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Math Help - Greatest integer function

  1. #1
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    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].

    Thank you, in advance
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  2. #2
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    Quote Originally Posted by Csou090490 View Post
    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
    Add "n" to all sides,
    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]
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  3. #3
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    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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