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Math Help - Constructing Z from N (integers from naturals)

  1. #1
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    Question Constructing Z from N (integers from naturals)

    Having some trouble understanding this...

    This method uses an equivalence relation defined on NxN such that: (a,b)~(c,d) iff a+d=c+b

    Note that we can subtract any k in N from both a and b to get an equivalent pair, providied k < min(a,b) (makes sense)

    It follows that every class [a,b] contains a special pair, in which atleast one of the components is equal to 1. Moreover there is only one such pair in each class:

    (1,b)=(1,d) => b=d
    (a,1)=(b,1) => a=b
    (a,1)=(1,b) => a=b=1

    What's going on here? Shouldn't the = signs be ~ signs?

    The special pairs thus form a transversal,

    T={(1,n+1): n in N} union {(1,1)} union {(n+1,1): n in N}

    Not sure what's going on here either???
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  2. #2
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    Can you possibility give us a reference on this?
    Is there a textbook or lecture notes?
    Frankly, I do not follow the construction.

    PS
    I have given thought to your posting. I had fallen to the sin of familiarity.
    I know a similar approach: (a,b)~(c,d) iff (a-b)=(c-d).
    But of course that is equivalent to the relation you posted.
    Have you shown that that is an equivalence relation on NxN?

    The equivalence class for [3,1] contains (4,2),(5,3),(9,7) etc.
    The equivalence class for [1,3] contains (2,4),(3,5),(7,9) etc.
    The equivalence class for [0,0] contains (2,2),(3,3),(7,7) etc.
    Thus we can identify n with [a,b] iff a-b=n.
    Thus we can identify -n with [b,a] iff a-b=n.
    Thus we can identify 0_z with [0,0].

    We define +_z as (a,b)+_z (c,d)=(a+c,b+d).

    Can you carry on from here?
    Last edited by Plato; June 1st 2007 at 03:38 PM.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Can you possibility give us a reference on this?
    Is there a textbook or lecture notes?
    Frankly, I do not follow the construction.

    PS
    I have given thought to your posting. I had fallen to the sin of familiarity.
    I know a similar approach: (a,b)~(c,d) iff (a-b)=(c-d).
    But of course that is equivalent to the relation you posted.
    Have you shown that that is an equivalence relation on NxN?

    The equivalence class for [3,1] contains (4,2),(5,3),(9,7) etc.
    The equivalence class for [1,3] contains (2,4),(3,5),(7,9) etc.
    The equivalence class for [0,0] contains (2,2),(3,3),(7,7) etc.
    Thus we can identify n with [a,b] iff a-b=n.
    Thus we can identify -n with [b,a] iff a-b=n.
    Thus we can identify 0_z with [0,0].

    We define +_z as (a,b)+_z (c,d)=(a+c,b+d).

    Can you carry on from here?
    Apologies, I think i'm at fault here. I only posted the very beginning of the construction. I kind of assumed it was a standard approach and other people would be familiar with it. I'll post it in its entirety now since I don't really follow the last 3 lines of your explanation. The construction I have here seems a lot more thorough but i wonder if it is essentially the same as yours?
    By the way the textbook i'm quoting from is 'Elements of Logic via Numbers and Sets' by D L Johnson. I could link you to some handwritten lecture notes online but i don't think they'd be very illuminating.


    This method uses an equivalence relation defined on NxN such that: (a,b)~(c,d) iff a+d=c+b

    Note that we can subtract any k in N from both a and b to get an equivalent pair, providied k < min(a,b)

    It follows that every class [a,b] contains a special pair, in which atleast one of the components is equal to 1. Moreover there is only one such pair in each class:

    (1,b)=(1,d) => b=d
    (a,1)=(b,1) => a=b
    (a,1)=(1,b) => a=b=1


    The special pairs thus form a transversal,

    T={(1,n+1): n in N} union {(1,1)} union {(n+1,1): n in N}

    Next, attempt to define the addition of classes in a natural way:

    [a,b] + [c,d] = [a+c,b+d]

    We need to check for independence of representatives in order that this operation be well defined. The effect of this operation on the members of T works out as follows:

    for all (a,b) in NxN, [a,b] + [1,1] = [a+1,b+1] = [a,b]

    so that [1,1] is an identity for this +. Next,

    [m+1,1] + [n+1,1] = [m+n+2,2] = [(m+n)+1,1]
    [1,m+1] + [1,n+1] = [2,m+n+2] = [1,(m+n)+1]

    so that the first and last sets of T both look like copies on N

    Finally,

    [n+1,1] + [1,n+1] = [n+2,n+2] = [1,1]

    so that [1, n+1] is the additive inverse of [n+1,1]


    Hope this makes sense. It seems to have similarities to yours. I understand the intention of the construction and what it arrives at seems to make sense, I think it's mainly just the initial steps i'm confused about.
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  4. #4
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    Quote Originally Posted by Obstacle1 View Post
    It seems to have similarities to yours. I understand the intention of the construction and what it arrives at seems to make sense, I think it's mainly just the initial steps i'm confused about.
    In fact, the approaches are identical. I think that your textbook uses a slightly different definition for the set of natural numbers. That is, I am accustomed to using N=\{0,1,2,3,\cdots\} whereas your textbook probability does not include 0 in N. So if you have a particular question, I may be able to answer it.
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  5. #5
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    Quote Originally Posted by Plato View Post
    In fact, the approaches are identical. I think that your textbook uses a slightly different definition for the set of natural numbers. That is, I am accustomed to using N=\{0,1,2,3,\cdots\} whereas your textbook probability does not include 0 in N. So if you have a particular question, I may be able to answer it.

    I think i've resolved all my problems now. One thing, what does a copy of N mean exactly?
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  6. #6
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    Quote Originally Posted by Obstacle1 View Post
    One thing, what does a copy of N mean exactly?
    You do know that we must show that addition is well defined?
    Also we must show that there is a well defined linear ordering.
    As well as a well defined muptiplicative operation.

    That is, each idea is a copy of the natural numbers.
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  7. #7
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    Perhaps this page on real analysis shall be useful.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    Perhaps this page on real analysis shall be useful.
    Actually that reference is not very helpful.
    In fact it may be misleading.

    The material that I have used with undergraduates is found in this text: An Outline of Set Theory by James M Henle.
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  9. #9
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    Quote Originally Posted by Plato View Post
    You do know that we must show that addition is well defined?
    Also we must show that there is a well defined linear ordering.
    As well as a well defined muptiplicative operation.

    That is, each idea is a copy of the natural numbers.
    Got it. Thanks
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