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Math Help - cancellation

  1. #1
    Junior Member
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    Nov 2009
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    cancellation

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    Any ideas to show that in integers, xz = yz \Rightarrow x=y, without division algorithm and without the corresponding property in natural numbers? (Note: z is nonzero.)

    In my course we are doing one of those "ground-up" constructions, so we started with a few axioms (Peano?), defined addition and multiplication of natural numbers, put the equivalence relation (a,b)~(c,d) if a+d=b+d on \mathbb{N} \times \mathbb{N}, then defined addition and multiplication of the resulting equivalence classes in what I assume is the standard way of doing that here, but its late and I'm tired so I don't want to type anymore. Maybe I'll have time to edit in a few hours after a nap.

    My attempt dead-ends me. I compute xz and yz in terms of the equivalence relation definitions (so x, y, and z are equivalence classes as above). I need to show that because xz=yz, it must be true that a+d = b+c (where [(a,b)]=x and [(c,d)]=y). Best I've been able to do so far is a few lines of manipulation that almost seems just random.
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    Got it. Had to do a li'l lemma first, which is what I was trying to avoid but gave up and did it another way.
    Last edited by cribby; April 26th 2010 at 03:03 PM.
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