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Show that R is a partial order on ℤ.

  1. Jul 25, 2014 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and solution are attached as TheProblemAndSolution.jpg.

    I will also copy down the problem and solution here.:
    Problem:
    Consider the set ℤ of integers. Define aRb by b = a^r for some positive integer r. Show that R is a partial order on ℤ, that is, show that R is (a) reflexive; (b) antisymmetric; (c) transitive.

    Solution:
    (a) R is reflexive since a = a^1.

    (b) Suppose aRb and bRa, say b = a^r and a = b^s. Then a = (a^r)^s = a^(rs).
    There are three possibilities: (i) rs = 1, (ii) a = 1, and (iii) a = -1.
    If rs = 1 then r = 1 and s = 1 and so a = b. If a = 1 then b = 1^r = 1 = a, and, similarly, if b = 1 then a = 1.
    Lastly, if a = –1 then b = –1 (since b ≠ 1) and a = b. In all three cases, a = b.
    Thus R is antisymmetric.

    (c) Suppose aRb and bRc say b = a^r and c = b^s. Then c = (a^r)^s = a^(rs) and, therefore, aRc. Hence R is transitive.

    Accordingly, R is a partial order on ℤ.

    2. Relevant equations
    Definitions of partial order, reflexive binary relations, antisymmetric binary relations, transitive binary relations.

    3. The attempt at a solution
    I don't get why if a = –1, then b = –1. If I assume that b ≠ 1, then I get it, but why is b = 1 forbidden?

    Any input would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Jul 25, 2014 #2

    HallsofIvy

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    If b= a "aRb= bRa" is trivial so you can assume [itex]b\ne a[/itex]. Since this is the case a= 1, you may assume that b is not 1.
     
  4. Jul 25, 2014 #3

    Orodruin

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    By your supposition, bRa, a = b^r for some r. This cannot be true if a = -1 and b = 1.
     
  5. Jul 25, 2014 #4

    s3a

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    Sorry, I double-posted.
     
  6. Jul 25, 2014 #5

    s3a

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    Orodruin, the thing is, if a ≠ b, then antisymmetry does not hold, and we can't assume that a = b for antisymmetry to hold, because we're trying to show that antisymmetry holds in the first place by showing that a = b must hold.

    HallsofIvy, are you saying I should assume a ≠ b, and that I should then show that the equations don't always hold if a ≠ b?

    I'm still confused about how I can say that –1 = (–1)^(positive_exponent), because positive_exponent must always be an odd integer for that to hold, but it could be an even integer too, so what do I do?
     
  7. Jul 25, 2014 #6

    s3a

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    Sorry, I double-posted.
     
  8. Jul 25, 2014 #7

    s3a

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    Never mind, I got it.

    Because a = b^s implies that –1 = 1^s is false, b ≠ 1.

    Thank you both.
     
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