# Thread: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

1. ## Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

Prove: [x]+[x+1/n].......[x+n-1/n]=[nx]

I know it is equal to n[x].but by putting value of 5.25 ,n[x] is not coming equal to [nx]

2. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

Originally Posted by AaPa
Prove: [x]+[x+1/n].......[x+n-1/n]=[nx]

I know it is equal to n[x] How do you know that? It is wrong:

$\displaystyle n = 2\ and\ x = 3.8 \implies \lfloor 3.8 \rfloor + \lfloor 3.8 + \dfrac{2 - 1}{2} \rfloor = \lfloor 3.8 \rfloor + \lfloor 4.3 \rfloor = 3 + 4 = 7 \ne 6 = 2 * 3 = 2 * \lfloor 3.8 \rfloor.$

.but by putting value of 5.25 ,n[x] is not coming equal to [nx]

$x = y + z,\ where\ y\ is\ an\ integer\ and\ \ 0 \le z < 1, \implies \lfloor x \rfloor = y.$

Do you know how to do a proof by weak mathematical induction? That looks to me to be the simplest approach.

3. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

Originally Posted by AaPa
Prove: [x]+[x+1/n].......[x+n-1/n]=[nx]

I know it is equal to n[x]
Why do you think so?

The following is not an induction approach.

Suppose $x=a+y$ where $a\in\mathbb{Z}$ and $0\le y<1$, i.e., $\lfloor x\rfloor=a$. Also let $k=0,\dots,n-1$ be such that
$$y+k/n<1,\text{ but }y+(k+1)/n\ge1.\qquad(1)$$
This means that $\lfloor z\rfloor=a$ for $z=x,\dots,x+k/n$ (the first $k+1$ terms of the left-hand side), but $\lfloor z\rfloor=a+1$ for $z=x+(k+1)/n,\dots,x+(n-1)/n$ (the last $n-k-1$ terms of the left-hand side). Thus, the left-hand side is $(k+1)a+(n-k-1)(a+1)= na+n-k-1$. Now derive lower and upper bounds on $ny$ from (1) and use it to find $\lfloor nx\rfloor$.

4. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

How did you say that that y+ k/n <1 ?

5. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

Originally Posted by AaPa
How did you say that that y+ k/n <1 ?
How? I typed it.

Are you asking why such nonnegative integer k exists? Take k = 0. Then $y+k/n=y<1$ by assumption about y. If already $y+1/n\ge1$, then the k that I am talking about is 0 because $y+k/n<1$, but $y+(k+1)/n\ge1$. If $y+1/n$ is still less than 1, then consider $y+2/n$ and so on. Eventually you will find such k because you start when the number (i.e., y) is < 1, and as you keep adding 1/n to it, eventually the sum will exceed 1. The idea is to denote by k + 1 the number of terms from the original sum that are rounded to $\lfloor x\rfloor$ (the rest n - k - 1 are rounded to $\lfloor x\rfloor+1$).

6. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

Hi,
I found this to be an interesting equation. So here's my solution:

7. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

Sorry for late reply. Wow. Thank you. The last part I solved on my own. Thanks again.:-)

8. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

In the first part of your answer I don't understand the last step where you put [nx] as upper limit.

9. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

I proved for $1\leq i\leq n$ that $\lfloor x+{n-i\over n}\rfloor$ is either 0 or 1 and is 1 precisely for $1\leq i\leq\lfloor nx\rfloor$. So

$$\sum_{i=1}^n\lfloor x+{n-i\over n}\rfloor=1+1+\cdots+1+0+0+\cdots+0$$

There are precisely $\lfloor nx\rfloor$ 1's in the above sum.

10. ## Re: Prove: [x]+[x+1/n].......[x+n-1/n]=[nx] where[x] is greatest integer function

Wow. Thanks a lot. I understand it completely. thank you thank you.

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# [x] [x 1/n] [x 2/n] ...... [x (n-1)/n] =[nx] where [.] is Greatest integer function

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