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Well-Ordering Principle on Natural Numbers

  1. Jan 13, 2012 #1
    This is my first proof and post. I'll eventually get better at tex.

    1. The problem statement, all variables and given/known data

    If [itex] n \in N[/itex], then [itex]n ≥ 0[/itex].

    Hint: [itex]N \subset N[/itex] (thus not any empty set) and has least member by the well-ordering principle.

    2. Relevant

    (i) [itex]0 \subset N[/itex]

    (ii) [itex]n+1 \in N[/itex] for all [itex] n \in N[/itex]

    (iii) [itex]n-1 \in N [/itex]for all [itex] n \in N [/itex]such that n≠0

    (iv) The well-ordering principle itself.

    3. The attempt at a solution

    As the hint suggests, I am supposed to prove this using the well-ordering principle.

    [itex]n-1 \in N[/itex] for all [itex] n \in N[/itex] such that n≠0. [itex] N \subset N[/itex] so N≠∅ and has a least member by well-ordering principle.

    I used (iii) [itex] n-1 \in N [/itex] for all n in N such that n≠0. But

    [itex](n=0) \in N[/itex] and [itex]-1 \notin N[/itex]. Hence, [itex] -1<n → 0≤n[/itex]

    I'm not sure if this is right though.

    Additionally, is there any way to write text inbetween {itex} text {/itex} such that notations appear in addition to original text?
     
    Last edited: Jan 13, 2012
  2. jcsd
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