# Homework Help: Zero issue

1. Jul 3, 2006

### EvLer

I am in discrete math class right now and trying to get the sets of numbers straight.
So, does the set of integers include 0? Is it ok to use 0 in proofs, that makes finding a counter-example a lot easier and disprove a statement about all integers.
Was just wondering if that is legal..... feedback on this is very much needed

Last edited: Jul 3, 2006
2. Jul 3, 2006

### shmoe

Yes, 0 is an integer. 0 may or may not be included in your courses definition of the natural numbers though.

3. Jul 3, 2006

### HallsofIvy

Are you sure you meant "integers" and not "natural numbers"?

0 certainly is a member of the "integers". When Peano constructed his axioms for the "natural numbers" he included 0 but today, the "natural numbers" is considered equivalent to "positive integers" which does not include 0.

Could you give an example of the "statement about all integers"? If, for example, it say "for all positive integers", then a counter-example involving 0 would not be valid.

Last edited by a moderator: Jul 3, 2006
4. Jul 3, 2006

### EvLer

the statement goes: prove for every integer n....
which i guess they do not mean natural numbers only, so I used 0 as a counter example. Yeah, I am careful with the "positive integers" which would exclude zero (thanks to HallsofIvy).

Thanks all for help everyone.

5. Jul 3, 2006

### shmoe

There's no universal definition of natural numbers today. Some authors include 0, some don't.

What is the rest of the statement? Is it true for the rest of the integers? If 0 is the only counter example, it may be an oversight (or maybe that's what they want you to be looking out for given your induction question).

6. Jul 3, 2006

### HallsofIvy

No, the problem does NOT go "prove for every integer n....
What was in place of ...? That's the crucial part! What were you asked to prove?

7. Jul 3, 2006

### EvLer

ok, the full problem is worded a bit different from what I said originally:
prove or disprove that sum of any 3 consecutive integers is even.

I disproved it like this in short:
given n,n+1,n+2 => 3(n + 1) and taking n = 0, sum is odd.

8. Jul 3, 2006

### shmoe

That looks fine by me then.

9. Jul 7, 2006

### Eryndel

It looks like that (dis)proof would work for any even numbers including 0. (2,3,4) would be odd, as would (4,5,6) and so on...

Just to pick nits.

10. Jul 8, 2006

### Gagle The Terrible

I would look at it this way : Given n, n+1, n+2, the sum (that I will note S) is 3(n+1).
We all agree on that. Now, if n is even (i.e. n = 2k)

S = 3(2k+1) = 6k + 3 = 0(mod2) + 1(mod 2) = 1(mod2)

if n is odd (i.e n = 2k+1)

S = 3(2k +2) = 6(k+1) will be odd or even depending on the choice of k

Hence , the sum of three any consecutive integers is not always even