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

[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.