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
• $n+\frac23\le x<n+1$.

Answer 2

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 $x$.

Answer 3

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 $x \in [0,\frac{1}{3})$.