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.

why doesnt n=2 work?

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

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.

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

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

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.

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. http://www.peachparts.com/shopforum/off-topic-discussion/302549-more-discrete-math-help.html

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.

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

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.

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.

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?

40+ years of math, a fair amount fairly advanced, the term "counterexample" doesn't "strike a bell".

Google tells me that counterexample is 3x more prevalent than disproof so you seem to be "right on the money".

The majority of my math has been application driven but the joy of just pursuing the rigor or intellectual discipline that abstract math contains still exists.

Glad to see you could contribute.

PS - I cannot see the bolded lines that you reference.

Matt is correct that let k = n is an incorrect substitution because the equation can't be shown to balance. n is already used on the left and I arbitrarily inserted it on the right.

And yes, a single counterexample does disprove a general theorem. However, relying on always being able to quickly find a specific integer that disproves a general theorem is a risky way of doing proofs.

It's not the approach I'd recommend counting on - or studying for - going into a final. What method did he use to find 3? Trial and error is great when it works, but is a horrible feeling in an exam when it doesn't.

What if the next question is "is there any solution set where the equation is true"? Would the answer be "not 3" or would there be more work involved?

This approach shows a problem with relying a specific counterexample being quickly identifiable (or unidentifiable). Using the approach of "the first step is to decide if the theorem can possibly be correct. Of course it cannot, because..."

What if you start with n=2? Can it possibly be correct? Yes. 8? Yes. 32? Yes. 488? Yes. 3? No.

What if there are no "disprove" questions on the final?

Thanks for telling me where it is.

Regarding the OP question of how to prepare for the exam. I'll suggest that developing the intellect is similar to developing muscle and that well structured mental exercise will literally both grow new brain cells and new inter-connections between cells.

So his (or her) best strategy is to develop healthy habits and to exercise their mind regularly.

Please post all the math remarks that you can. It's enjoyable to encounter the diversity of topics in OD.

Take care.

Thanks for your help. All I can say is this class has been the most confusing and difficult to date

Go study Physical Chemistry for a year; spend 4 months with the Schrodinger equation. I still hurt 35 years later.

Best to you with the exam.