Is n(0) Always Equal to 0? A Debate on Friend's Creative Proof

[Nano-Passion]

When I said that your proof (the picture in post #17) wasn't valid, I meant that it was completely wrong. 0 points, may God have mercy on your soul and all that. I just didn't want to be rude. (Sorry, I couldn't resist the Billy Madison reference. It was definitely not meant as an insult, just a little joke).

To a mathematician, your proof looks like "since the next few months are in the year 2012, it must be 2012 right now". The things you said about the numbers 1,2,3,4 don't imply anything useful about the number 0. It is, but what exactly it means depends on what axioms you take as the starting point. It's especially important to be careful with this when you're talking about zero terms. What does it mean to add zero copies of x together? It's reasonable to define it by saying that the notation 0x denotes 0. But then there's nothing to prove. His proof is not correct. I stopped reading it at "0+0=0. ergo, n(0)+n(0)=0". It's not clear to me what he's thinking here, and in my opinion, that's enough to make a proof invalid. But it looks like he's doing something seriously wrong. For example, if the idea is to multiply the first equality by n, then the result is n0+n0=n0, not n0+n0=0. So it looks like he's using what he's trying to prove.

My proof was a more rigorous (or at least more explicit) version of DivisionByZro's proof. I just listed the axioms and explained which one I was using at each step. He didn't explain why you can cancel a term from each side, e.g. that x+z=y+z implies x=y. That can be proved as a separate theorem, but you can also just add -z and use the axioms.

But then saying 0x denotes 0 is still an assumption that needs proof as much as saying that multiplication can be described as repeated addition. What is the explicit definition of multiplication? I thought it would be something agreed on.

Not quite. My point was that it depends on context. The context in which case your proof could be formalized is a construction of N where you have defined multiplication in terms of repeated addition. This is most likely not the case, so in all likelihood, your proof is incorrect. It is much more likely that you are working with a synthetic approach to number systems, where you assume all of the basic properties that you want the number system to have, and then you need to prove everything from there.
It really depends on the context. But making your argument more rigorous would take a lot more work than you want since you would have to work with a construction of N from ∅. Also, it is worth noting the following: Just because you point out a pattern for a few numbers, does not mean that pattern continues. For example, you noted that 2*3 = 6 and 2 + 2 + 2 = 6 and 3 + 3 = 6 and then generalized this to make the claim that n*a is n added a times or a added n times. You would have to show that this holds for all pairs of natural numbers for this to work, and doing this would have to rely on an explicit definition of multiplication.

But I thought that multiplication is an explicit definition of repeated additions. How else can you explicitly define multiplication?

 

[jgens]

But then saying 0x denotes 0 is still an assumption that needs proof as much as saying that multiplication can be described as repeated addition.

Fredrik never said claimed that we can say 0*x = 0 without proof. This is why he gave a proof of the result using the ring axioms.

It is also important to note that Fredrik is not just criticizing you using multiplication as repeated addition. He is criticizing the way you come to the conclusion that n*m is n added m times or equivalently m added n times (and rightly so). You gave a few examples where "n*m is n added m times or equivalently m added n times" but that does not mean this holds in general. You would have to prove it. So to a mathematician, you really have not shown anything.

There is a context in which you can formalize the notion in your proof and make everything rigorous. You are not working in this context however, so it is a moot point.

But I thought that multiplication is an explicit definition of repeated additions. How else can you explicitly define multiplication?

Is that how you would define multiplication in R? What would √3 * √2 be in terms of adding things then? The point is thinking of multiplication in terms of repeated addition is really only well-defined in N, Z. And these are not the only multiplication operations we can define either!

You are not doing a constructive proof where you have information like "multiplication is repeated addition". All you have are certain properties that a structure satisfies. You need to use these properties.

 

[Fredrik]

But then saying 0x denotes 0 is still an assumption that needs proof as much as saying that multiplication can be described as repeated addition.

I should perhaps have elaborated a bit. The claim that "0x=0 for all real numbers x" certainly needs to be proved, because the notation "0x" already has a meaning when x is a real number. However, if you consider an arbitrary field* F. Then nx, where n is an integer, isn't defined until we have chosen a definition, and it's certainly OK to define it by 0x=0, nx=x+...+x, and (-1)x=-x for all x in F. No proof is needed here, because now the "product" of an integer and a member of the field is defined by these statements.

The only reason I know why one might want to do something like this, is that it's a convenient notation when we want to prove that every ordered field has a subfield that's isomorphic to the field of rational numbers. Edit: I added the word "ordered" to this statement after jgens corrected it below.

*) If you're not familiar with the term "field", think of it as something just like the real numbers, except that it may not be equipped with an order relation. In other words, a field has all the properties of the real numbers that I mentioned in post #21.

But I thought that multiplication is an explicit definition of repeated additions. How else can you explicitly define multiplication?

It's usually not defined explicitly. I would define ℝ ("the" field of real numbers) as any Dedekind-complete ordered field. This makes sense because a) any two Dedekind-complete ordered fields are isomorphic (so it doesn't matter which one we use), and b) it can be proved that a Dedekind-complete ordered field exists. If we take this as the definition, multiplication isn't defined explicitly because the multiplication operation is part of what makes a field a field. (Something without a multiplication operation wouldn't be considered a field).

The multiplication operation can be defined explicitly, but only if we first define the set of real numbers explicitly. This is the sort of thing we have to do to prove the existence of a Dedekind-complete ordered field.

 

[jgens]

The only reason I know why one might want to do something like this, is that it's a convenient notation when we want to prove that every field has a subfield that's isomorphic to the field of rational numbers.

Every ordered field has a subfield isomorphic to Q

 

[Fredrik]

D'oh. Thanks.

 

[Nano-Passion]

This is a disappointment, I guess I got to be more careful with my mathematical assumptions. It wouldn't make sense to define multiplication as repeated addition because it doesn't work for irrational number. It doesn't make that much sense to add the number an irrational amount of times.

Is there hope for my 'proof' to least work for strictly integers?

 

[Fredrik]

Is there hope for my 'proof' to least work for strictly integers?

Not in its current form, because of how you're "extrapolating" a property of 1,2,3,4 to 0. One idea that could work is to define nx=x+...+x (with n terms in the sum) and (-n)x=-(nx) for n=1,2,3,..., and then prove that if we don't define 0x=0, then it's impossible to get simple rules like (n+m)x=nx+mx to hold for all integers n,m.

However, I think this would be a weird way to do things. We would essentially be leaving 0x undefined only so that it will be one of the things left to prove. If we define nx for all integers n≠0, we might as well just define it for n=0 as well, and then prove that those simple rules are satisfied.

 

[jgens]

Is there hope for my 'proof' to least work for strictly integers?

Before I answer this question I am going to give you some advice: I would give up on trying to make your proof work out for now. We all get attached to certain ideas, especially when we think that our ideas provide a novel or simple way at looking at things. Sometimes, our ideas just don't work out. No matter how clever our idea might have seemed, sometimes it just can't do the job. In these cases, you just need to let it go and move on. In this instance, I think you should just learn the proof that Fredrik posted and move on.

That said, there is a way to make the idea in your 'proof' correct in \mathbb{N}. We could do it by defining \Sigma recursively as follows:
\sum_{i=1}^0 k = 0
\sum_{i=1}^n k = k + \sum_{i=1}^{n-1} k
where n \in \mathbb{N}. Then we define the multiplication operation \cdot:\mathbb{N} \times \mathbb{N} \to \mathbb{N} as follows:
\cdot(n,m) = \sum_{i=1}^n m
Then we can prove that \cdot(n,0) = 0 by induction and the equality \cdot(0,m) = 0 follows by the definition of the empty sum. I might be missing some details, but this is essentially the idea.

It is worth noting however, that if we utilize this definition of multiplication that we know almost nothing about the multiplication operation. In fact, we have not shown that this multiplication operation is associative, commutative, distributes over addition, has 1 as a multiplicative identity, etc. So, do you see why this method of proof is far from ideal?

 

[Nano-Passion]

Okay, thanks Fredrik and jgens for your patience. That helped, I'll post some other proofs here later on . I understand now that in mathematics things need to be more defined and structured.