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I am working on an assignment and have to write a proof for the following:

$$\lfloor 4x \rfloor = \lfloor x \rfloor + \lfloor x+\frac{1}4 \rfloor + \lfloor x+\frac{1}2 \rfloor + \lfloor x+\frac{3}4 \rfloor$$

I have to do by proof by case I know the cases to be as follows

$$0 \le x-\lfloor x \rfloor \lt \frac 1 4$$ $$\frac 1 4 \le x-\lfloor x \rfloor \lt \frac 1 2$$ $$\frac 1 2 \le x-\lfloor x \rfloor \lt \frac 3 4$$ $$\frac 3 4 \le x-\lfloor x \rfloor \lt 1$$

The problem is I don't understand how I can express this.

Case 0
In this case I know that it would be $4n$ here

Case 1
In this case I know that it would be $4n+1$

Case 2
Here it would be $4n+2$

Case 3
Now the last case would end up being $4n+3$

Can someone help me express this?

wolfcall
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1 Answers1

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One way to tackle these sorts of proof is to give a name to the leftover bit: $\epsilon = x-\lfloor x \rfloor$. [I'm sure there is a slicker way to prove this, but here's a straight forward attempt...]

Then, for example, in case 1:

$$0 \le x-\lfloor x \rfloor = \epsilon \lt \frac 1 4$$ Clear the fraction... $$0 \le 4x-4\lfloor x \rfloor = 4\epsilon \lt 1$$ Then $$\lfloor 4x \rfloor = \lfloor 4\lfloor x \rfloor +4\epsilon \rfloor = 4\lfloor x \rfloor$$ since $4\lfloor x \rfloor$ is an integer and $0 \leq 4\epsilon \lt 1$. $$\lfloor x \rfloor + \lfloor x+\frac{1}4 \rfloor + \lfloor x+\frac{1}2 \rfloor + \lfloor x+\frac{3}4 \rfloor = 4\lfloor x \rfloor$$ since $\lfloor x \rfloor \leq x < x + \frac 1 4 < x + \frac 1 2 < x + \frac 3 4 = \lfloor x \rfloor + \epsilon + \frac 3 4$ where $0 \leq \epsilon + \frac 3 4 < \frac 1 4 + \frac 3 4 =1$ so all get rounded down to $\lfloor x \rfloor$.

I hope this helps get you started on the other cases! :)

Bill Cook
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