Identity or an equation?

Question

One of my friends send me an equation $$\lfloor x\rfloor-\left\lfloor \frac x3\right\rfloor=\left\lfloor\frac{\lfloor x\rfloor+1}{3}\right\rfloor+\left\lfloor\frac{\lfloor x\rfloor+2}{3}\right\rfloor$$ I put it some different numbers, and it works fine like an identity. I don't know if this is an identity. If it is an identity, how can start to prove it?

Answer 1

I think there is an identity like this $$\lfloor x\rfloor=\lfloor\frac{\lfloor x\rfloor+0}{3}\rfloor+\lfloor\frac{\lfloor x\rfloor+1}{3}\rfloor+\lfloor\frac{\lfloor x\rfloor+2}{3}\rfloor$$ if its true !? easy proof for $\lfloor x\rfloor-\lfloor \frac x3\rfloor=\lfloor\frac{\lfloor x\rfloor+1}{3}\rfloor+\lfloor\frac{\lfloor x\rfloor+2}{3}\rfloor$ is $$\left(\lfloor\frac{\lfloor x\rfloor+0}{3}\rfloor+\lfloor\frac{\lfloor x\rfloor+1}{3}\rfloor+\lfloor\frac{\lfloor x\rfloor+2}{3}\rfloor\right)-\lfloor \frac x3\rfloor=\lfloor\frac{\lfloor x\rfloor+1}{3}\rfloor+\lfloor\frac{\lfloor x\rfloor+2}{3}\rfloor\\$$

Answer 2

Evidently it is sufficient to consider 3 cases 1) $0\le x<1,$ 2) $1\le x<2$ and 3) $2\le x<3.$

Verification is trivial so this is an identity.

Answer 3

If suffices to show that the identity holds for integer $x$.

$$n=\left\lfloor\frac{n}3\right\rfloor+\left\lfloor\frac{n+1}3\right\rfloor+\left\lfloor\frac{n+2}3\right\rfloor.$$

The property is true for $n=0,1,2$ hence for $3k+0,3k+1$ and $3k+2$.