Floor function equality

So the problem is the following

"Does the equality $\lfloor$nx$\rfloor$=n$\lfloor$x$\rfloor$ hold for integers n$\le$2"

I have been trying to look at the Hermite's identity to see if that could help but I couldn't get anywhere with that and I'm confused at to what approach I could use to simplify the problem. I know that you can for example use a=$\lfloor$x$\rfloor$ and use the doubleinequality a$\le$x$\lt$a+1 to argue different cases. Could someone help to shed some light at to what of the basics I'm missing to solve this problem?

• Did you mean $n\ge 2$? – Mike Earnest Aug 1 '17 at 14:26
• Do you have any restriction on $x$? If not the equality does not hold, take $n = 2$ and $x = 1/2$ for instance. – Mariuslp Aug 1 '17 at 14:29

We can rewrite $x=\lfloor x \rfloor + y$ where $0 \leq y < 1$.
The equation becomes $$\lfloor n \lfloor x \rfloor + ny \rfloor = n \lfloor x \rfloor$$
$$n \lfloor x \rfloor + \lfloor ny \rfloor = n \lfloor x \rfloor$$
So we see that the equality holds iff $\lfloor ny \rfloor = 0$. However, this is not always the case - consider $y = \frac{1}{2}, n=2$.