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Mathematics 8 Online
OpenStudy (rational):

Help with easy proof Prove or disprove : http://prntscr.com/3h1096

OpenStudy (rational):

\((2*3) ! \ne 2! 3!\) \((2+3)! \ne 2! + 3!\) can this be called a legitimate proof ?

OpenStudy (anonymous):

It depends what level of proof your teacher/professor is requesting. If you are used to writing more formal proofs, then I would say no. However, simply showing a counterexample is a legitimate proof in most cases. I would say you disproved both statements.

OpenStudy (anonymous):

its a counter example for disprove and yes its a legitimate proof

OpenStudy (rational):

Yep ! how should i set it up for a more formal/elegant proof ? where should I start... any ideas ? basically I could not think of any other ways to approach this as of now...

OpenStudy (anonymous):

if there is a counter example then no need to prove or disprove like in formall, just say disprove and here is a counter example

OpenStudy (rational):

Bswan i think so... counter example is considered a legitimate proof..

OpenStudy (anonymous):

only in disprove cases :)

OpenStudy (rational):

just wondering if there are any other neat ways to think about this...

OpenStudy (rational):

ofc yes :) we cant say "its ridiculous, so not possible" in a proof lol :P

OpenStudy (anonymous):

you mean like real disprove :D but why to bother ur self with that hah but i should give it a try , ill check it again nd tell you

OpenStudy (rational):

not just for this problem... i want to know in general :) thank you, il wait for your reply :)

OpenStudy (amistre64):

counterexample comes to mind

OpenStudy (rational):

only counter example is the known way for doing these kind of proofs i guess...

OpenStudy (amistre64):

one counter is: (2*5)! = 10.9.8.7.6.5.4.3.2.1 2! * 5! = 5.4.3.2.1 * 2.1

OpenStudy (amistre64):

10.9.8.7.6.5.4.3.2.1 =? 5.4.3.2.1.2.1 10.9.8.7.6 =? 2.1, doesnt look promising lol

OpenStudy (rational):

lol yesss i got counter examples for both (mn)! and (m+n)! m = 2, n = 3 it should be QED

OpenStudy (amistre64):

QED is for when you prove something true.

OpenStudy (amistre64):

there is no QED for this proof :)

OpenStudy (rational):

ooops !

OpenStudy (anonymous):

@rational from the beginning, you successfully disproved the statement. What I meant by "more formal" is guiding the reader to your conclusion, not just stating that \(6!\neq2!3!\) but this is purely technical. You have shown the statement is not true by providing a counterexample.

OpenStudy (rational):

got you :) \(((2*3)! = 720) \ne (12 = 2!3!) \implies (mn)! \ne m!n!\) \(((2+3)! = 120) \ne (8= 2! + 3!) \implies (m+n)! \ne m!+n!\) so the given statements are not true in general

OpenStudy (anonymous):

If you're asking whether or not there is a more elegant way, I think not. Elegance is simplicity, and a counterexample is the simplest way to disprove a statement. However, it is different if you are trying to prove a statement TRUE. In that case, you need a more general approach. That is when you have to think about every possible case. Here though, you're proving the statement false. So a counterexample will do just fine :)

OpenStudy (rational):

thanks @dummyguy @BSwan @amistre64 :)

OpenStudy (anonymous):

lol and i was thinking about this while washing the dishes to give a reply xD glad u got it :)

OpenStudy (anonymous):

what is this subject ?

OpenStudy (rational):

number theory...

OpenStudy (anonymous):

cool :)

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