I'm pulling out my hair with these proofs!!

[daon]

I have spent about 3 hours stuck on this one problem, and I have made little progress...

"if m is an integer, one and only one of these are true:

m is a natural, -m is a natural, m=0"
..............

I have to prove this given that very basic algebra is understandable, and that the properties of the natural numbers have been defined as:

1. 1 is a natural number
2. if n is a natural, then n+1 is a natural
3. 0 is not a natural
4. if n is an integer s.t. n is not zero, then either n is a natural or -n is a natural.
(note that 4 is not an exclusive or)

I am pretty good at logic, but I can't find a way to prove this. I know I need to prove:

(a) If m is a natural , then -m is not and m is not 0.
(b) If -m is a natural , then m is not and m is not 0.
(c) If m=0, then m is not and -m is not

I have proven part c, but am stuck on a and b. I have gotten as far as proving that if m is natural then m cannot be 0, but cant figure out how to say if m is a natural then -m is not a natural.

Can anyone help?

 

[stapel]

I think you're making this too complicated. Just work by cases and by following the definitions:

. . . . .Let m be an integer.

. . . . .Either m = 0 or it doesn't.

. . . . .If m = 0, then you're done. Suppose it doesn't.

. . . . .Then, by definition of the integers, m < 0 or m > 0.

. . . . .Suppose m > 0. Then m is also a natural number,
. . . . .by definition of "integer" and "natural number".

And so forth.

Eliz.

 

[Matt]

4. if n is an integer s.t. n is not zero, then either n is a natural or -n is a natural.
(note that 4 is not an exclusive or)

If (4) is not an exclusive or, then there is a nonzero number n for which n and -n are both natural. But this does not make any sense; negative numbers are not natural.

My point: (4) is an exclusive or.

 

[daon]

Yes, this was believed to be necessary for these proofs by all of my class. Our professor has said that it is indeed not an exclusive or because of the way we defined natural numbers. We have not yet proven that N is a subset of the integers such that any n in N has to be greater than zero. We have only said that 1 is a natural number and 1+1 is a natural number...etc

My professor is very keen on details and we can't just use common sense to say "oh, its only positive integers"

 

[daon]

stapel, > and < haven't yet been introduced at this point in the proofs.

edit: By definition in our class, we can have 1, 2, 3, etc, and the additive inverse of these numbers, but not negative numbers.

 

[Matt]

Yes, this was believed to be necessary for these proofs by all of my class. Our professor has said that it is indeed not an exclusive or because of the way we defined natural numbers. We have not yet proven that N is a subset of the integers such that any n in N has to be greater than zero. We have only said that 1 is a natural number and 1+1 is a natural number...etc

My professor is very keen on details and we can't just use common sense to say "oh, its only positive integers"

Okay, I understand your point. Let's restrict ourselves to the definition of natural numbers that you provided above.

I know I need to prove:

a) If m is a natural , then -m is not and m is not 0.

I have gotten as far as proving that if m is natural then m cannot be 0, but cant figure out how to say if m is a natural then -m is not a natural.

It would help if you could prove that the sum of two natural numbers is again natural (using the properties of natural numbers that you listed above). Then if both m and -m were natural, then their sum would have to be natural.

 

[daon]

I really like that Idea. Hopefully we can use proofs BWOC at this point. Though I don't see why not as we have just covered negations.

Basically under the assumption that m was a natural number (by the hypothesis), I assumed -m also to be a natural number, and showed that their sum (0) was a natural number through logic. This proved to be a contradiction, so can I assume that the opposite is true?

 

[pka]

daon, perhaps as Matt suggests there other theorems about integers and/or natural numbers with respect to operational closure.
For example, using 1., 2, and 3., you can see that the integer −1 is not natural. Now how far does your given system of axioms and definitions allow us to go with −2?
You can see that the tools you have are not obvious from what you have posted.
Do you have some sort of induction at this point?

 

[daon]

pka, yes I understand what you and matt are saying. This is the first time I've had to pay so much attention to detail, so it is a bit difficult for me. And yes, I have proved mathematical induction. How can it be applied to this problem?

Thanks to Matt's help, by way of contradiction I can prove that: Given m is an integer and m is not 0, either m or -m is a natural number (exclusively). Might there be another method to do this?

Also, if I have shown that -m is not zero, how can I prove that m is also not zero. I don't know if we can assume that "Equals" and "not equals" have the same properties (i.e. adding to both sides, etc). This is becomming an issue in part b of what I originally posted.

 

[daon]

I have attatched what I have done so far. I believe the proofs of (a) and (c) are correct. If not, someone please correct me. I also have no idea where to continue on part b as you can see from the picture. Let me know if you can help, I'll be working on it.

Thanks again.

 

[pka]

Assuming that you know what theorems, axioms, and definitions you may use, you work is logically correct.