Proofs (1 Viewer)

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I am not to sure how to go about this

 

[SadCeliac]

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I am not to sure how to go about this

I'm gonna try using contrapositive, if that doesn't work, I'll try contradiction - give me like five mins lol

edit: I HAVE NO CLUE HOW TO DO THIS


@synthesisFR @carrotsss calling in the backup

 

[cossine]

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I am not to sure how to go about this

This is Wilson's Theorem

https://en.wikipedia.org/wiki/Wilson's_theorem

 

[Sam14113]

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I am not to sure how to go about this

well first off the question is wrong (p = 1 is a counterexample)

but if we be generous and change ‘prime’ to ‘not composite’ then yes SadCeliac is right go by contrapositive: p is composite => p does not divide that thing

so p is composite, we can write it as p=ab, where 1<a<p

so (p-1)! is a multiple of a

so (p-1)! + 1 is not a multiple of a and so not a multiple of p.

that’s the basic outline - you just need to write it better.

Does that help?

also for the record this is not Wilson’s thm - Wilson’s thm is that AND that the converse is true. Proving the converse I believe is a much harder exercise [maybe can be done with strong induction? Not sure - haven’t tried] anyway this paragraph is irrelevant for 4U don’t worry about it

 

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Ohh okay just to make sure (p-1)! is a multiple of a because there is a factor of a that makes it divisible by (p-1)!?

 

[Sam14113]

Ohh okay just to make sure (p-1)! is a multiple of a because there is a factor of a that makes it divisible by (p-1)!?

Exactly - I'll just clarify further but I think you got it.

Because (p-1)! = (p-1)(p-2)...(a+1)(a)(a-1)...(3)(2)(1), which has a factor of a

 

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ahh right that makes sense tysm