furthermore $[y] = [[y] + \{y\} + \{x\}]= [y + x - [x]]$ so $[x + y]=[y + x - [x]] + [x]$.

Case 2: $1 \le \{x\} + \{y\} < 2$

Then $[x + y] = [[x]+ [y] + \{x\} + \{y\}] = [[x] + [y]] + 1 = [x] + [y]+ 1$

And note $[y + x - [x]] = [y + \{x\}] = [[y] + \{y\} + \{x\}] = [[y] + 1] = [y]+1$.

So $[x +y] = [x]+[y] + 1 = [x] + [y + x - [x]]$.

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