[Math] Floor and ceiling function proof

Question

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.

Answer 1

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
• $n+\frac23\le x<n+1$.