proof using well ordering principle

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Econowizmeister
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#1
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https://imgur.com/a/WxPs089
Here is the proof.

I don't understand the bit where it says 'if n0 is not a prime, then n0=n1 x n2. n1 and n2 are both greater than 1 and less than n0. But then neither can be in S'.

Why cant n1 or n2 be in S? S is just defined as all the elements greater than 1 and not divisible by a prime. How do we we know n1 and n2 are divisible by a prime?
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Econowizmeister
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#2
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(Original post by Econowizmeister)
https://imgur.com/a/WxPs089
Here is the proof.

I don't understand the bit where it says 'if n0 is not a prime, then n0=n1 x n2. n1 and n2 are both greater than 1 and less than n0. But then neither can be in S'.

Why cant n1 or n2 be in S? S is just defined as all the elements greater than 1 and not divisible by a prime. How do we we know n1 and n2 are divisible by a prime?
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_gcx
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n_0 is the least element of S, so we can't have any elements less than n_0 in S. We know that n_1 and n_2 are divisible by primes: since they're less than n_0 and hence not in S.

Strong induction is the same idea (just a different explanation of the same thing with WOP) and might be clearer.
Last edited by _gcx; 1 month ago
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Econowizmeister
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(Original post by _gcx)
n_0 is the least element of S, so we can't have any elements less than n_0 in S. We know that n_1 and n_2 are divisible by primes: since they're less than n_0 and hence not in S.

Strong induction is the same idea (just a different explanation of the same thing with WOP) and might be clearer.
ohhhh wow i completely forgot we defined n_0 as the minimum of the set. Thank you so much
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