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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!

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

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

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

or

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

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

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  • $\begingroup$ 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

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