Proof about floor function: $\lfloor x\rfloor+\lfloor x + \frac{1}{n} \rfloor+\cdots + \lfloor x + \frac{(n-1)}{n} \rfloor = \lfloor nx \rfloor$

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Denoting the left hand side with $f(x)$, the right hand side with $g(x)$, show that $f(x+\frac1n)=f(x)+1$ and $g(x+\frac1n)=g(x)+1$. Also show $f(x)=0=g(x)$ for $0\le x<\frac 1n$. Then show by backward and forward induction on $k\in \Bbb Z$ that $f(x)=g(x)$ for all $x$ with $\frac kn\le x<\frac{k+1}n$.

Alternatively, write $x=k+y$ with $k=[x]\in\Bbb Z$ and $y\in[0,1)$, then write $ny=m+z$ with $m=[ny]\in\{0,1,\ldots,n-1\}$ and $z\in[0,1)$ and see what happens if you plug in $x=k+\frac1n m+\frac1nz$.

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Updated on August 17, 2020

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  • Matt
    Matt about 3 years

    How can we prove that $$\left\lfloor x\right\rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \cdots + \left \lfloor x + \frac{(n-1)}{n} \right\rfloor = \left\lfloor nx \right \rfloor$$ Where $\lfloor x\rfloor$ denotes greatest integer less than or equal to $x$. I was able to prove it when $n = 2$ or $3$. Please see this link.

    But I can't prove it generally. I tried it using Principle of mathematical Induction but couldn't. Can someone prove it using some other way using properties of greatest integer function?

    • Martin Sleziak
      Martin Sleziak about 7 years
      This is the same question as math.stackexchange.com/questions/5650/…
    • Matt
      Matt about 7 years
      The answers given there are really hard for me to understand as they are not completely elaborated. Hence I reasked it.
    • Martin Sleziak
      Martin Sleziak about 7 years
      Raghav Singal: That does not change the fact that you have asked a duplicate question. Moreover, you now say that you have seen the linked question and you did not mention that in your post at all. The right thing to do in your situation would be clearly explain which part of older answers you do not understand and ask about that part, see meta. Or perhaps ask in chat.
    • Martin Sleziak
      Martin Sleziak about 7 years
      BTW here is one more post about the same problem: math.stackexchange.com/questions/403879/…
    • Matt
      Matt about 7 years
      Thanks for the help, this answer was really simple and elaborate, Now I get it.
  • Matt
    Matt about 7 years
    How is $f(x)=0$ for $0\le x<\frac 1n$?
  • Matt
    Matt about 7 years
    I also can't understand how to use backward and forward induction and I am not sure even if I know what it is. Can you please give a simpler answer?