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Math Help - Real Analysis: proving the triangle inequalities

  1. #1
    Newbie mjjoga's Avatar
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    Real Analysis: proving the triangle inequalities

    I am trying to prove one of the triangle inequalities.

    [] are going to be my absolute value symbols, sorry, i don't have mathematica on this computer with the symbols.

    I am proving that [[x]-[y]]<=[x-y].

    i tried this: we know -[x]<=x<=[x] and -[y]<=y<=[y]
    if i subtract these i get: -[x]--[y]<=x-y<=[x]-[y]
    then you simplify and get -([x]-[y])<=x-y<=([x]-[y])
    so, [x-y]<=[[x]-[y]].
    I thought i was on the right track but then i realized that the inequality was pointing in the wrong direction. Did i make some computational error or am i going about this the wrong way? I appreciate any hints on where I might start. thanks!
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  2. #2
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    -|x|<=x<=|x|
    -|y|<=-y<=|y|
    deduce by adding: |x-y|<=|x|+|y|
    |x-y|-|y|<=|x|
    insert x=y-z and you get:
    |z|-|y|<=|y-z|=|z-y|
    Q.E.D
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    |x| = |x - y + y| \le |x - y| + |y|

    \implies |x - y| \ge |x| - |y| ...........(1)


    Also, |y| = |x - (x - y)| \le |x| + |x - y|

    \implies |x - y| \ge |y| - |x| ............(2)


    By (1) and (2) we see that

    -|x - y| \le |x| - |y| \le |x - y| and the conclusion follows (by the fact that -b \le a \le b \implies |a| \le b)
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