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Math Help - Reflexive + lineair

  1. #1
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    Reflexive + lineair

    Hey all, I got to proof



    (still my backslash doesn't work, therefor no LaTeX )

    I see that there is a n so that it holds for all x and for all y (n = 1)

    Anyone?
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  2. #2
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    I see that there is a n so that it holds for all x and for all y (n = 1)
    Do you mean that y=x^1 for all x and y?

    Well, here is how to show antisymmetry. Assume xRy and yRx, i.e., y=x^n and x=y^m for some natural m, n. Then y=y^{nm}, so by the assumption in the problem statement y=1 or nm=1. If y=1, then x=y^m=1, so x=y. If nm=1, then n=m=1, so again x=y.

    If you are having difficulties with other parts, could you describe what they are?
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  3. #3
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    I forgot to add that it's a binary ordening.. Does tht make any difference to this?

    If you have to pick a n out of the pos. nat. numbers so that that holds, 1 should be enough?

    Therefore, x would be y, and because xRy, xRx / yRy would be true.

    And if I remember right that's the definition of a reflexive ordening, or am I thinking too easy?
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  4. #4
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    I forgot to add that it's a binary ordening
    A reflexive (which is the same thing as non-strict, or weak) partial order is a binary relation, but I am not sure what "binary ordering" is.

    If you have to pick a n out of the pos. nat. numbers so that that holds, 1 should be enough?

    Therefore, x would be y, and because xRy, xRx / yRy would be true.
    So that what holds?

    Could you write precisely the statements that you have to prove and that you claim are true? What I mean in particular is that in a meaningful proposition every variable must be properly introduced. To introduce a variable one uses phrases like "Let us fix some natural number n", or "Consider x such that P(x) holds", or "let y = x^n+z^n". So every variable that you mention either in a claim that you need to prove or in a proof of that claim must be introduced in one of these ways. In your quote above, I am not sure what x and y are. Probably for some x and y what you say is true, for others it may be false.

    If I were to guess, you are saying that R is reflexive for the following reason. Fix any natural number x. Then x=x^1, so xRx. Since this holds for any x, R is reflexive. This is correct, but it needs to be said more precisely.

    And if I remember right that's the definition of a reflexive ordening, or am I thinking too easy?
    A reflexive relation R has indeed only one property: for every x, xRx. However, a reflexive order is a relation that has three properties; they are described in the link above.
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