This is my first proof and post. I'll eventually get better at tex.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homework Help: Well-Ordering Principle on Natural Numbers

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