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Math Help - Distance - absolute value proofs

  1. #1
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    Distance - absolute value proofs

    Hey all, how would one go about proving the following?

    abs|x - z| <= abs|x - y| + abs|y - z|

    abs|x - y| >= abs| abs|x| - abs|y| |


    where we know the following,

    N (naturals), considered as a subset of R (reals), is not bounded above.
    Z (integers) is not bounded above.
    Z (integers) is not bounded below.
    For each epsilon > 0, there exists n in N (naturals) such that 1/n < epsilon

    Absolute value is defined to be x if x >= 0 and to be -x if x < 0.

    Also, x<y if and only if x^2 < y^2
    abs|x| < abs|y| if and only if x^2 < y^2

    For all x,y in R (reals):

    abs|x| = 0 if and only if x = 0
    abs|xy| = abs|x| * abs|y|
    - abs|x| <= x <= abs|x|
    abs|x+y| <= abs|x| + abs|y|
    If -y < x < y then |x| < |y|

    Also, if you let x be real such that 0 <= x <= 1, and let m,n be natural such that m >= n, then x^m <= x^n

    Thank you!
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    First I would prove the triangle inequality for two elements

    |a+b| \le |a|+|b|

    By the absolute value property we have the

    -|a|\le a \le |a| and -|b|< b < |b|
    add these two equations together to get

    -(|a|+|b|) \le a+b \le (|a|+|b|)
    Now by definition of the absolute value again you get

    |a+b| < |a|+|b|

    Now here is a hint for the 2nd part

    |x-z|=|x-y+y+(-z)|
    Can you finish from here?
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