# Homework Help: Etiquette for Proofs

1. Mar 16, 2006

### masterofthewave124

For proofs, can we take for granted that an even number x an odd number is even?

I'm supposed to prove that for ever natural number, n, n^2 + 2 is even.

Proof:

n^2 + n
= n(n+1)

Since n and n + 1 and two consecutive integers, one must be even and one must be odd so there product must be even.

If something this basic is not required to prove, out of curiousity what would be the proof? And what is a good general rule of thumb for deciding what to prove or support in your proofs?

2. Mar 16, 2006

### 0rthodontist

n^2 + 2 is not the same as n^2 + n. I assume you mean n^2 + 2.

You can prove that the product of an even and an odd integer is even by writing the odd integer as 2k + 1 and the even integer as 2j for arbitrary integers j and k. You are looking to put the expression in the form 2m for some number m.

In general, I find that I am successful when I focus on the level of the class. If the class is somewhat informal and the proofs are difficult to figure out, then I probably wouldn't prove little details like that since the important thing is the concept. But since you are doing such a short proof it couldn't hurt to prove stuff like that.

3. Mar 16, 2006

### masterofthewave124

Yeah, my initial post had a typo in it and the question indeed wanted a proof for n^2 + n.

Sorry for the confusion.

So,

As any even integer can be represented as 2n and any odd integer can be represented by 2n + 1;
2n(2n + 1) is obviously divisible by 2, and thus is even.

Is that a sufficient proof?

Last edited: Mar 16, 2006
4. Mar 16, 2006

### 0rthodontist

No, the n is not necessarily the same for the odd number and the even number. You only proved that any consecutive pair of integers of which the smaller one is even has an even product. Write one number as 2n+1 and the other as 2m.

5. Mar 16, 2006

### masterofthewave124

Even if the numbers were 2m and 2n+1,

2m(2n+1) is divisible by 2 so what's the difference?

6. Mar 16, 2006

### 0rthodontist

Try listing the first few numbers of the form 2n(2n + 1) in order, and then try listing the first few numbers of the form 2m(2n + 1) in order. You'll see that the lists are not the same.

Essentially when you have two numbers p and q, and you let p = 2n, then you have now bound n in the definition of p. If you then say let q = 2n + 1, then you're saying it's the _same_ n--you are necessarily saying q = p + 1, which is not sufficiently general if you want to prove your result for _any_ numbers p, q of which one is odd and the other is even.

7. Mar 17, 2006

### rhj23

true, but he corrected himself in his last post

8. Mar 17, 2006

### VietDao29

0rthodontist, I am very sorry to say this, but your posts in this thread do not make much sense to me. May be I am missing something, but here are my points:
First, in the first post of this thread, the OP provided a complete valid proof, by saying that:
n2 + n = n(n + 1).
Hence, the product of 2 consecutive numbers is always a multiple of 2, since at least one of them must be an even number.
------------------
No, n2 + 2 will be odd if n is odd, hence, you cannot assume that.
Now that you really lost me...
I am completely lost...
I must be missing something, aren't I?

9. Mar 17, 2006

### rhj23

<leaps to orthodontist's defense>

nothing said so far by orthodontist has been incorrect - just his last post was pretty irrelevant.

Proposition: for every n a natural number, n^2 + n is even.

Proof: n^2 + n = n(n+1), so it remains to prove that the product of consecutive integers is even.

Case 1: n even. then n = 2m. 2m(2m+1) = 2 (2m^2 + m) and therefore even.

Case 2: n odd. then n = 2m + 1. (2m+1)(2m+2) = 2 (2m^2 + 3m + 1) and therefore even. //

10. Mar 17, 2006

### 0rthodontist

He did produce a correct proof, but his question was does he have to, in support of his proof, prove that the product of an even number and an odd number is an even number. My answer is that given the level of the original proof it couldn't hurt, and I believe I am correct in this because of the difficulty he turned out to have proving that latter proposition.

Ah, yes, that was a typo.

I was explaining to him why his attempted proof that the product of an even number and an odd number is an even number, is not correct because it used the same integer n for both the 2n with the even number and 2n+1 with the odd number.

No, he was still confused. His last post was
indicating that he still did not understand the difference between 2n(2n+1) and 2m(2n+1) or the reason for using two different numbers m and n.

11. Mar 17, 2006

### Curious3141

(and "numbers" should preferably be replaced by "integers")

12. Mar 17, 2006

### Muzza

Both are correct.

13. Mar 17, 2006

### masterofthewave124

Thanks guys!

My mistake was assuming the first number (2n) in the two number consecutive sequence was even and hence the succeeding number must have been 2n+1. I did not account for the possibility that the first number could be odd and thus the sequence would have been 2n+1(2n+2). The two cases rhj23 provided really cleared things up.

14. Mar 17, 2006

### 0rthodontist

You could, if you wanted to, do it that way by considering the two cases of n being even and n being odd. The way you were originally doing it is cleaner in my opinion because it doesn't have to break it up into cases.

It really is important that you understand why when proving that the product of an even integer and an odd integer is even, you cannot write the odd integer as 2n+1 if you write the even integer as 2n.

15. Mar 17, 2006

### rhj23

yeah, it's important to get that distinction (even though the proof he wanted only required consecutive integers, so it wasn't important)

the only reason i went through the proof in a fairly condescending way, and split it into cases, was to make sure it was clear what there was to prove

16. Mar 17, 2006

### Curious3141

No. "At least" basically means "not less than". Which in this case where we're dealing with two consecutive integers can mean "one" or "both". Both obviously cannot be even, so that implication is false.

"Exactly" one is perfectly correct. Either the first or the second, but NOT both is even.

17. Mar 17, 2006

### shmoe

Saying "at least one of n and n+1 is even" is absolutely correct- 1 is "not less than" 1. Not having the tightest range does not make a statement false.

Last edited: Mar 17, 2006
18. Mar 17, 2006

### Curious3141

It makes it at the least redundant, since it admits of a superfluous (and in this instance, erroneous) case. I didn't say the statement itself was false, I said the particular implication (that both can be even) is false. I consider the statement to be potentially misleading.

I can make a case for the statement being misleading. Let's consider the generally understood usage of the phrase "at least". We see lots of this in the combinatorics problems posted on this site. When we see a problem asking for "Out of all possible selections of m objects out of n, how many would be expected to contain *at least* one object of type "O" ?", we are looking for the cardinality of a solution set that contains all possibilities satisfying the requirement and no other viz. selections containing 1 "O", 2 "O"s....m "O"s. If a student gave an answer for 1 "O" it would be wrong. We cannot justify that wrong answer saying that his solution satisfied the criterion of "at least one "O" " and therefore should be accepted, now, can we ?

In a similar fashion, we shouldn't accept the way this is worded. The truth of the matter is only one of the numbers is even. This *satisfies* the criterion "at least one" but the criterion admits of another, wrong, possibility. If someone who doesn't know anything about number theory looks at a statement saying "at least one of two numbers is even", he would think that meant "one or both numbers" which would be uninformative at best, misleading at worst.

Last edited: Mar 17, 2006
19. Mar 17, 2006

### 0rthodontist

Actually I would think of saying "exactly one" as being slightly redundant, because knowing that there is exactly one even number is more knowledge than knowing that there is at least one number. "Exactly one" contains more information than "at least one." And "at least one" is all that's necessary so why say more?

20. Mar 17, 2006

### Curious3141

I think at the end of the day, it boils down to how precise you'd like your language to be. "Exactly one" certainly contains more information, but it also allows one to draw a completely accurate conclusion about the numbers. "At least one" contains less information, but one of the conclusions that can be drawn from the statement is false.

From an epistemological POV, it is not significantly harder to observe that "exactly one" is a correct descriptor than it is to observe "at least one". Why not be precise ?

21. Mar 17, 2006

### 0rthodontist

Either way has its arguments in favor or against, and both ways are correct. A close reader would not be mislead by "at least one." And saying "exactly one" could mislead a careless reader into thinking that the conclusion that the product is even would not be valid unless exactly one of the numbers is even--again, a close reader would avoid this.

An absolutely clear phrase would be something like, "...at least one--in fact exactly one--..." to make it clear that it's the "at least" that the implication is drawn from and not the "exactly," but also to note that "at least" is not the most specific phrase available.

22. Mar 17, 2006

### Curious3141

OK, I see your point. I think an even clearer way would be "...the product of two numbers is even if at least one of the numbers is even. Exactly one of n, n+1 is even..."

Better ?

23. Mar 18, 2006

### shmoe

This would be wrong because of a failure to enumerate all the admissible cases, not from any misinterpretation or ambiguity of "at least one" that I see.

If I asked you for all the subsets of {1,2,3} that contained at least one even integer and you left out {1,2}, {1,3}, and {1,2,3} you'd be wrong.

We only care that one (or more) of {n,n+1} is even so that their product is even. There's no need to be more precise than knowing "at least one" is even.

Are you mislead if I say $$1<\pi$$ because it doesn't rule out the possibility that $$\pi=2$$?

Last edited: Mar 18, 2006
24. Mar 20, 2006

### Curious3141

My main gripe was with the fact that one could be more precise in that statement without any more effort. I never argued that the statement was wrong, since "at least" is essentially an OR statement (one is even OR both are even) with one argument the former being true. I'm not arguing with any of your points, really.

My example with the combinatorics question was not a very good one, and I see your point. Essentially the question is tacitly asking for the entire "truth table" that would make the "at least" (OR) statement true. Make any sense ?

Little error : {1,3} is not a solution.

25. Mar 20, 2006

### Muzza

You sure did - look at post #16.