I have the following to prove:

$$\lfloor 3x\rfloor = \lfloor x\rfloor + \left\lfloor x+\frac 13 \right\rfloor + \left\lfloor x+\frac 23 \right\rfloor $$ The definition of a floor function is: $ \lfloor x \rfloor = n \le x \lt n+1 $

So my first instinct was to do $ \lfloor 3x\rfloor=3 \lfloor x\rfloor $ and then let $n=\lfloor x\rfloor$ so basically we get $3n$. But if I were to replace both sides of the equation with $n=\lfloor x\rfloor$ I get : $$3n = n + \left\lfloor x + \frac 13 \right\rfloor + \left\lfloor x+\frac 23 \right\rfloor =3n+\frac 33=3n+1$$

But I don't know what to do with this and I'm not sure if this is a formal way of doing these types of proofs.

  • $\begingroup$ You probably just want to handle this by cases. You can't say that $\lfloor{3x}\rfloor=3\lfloor{x}\rfloor$, as that's not true (try $x=1/2$, say). $\endgroup$
    – mjqxxxx
    Commented Nov 18, 2013 at 18:21
  • $\begingroup$ en.wikipedia.org/wiki/Hermite%27s_identity $\endgroup$ Commented Nov 18, 2013 at 18:24

3 Answers 3


It’s not generally true that $\lfloor 3x\rfloor=3\lfloor x\rfloor$; try $x=\frac13$, for example. One very straightforward approach is to let $n=\lfloor x\rfloor$, so that $n\le x<n+1$, and consider three cases:

  • $n\le x<n+\frac13$;
  • $n+\frac13\le x<n+\frac23$; and
  • $n+\frac23\le x<n+1$.
  • $\begingroup$ I'm not sure where you got the three cases from. How were they derived from the definition? $\endgroup$
    – Dimitri
    Commented Nov 18, 2013 at 18:31
  • 2
    $\begingroup$ @Dimitri: By considering how $\lfloor 3x\rfloor$ is related to $\lfloor x\rfloor$. As $x$ moves up from $n$, $\lfloor 3x\rfloor$ takes a sudden jump at $x=n+\frac13$ and again at $x=n+\frac23$. $\endgroup$ Commented Nov 18, 2013 at 18:34
  • $\begingroup$ I see now. In this case, how would one begin the proof. I'm understanding what is happening, but don't how to show it. $\endgroup$
    – Dimitri
    Commented Nov 18, 2013 at 19:39
  • 1
    $\begingroup$ @Dimitri: Take the middle case: $3n+1\le 3x<3n+2$, so $\lfloor 3x\rfloor=3n+1$. Now calculate $\lfloor x\rfloor$, $\left\lfloor x+\frac13\right\rfloor$, and $\left\lfloor x+\frac23\right\rfloor$ in terms of $n$ and verify that you get $3n+1$. The other two cases are done similarly. $\endgroup$ Commented Nov 18, 2013 at 19:42

Both sides are $0$ at $x=0$ (can you check this?), they are constant on every interval $[n/3,(n+1)/3)$ (can you check this?) and they jump by $+1$ when $x$ increases through a point $n/3$ with $n$ an integer (can you check this?). Hence they coincide for every $x$.


let $l(x) = \lfloor 3x\rfloor$, $r(x) = \lfloor x\rfloor + \left\lfloor x+\frac 13 \right\rfloor + \left\lfloor x+\frac 23 \right\rfloor$.

Note that if $n \in \mathbb{Z}$, we have $l(x+ n \frac{1}{3}) = l(x)+n$ and $r(x+ n \frac{1}{3}) = r(x)+n$.

Hence we need only show $l(x) = r(x)$ for $x \in [0,\frac{1}{3})$.

It is easy to see that $l(x) = 0, r(x) = 0$ for $x \in [0,\frac{1}{3})$.


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