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  #1  
Old 07-28-2011, 11:09 AM
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more discrete math help

I have my final on Monday so I need some help.

I want to make sure this makes sense to those out there that know the subject

Prove of Disprove that n^3+2n is divisible by 2 ; n >= 2 (n is greater than or equal to 2)

I want to disprove that the equation is divisible by 2 but I think my method I have now wont fly.

n = 3

3^3 + 2(3) = 33 which is not divisible by 2 thus disproving it. But that is only one case, which is why I think it will not fly.

help? I believe induction might need to be used.
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  #2  
Old 07-28-2011, 11:58 AM
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why doesnt n=2 work?
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  #3  
Old 07-28-2011, 12:11 PM
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n = 2 does work but I need to disprove this equation some how. and my method I have probably isn't the best.
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  #4  
Old 07-28-2011, 02:58 PM
sjh sjh is offline
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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.
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  #5  
Old 07-28-2011, 03:01 PM
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I was thinking it was a trick and I'd have to use induction
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  #6  
Old 07-28-2011, 05:08 PM
sjh sjh is offline
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If you could that would be great but once you find ANY disproof your done.
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  #7  
Old 07-28-2011, 09:28 PM
Yak Yak is offline
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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; 07-28-2011 at 11:30 PM. Reason: technical fix
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  #8  
Old 07-28-2011, 09:53 PM
sjh sjh is offline
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Quote:
Originally Posted by Yak View Post
A counterexample may be sufficient, but since you were working with modular 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.
I do not understand the transition from step 1 to 2 but I do not doubt it.

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.
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  #9  
Old 07-28-2011, 11:21 PM
Yak Yak is offline
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Quote:
Originally Posted by sjh View Post
I do not understand the transition from step 1 to 2 but I do not doubt it.

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.
I had to look it up to refresh my memory from an earlier post. See #15 here for the question and symbols. more discrete math help

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.


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Old 07-29-2011, 01:43 AM
sjh sjh is offline
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Quote:
Originally Posted by Yak View Post
I had to look it up to refresh my memory from an earlier post. See #15 here for the question and symbols. more discrete math help

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.


I understand Yak. I've encountered Mod math (slightly) for over 40 years. But I could not (and still cannot) make the step without greater clarity and understanding than I currently possess.

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.
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  #11  
Old 07-29-2011, 10:50 AM
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yes, that is really what I needed to probably show. Sure n=3 doesn't work but what about n=4, 5, 6, etc?
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  #12  
Old 07-29-2011, 10:55 AM
sjh sjh is offline
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Quote:
Originally Posted by Matt L View Post
It's very nice, but it is also quite wrong.

His (invalid) proof seems to show that the relationship holds for no n>=2. It definitely holds for any n that is even. Yet, even for an even n, n^3 is not 0.
Steps 2 forward are fine I think. The transition from 1 to 2 is beyond me.

And Don. Just show 1 disproof and your done.
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  #13  
Old 07-29-2011, 11:14 AM
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Fine.... lol

It just seems to easy for it to be correct
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  #14  
Old 07-29-2011, 11:17 AM
sjh sjh is offline
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Originally Posted by TheDon View Post
Fine.... lol

It just seems to easy for it to be correct
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.
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  #15  
Old 07-29-2011, 12:01 PM
Yak Yak is offline
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Quote:
Originally Posted by Matt L View Post
Here's the problem:



For the congruence to not hold, you show that it can't hold for ANY value of k. You show that it doesn't hold for a particular value, so nothing is proved.

And a single counterexample really is sufficient to disprove the claim. Providing more will not amuse the professor.
I understand the k = n is a particular value. I'm not sure I understand the precursor due to the multiple negatives in the statement (...not hold...can't hold). If the issue is improper use of the modulus, that may be correct.

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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