I have to prove that


For any positive integers $x, y$.

Can anyone help me?


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  • $\begingroup$ Have you checked a few cases? Seen what happens? $\endgroup$ – Arthur Sep 7 '16 at 7:17
  • $\begingroup$ The statement is false for $x=1,y=\frac12$ $\endgroup$ – 5xum Sep 7 '16 at 7:19
  • $\begingroup$ Do you have to prove this for integer $x, y$? $\endgroup$ – 6005 Sep 7 '16 at 7:22
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    $\begingroup$ You probably want to specify that $x,y\in\mathbb{Z}^{+}$. $\endgroup$ – barak manos Sep 7 '16 at 7:22
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    $\begingroup$ @6005: Not true for all $x,y\in\mathbb{Z}^{-}$ (i.e., false for some $x,y\in\mathbb{Z}^{-}$). $\endgroup$ – barak manos Sep 7 '16 at 7:24

For this to be true, you need to specify that $\boldsymbol{x,y \in \mathbb{Z}}$ and $\boldsymbol{y > 0}$.

Then: if $$ \left\lceil \frac{x}{y} \right\rceil = m $$ then $$ m - 1 < \frac{x}{y} \le m $$ which means $$ my - y < x \le my. $$ which we can rewrite $$ my -y \le x - 1 < my $$ since $x$ is an integer and both bounds are integers.

Can you finish? You need to reach the conclusion that $$ \left\lfloor \frac{x-1}{y} \right\rfloor = m - 1. $$

  • $\begingroup$ oh i see, sorry $\endgroup$ – JonMark Perry Sep 7 '16 at 7:54

As @6005 clearly said, you need to agree that the $x,y$ are positive integers with the constraint $y \gt 0$.

This is a hint regarding the right side of the inequality:

$$\left\lceil\frac{x}y\right\rceil=\left\lfloor\frac{x-1}y\right\rfloor+1=\left\lfloor\frac{x-1}y+1\right\rfloor=\left\lfloor\frac{x-1+y}y\right\rfloor=\left\lfloor\frac{x}{y} + \frac{y-1}y\right\rfloor=\left\lfloor\frac{x}{y} + 1 - \frac{1}y\right\rfloor$$

  • $\begingroup$ The statement $\lceil x/y \rceil = \lfloor x/ y + 1\rfloor$ is only true if $x/y$ is not an integer. $\endgroup$ – 6005 Sep 7 '16 at 7:47
  • $\begingroup$ Your modification did not fix it :) $\endgroup$ – 6005 Sep 7 '16 at 7:49
  • $\begingroup$ @6005 you were right, thank you for the feedback! I have removed that part of the explanation, I am keeping the part that can help to study what happens with the right side term. $\endgroup$ – iadvd Sep 7 '16 at 7:54
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    $\begingroup$ Well you were almost there before, if you just split into cases based on whether or not $\frac{x}{y}$ was an integer. $\endgroup$ – 6005 Sep 7 '16 at 7:55
  • $\begingroup$ @6005 good hint for the OP, thank you again! $\endgroup$ – iadvd Sep 7 '16 at 7:56

Let $x=py+q$, with $0\le q<y$.

Then if $q=0$, $$\left\lceil\frac{x}y\right\rceil=p,\left\lfloor\frac{x-1}y\right\rfloor+1=p+\left\lfloor\frac{y-1}y\right\rfloor+1=p.$$ and if $q>0$,



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