I have little to no experience working with floor inequalities so I am kind of stuck on this one. It seems pretty intuitive though.

So basically I want to show that

$$\left\lfloor\frac{a}{b}\right\rfloor \geq \left\lfloor\frac{a}{b+1}\right\rfloor$$ where $a$ and $b$ are strictly positive integers.

This is true for $b \geq a$, so I'm looking at the $b < a$ case but have no idea where to start. Primarily the denominators are different and I have no idea how to get it out of the floor function. Thanks!

  • $\begingroup$ Ok thanks guys. I figured out a formal proof by contradiction. Since we know $\frac{a}{b} > \frac{a}{b+1}$, assume $\lfloor\frac{a}{b}\rfloor < \lfloor\frac{a}{b+1}\rfloor$. Then $\lfloor\frac{a}{b}\rfloor < \frac{a}{b+1}$ which is a contradiction with $\frac{a}{b} > \frac{a}{b+1}$. $\endgroup$ – hirantal91 Jul 31 '14 at 9:13

I think it's obvious because $$\frac{a}{b}\gt\frac{a}{b+1}\Rightarrow\left\lfloor\frac ab\right\rfloor\ge\left\lfloor\frac {a}{b+1}\right\rfloor.$$

  • $\begingroup$ So I get that it's obvious that $\frac{a}{b} > \frac{a}{b+1} \rightarrow \frac{a}{b} \geq \lfloor\frac{a}{b+1}\rfloor$. But how did you put a floor on the left side? The implication isn't clear. $\endgroup$ – hirantal91 Jul 31 '14 at 9:07
  • $\begingroup$ @onionguy4: First of all, you can see that $a/b\gt a/(b+1)$ holds for any $a\gt 0,b\gt 0$, can't you? Then, consider the meaning of $\lfloor x\rfloor$. $\endgroup$ – mathlove Jul 31 '14 at 9:10
  • $\begingroup$ Thanks! I know it makes sense when you think about it but I can't seem to express it formally besides my contradiction method. $\endgroup$ – hirantal91 Jul 31 '14 at 9:20
  • $\begingroup$ @onionguy4: You are welcome. Since $\lfloor x\rfloor$ is the largest integer not greater than $x$, it's obvious as I wrote. Please note that $\gt$ and $\ge$. $\endgroup$ – mathlove Jul 31 '14 at 9:24

For $x,y \in \mathbb{R}^{2}$ such that $x \leq y$, $\lfloor x \rfloor \leq \lfloor y \rfloor$.

  • $\begingroup$ They aren't integers though $\endgroup$ – hirantal91 Jul 31 '14 at 9:06
  • $\begingroup$ @onionguy4 : You're right, that's a typo ! It should be $\mathbb{R}$ instead of $\mathbb{Z}$. I will update my answer ! Thanks !! $\endgroup$ – jibounet Jul 31 '14 at 9:16

Note that

$$\frac{a}{b} > \frac{a}{b+1}$$

however when using the floor function we round to the nearest whole integer below so therefore there are two cases:

  1. $$\lfloor{\frac{a}{b+1}}\rfloor = \lfloor \frac{a}{b} \rfloor - 1 $$


$$\lfloor{\frac{a}{b+1}}\rfloor = \lfloor \frac{a}{b} \rfloor $$

A more formal answer is that the floor function is monotonically increasing and therefore preserves ordering between fractions up to an equality sign

  • $\begingroup$ 1. isn't true. For example, a = 10,b = 2. The left side is 3 while the right side is 5. But I kind of see the logic. Will think through it a bit. $\endgroup$ – hirantal91 Jul 31 '14 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.