Proving a floor sum.

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

Prove the following:



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


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

 a,bZ+gcd(a,b)=1    nab∉Z    0<n<b  nZ\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
4 years, 6 months ago

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

k=1b1kab=abj=1b1jk=1b1{kab}=a(b1)2k=1b1{kab}.\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\}.



Here no term can be zero since na/bna/b is not integer for 1n<b1\le n<b provided that (a,b)=1(a,b)=1. So k[1,(b1)/2]\forall k\in[1, (b-1)/2]

{kab}+{akab}=1    k=1(b1)/2({kab}+{akab})=k=1(b1)/21=b12\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.

Jubayer Nirjhor - 4 years, 6 months ago

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Very well done....

Vighnesh Raut - 4 years, 6 months ago

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Consider b=8b=8 and a=9a=9. We have gcd(a,b)=1\gcd(a,b)=1 but (b1)2\frac{(b-1)}{2} is not a positive integer 1\geq 1. Then, what does the sum k=1(b1)/2(1)\displaystyle\sum_{k=1}^{(b-1)/2}(1) mean?

Prasun Biswas - 4 years, 6 months ago

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The sum is to b1 b-1, as in (b1)ab \lfloor (b-1) \frac{ a}{b} \rfloor , and not to b12 \frac{ b-1}{2} as you had it.

Calvin Lin Staff - 4 years, 6 months ago

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Warning: there is a huge hint below.

na(modb)na (\mod b) is pairwise distinct for n=1,2,...,bn=1, 2, ..., b. Proof: The opposite is false by Pigeonhole Principle. Hence if we limit nn to 1,2,3,...,b11, 2, 3, ..., b-1 the possible remainders are 1,2,...,b11, 2, ..., b-1, once each.

From here we can find the sum of the b1b-1 fractional parts of nab,n=1,2,...,b1n•\frac {a}{b}, n=1, 2, ..., b-1. We know their sum, so we know the sum of their floors.

Joel Tan - 4 years, 6 months ago

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

Hint: Heads and tails.

Calvin Lin Staff - 4 years, 6 months ago

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