# Proving a floor sum.

Given that $$a,b\in\Bbb{Z}^+$$ and $$\gcd(a,b)=1$$.

Prove the following:

$\left\lfloor\frac{a}{b}\right\rfloor+\left\lfloor2\cdot\frac{a}{b}\right\rfloor+\left\lfloor3\cdot\frac{a}{b}\right\rfloor+\ldots+\left\lfloor(b-1)\cdot\frac{a}{b}\right\rfloor=\frac{(a-1)(b-1)}{2}$

Clarifications:

• $$\lfloor\cdot\rfloor$$ denotes the floor function (greatest integer function).
• $$\gcd(x,y)$$ denotes the greatest common divisor of $$x$$ and $$y$$.

Hint:

It'd be helpful to prove this first before attempting to prove the main problem.

$\forall~a,b\in\Bbb{Z}^+\mid\gcd(a,b)=1\implies n\frac{a}{b}\not\in\Bbb{Z}~~\forall~~0\lt n\lt b~\land~n\in\Bbb{Z}$

Note by Prasun Biswas
3 years, 5 months ago

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Use the fact that $$\lfloor x\rfloor=x-\{x\}$$ to rewrite as

$\sum_{k=1}^{b-1}\left\lfloor\dfrac{ka}{b}\right\rfloor=\dfrac a b\sum_{j=1}^{b-1} j-\sum_{k=1}^{b-1}\left\{\dfrac{ka}{b}\right\}=\dfrac{a(b-1)}{2}-\sum_{k=1}^{b-1}\left\{\dfrac{ka}{b}\right\}.$

Now

$\sum_{k=1}^{b-1}\left\{\dfrac{ka}{b}\right\}=\sum_{k=1}^{(b-1)/2}\left(\left\{\dfrac{ka}{b}\right\}+\left\{\dfrac{(b-k)a}{b}\right\}\right)=\sum_{k=1}^{(b-1)/2}\left(\left\{\dfrac{ka}{b}\right\}+\left\{a-\dfrac{ka}{b}\right\}\right).$

Here no term can be zero since $$na/b$$ is not integer for $$1\le n<b$$ provided that $$(a,b)=1$$. So $$\forall k\in[1, (b-1)/2]$$

$\left\{\dfrac{ka}{b}\right\}+\left\{a-\dfrac{ka}{b}\right\}=1\implies \sum_{k=1}^{(b-1)/2}\left(\left\{\dfrac{ka}{b}\right\}+\left\{a-\dfrac{ka}{b}\right\}\right)=\sum_{k=1}^{(b-1)/2}1=\dfrac{b-1}{2}$

and we're, um, done.

- 3 years, 5 months ago

Consider $$b=8$$ and $$a=9$$. We have $$\gcd(a,b)=1$$ but $$\frac{(b-1)}{2}$$ is not a positive integer $$\geq 1$$. Then, what does the sum $$\displaystyle\sum_{k=1}^{(b-1)/2}(1)$$ mean?

- 3 years, 5 months ago

The sum is to $$b-1$$, as in $$\lfloor (b-1) \frac{ a}{b} \rfloor$$, and not to $$\frac{ b-1}{2}$$ as you had it.

Staff - 3 years, 5 months ago

Very well done....

- 3 years, 5 months ago

Warning: there is a huge hint below.

$$na (\mod b)$$ is pairwise distinct for $$n=1, 2, ..., b$$. Proof: The opposite is false by Pigeonhole Principle. Hence if we limit $$n$$ to $$1, 2, 3, ..., b-1$$ the possible remainders are $$1, 2, ..., b-1$$, once each.

From here we can find the sum of the $$b-1$$ fractional parts of $$n•\frac {a}{b}, n=1, 2, ..., b-1$$. We know their sum, so we know the sum of their floors.

- 3 years, 5 months ago

There's an easier approach (though essentially based on the same idea).