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I would like to know if the following statement about x,y and z is true:

$$x=\lfloor\frac{y}{z}\rfloor \iff z=\lfloor\frac{y}{x}\rfloor$$

I think it is true but am having a hard time wrapping my head around it.

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    $\begingroup$ Try $\,y=1,z=2\,$ for example. $\endgroup$ – dxiv Oct 2 '18 at 23:47
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    $\begingroup$ Let $x=1, y=3, z = 2$. $\endgroup$ – Mark Oct 2 '18 at 23:48
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    $\begingroup$ Try any example where $z$ is not an integer. $\endgroup$ – Gerry Myerson Oct 2 '18 at 23:53
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If $x = [\frac yz]$ then $x$ is an integer and

So $x \le \frac yz < x+1$ =

So $xz \le y < xz + z$ (assuming $z > 0$)

The means $z \le \frac yx < z + \frac zx$ (assuming that $x > 0$).

But there is no reason that $z$ must be an integer.

Suppose $y = 3$ and $z = \frac 34$ then $\frac yz = \frac 3{\frac 34} = 4$.

And let $x = 4$ so we have $x = [\frac yz]$ as $4 = [\frac 3{\frac 34}]$.

Now $\frac yx = \frac 34$ and $[\frac yx] =[\frac 34] = 0$.

And $y = \frac 34 \ne 0$.

So no, this is not true.

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Perhaps a more technical way:

Let $\frac yz = x + r$ where $x$ is an integer and $0 \le r < 1$. Then $x = [\frac yz]$.

If $x \ne 0$ then $\frac yx = z + r\frac zx$

Now at this point we have no reason to assume either $z$ is an integer, or that $0\le r\frac zx< 1$.

We can easily come up with counter examples. An example where $z$ isn't an integer; or an example where $r\frac zx \ge 1$ or $r\frac zx < 0$.

For example: if $r = \frac 12$ and $\frac zx = 2$ and so $\frac yz =\frac y{2x} = x + \frac 12$ or $y=2x^2 + 1$. Say, $x = 1$ and $z =2$ and $y = 3$.

Then $x =1 = [\frac 32]=[\frac yz]$ but $z = 2$ and $[\frac yx ] =[\frac 32] = 1$.

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