# is this statement about x, y, and z true?

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.

• Try $\,y=1,z=2\,$ for example. – dxiv Oct 2 '18 at 23:47
• Let $x=1, y=3, z = 2$. – Mark Oct 2 '18 at 23:48
• Try any example where $z$ is not an integer. – Gerry Myerson Oct 2 '18 at 23:53

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 $$[\frac yx ] =[\frac 32] = 1$$.