How does one terminate a real number from a floorfunction expression?

Question

Hey I'm struggling to see as to what simplification that would be proper to use in the following problem.

"Show that for all real x the following applies

$$\lfloor x-\frac13 \rfloor+1+\lfloor x+\frac13\rfloor=\lfloor 3x \rfloor-\lfloor x\rfloor$$

In the textbook the only "tips" I've gotten is that in some problems it's is advised to use a=$\lfloor$x$\rfloor$ and use the double inequality $a\le x\lt a+1$ to argue different cases but that doesn't really help me here. What's really bugging me is how I can "remove" the 1 in the left expression. Any ideas or thoughts? :-)

Answer 1

I would advise you to separate three cases:

• $x\in \left[a, a+\frac13\right)$
• $x\in \left[a+\frac13, a+\frac23\right)$
• $x\in \left[a+\frac23, a+1\right)$

Then write $x=a+\delta$ where $\delta\in (0,1)$

That way, in the first case, you know that $\delta\in (0,\frac13)$ which means that $\lfloor 3x\rfloor = \lfloor 3a + 3\delta\rfloor = 3a$. You also know that $\lfloor x + \frac 13\rfloor = a$ and $\lfloor x-\frac13\rfloor = a-1$, so the inequality becomes easy to prove.