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Prove that if $[.]$ is Greatest integer function then $$\left[\frac{[x]}{n}\right]=\left[\frac{x}{n}\right]$$ $\forall$ $n \in \mathbb{Z}$

My Try is i have used both LHS and RHS:

Case $1.$ if $x \in \mathbb{Z}$ proof is trivial

Case $2.$ if $x=q+f$ where $q \in \mathbb{Z}$ and $ f \in (0 \:\: 1)$

Then $$\left[\frac{x}{n}\right]=\left[\frac{q+f}{n}\right]$$

Now by euclid's algorithm for some positive integer $p$ we have $$q=np+r$$ where $0 \le r \lt n-1$

So $$\left[\frac{x}{n}\right]=\left[\frac{q+f}{n}\right]=\left[\frac{np+r+f}{n}\right]=p+\left[\frac{r+f}{n}\right]$$

Now since $$0 \le f \lt 1$$ and $$0 \le r \lt n-1$$ we have

$$ 0 \le r+f \lt n$$

So $$\left[\frac{r+f}{n}\right]=0$$

hence

$$\left[\frac{x}{n}\right]=p \tag{1}$$

Now

$$\left[\frac{[x]}{n}\right]=\left[\frac{[q+f]}{n}\right]=\left[\frac{q}{n}\right]=\left[\frac{np+r}{n}\right]=\left[p+\frac{r}{n}\right]=p+\left[\frac{r}{n}\right]$$

Now since

$$0 \le r \lt n-1$$ we have

$$\left[\frac{r}{n}\right]=0$$ So

$$\left[\frac{[x]}{n}\right]=p \tag{2}$$

From $1$ and $2$ we have the required proof.

is there a better way please share

Ekaveera Gouribhatla
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    A small issue: We don't necessarily have $0 \leqslant r < n-1$, it is possible that $r = n-1$. So $0 \leqslant r \leqslant n-1$. Still, $0 \leqslant r+f < n$ follows, so the argument works. – Daniel Fischer Jul 06 '17 at 11:59
  • A not-so-small issue: The proof only works for $n > 0$. For $n < 0$, the assertion need not hold. Consider $n = -3$ and $x = -2.9$. Then $$\biggl\lfloor \frac{x}{n}\biggr\rfloor = 0 < 1 = \biggl\lfloor \frac{\lfloor x\rfloor}{n}\biggr\rfloor.$$ – Daniel Fischer Jul 06 '17 at 12:14

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