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#1




more discrete math help

#2




why doesnt n=2 work?
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#3




n = 2 does work but I need to disprove this equation some how. and my method I have probably isn't the best.

#4




To disprove
Hello,
While there is preference for using formal proofs to demonstrate the validity of a concept a disproof is proved by stating any example. Therefore you have provided an adequate disproof. To prove a general (and somewhat complex) statement would probably take more ability than is present on this board or at many institutions of higher leaner. Fortunately your instructor gave you an easy way out. 
#5




I was thinking it was a trick and I'd have to use induction

#6




If you could that would be great but once you find ANY disproof your done.

#7




A counterexample may be sufficient, but since you were working with congruent arithmetic:
n^3 + 2n ~ 0mod2 (divisible by 2, no remainder, ~ means congruent) n^3 + 2n = 2k n^3 + 2n  2k = 0 let k = n n^3 + 2n  2n = 0 n^3 = 0 Since n^3 cannot = 0 when n >=2, it's disproven. Last edited by Yak; 07282011 at 11:30 PM. Reason: technical fix 
#8




Quote:
Once that is accomplished the rest of your proof is simple, clear and well done. My career has not involved mod/etc except tangentially but for some reason I'v always enjoyed geometric proofs. 
#9




Quote:
Step 1 is basically writing the text "is divisible by 2 without remainder" in congruent format. Step 2 is solving that: There must be some integer (k) when multiplied by 2 that equals n^3 + 2n. 
#10




Quote:
It's not a subset of math that I learned and I've been prettied drugged (finally off those) and infirmed for almost 2 years. Nice to see someone provide a general instead of a specific disproof. 
#11




yes, that is really what I needed to probably show. Sure n=3 doesn't work but what about n=4, 5, 6, etc?

#12




Quote:
And Don. Just show 1 disproof and your done. 
#13




Fine.... lol
It just seems to easy for it to be correct 
#14




Those are often the best test problems.
Give a student 2 pages for an answer and 3 paragraph question that can be answered in one brief set of equations. A good way to find out if they know the material. 
#15




Quote:
However k = n is no more particular than n = 3, so if k = n proves (or disproves?) nothing I don't understand how n = 3 could prove anything. A problem is that n^3 + 2n IS divisible by 2 for all even n; it is NOT divisible by 2 for all odd n (even x even x even) + (2 x even) = even (odd x odd x odd) + (2 x odd) = odd So the question "what about n = 2, 4, 6..." is valid. The equation is true for n = 2a (even numbers), it is not true for n = 2a+1 (odd numbers). So you could write the equation that way and have at it. If picking a specific integer where the equation is not true suffices for a rigorous proof, that sets a slippery precedent for picking a specific a specific integer where the equation IS true as a proof. It does show the equation is incorrect, but it doesn't show why  and isn't that the point of a proof? 
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