1. ## Greatest integer function

Hello, I am new here and was wondering if anyone can help me with this problem:

Prove that the greatest integer function has the following properties
A) [x+n]= [x]+n for every integer n.
B)

[-x]= -[x] if x is an integer
-[x]-1 otherwise.

The integer n is called the greatest integer of x and is denoted by [x].

2. Originally Posted by Csou090490
Hello, I am new here and was wondering if anyone can help me with this problem:

Prove that the greatest integer function has the following properties
A) [x+n]= [x]+n for every integer n.
The defintion of the greatest integer function is.

[x] is the integer such that.
x-1<[x]<=x.

It can be shown that this definition is well-defined. Meaning, the greatest integer always exists and is unique. If you wish I can show that to you.

Thus we claim that,
[x+n]=[x]+n.

Meaning,
x+n-1<[x+n]<=x+n.......(1)
And we can to show that,
x+n-1<[x]+n<=x+n.......(2)
Satisfies the same inequality.
Then by well-defineness we have that.
[x+n]=[x]+n

We need to show (2) because (1) is the definition (nothing to prove).

Note that,
x-1<[x]<=x
x-1+n<[x]+n<=x+n
And [x]+n is an integer because [x] and "n" are both are and integers are closed under addition.
Thus,
[x]+n=[x+n]

3. Here is another proof. Perhaps your definition is something like this: if x is a real then [x] is the integer such that [x]<=x<x+1 (BTW this can be proved to exist using completeness of the real numbers). If n is an integer then so is [x]+n. By definition [x+n]<=x+n, and thus we know [x]+n<=[x+n] but if [x]+n<[x+n]<=x+n then [x]<[x+n]-n<=x. But [x+n]-n is an integer so [x+n]=[x]+n

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# greatest integer function property

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