A proof of $n*0=0$?

Question

The only proof I've seen for this assumes that $0$ follows all the rules of arithmetic. How can we make that assumption when dividing by $0$ is a problem? I know that some people don't agree that all of the numbers follow the rules for arithmetic; for example, people say that the proof of $.99999...=1$ is invalid because arithmetic can't deal with these "infinite numbers".

Answer 1

We take:

$$0=0$$ by zero property of addition: $$0+0=0$$ by definition of multiplication: $$a\cdot(0+0)=a\cdot0$$ by distributive law: $$a\cdot0 + a\cdot0 = a\cdot0$$ by cancellation law: $$a\cdot0=0$$

The cancellation law isn't under the field axioms and requires a proof for the above to be complete. Here's a proof:

We want to prove that if $a+c=b+c$, then $a=b$.

by the additive inverse property, we have an $c^{-1}$ such that $c^{-1}+c=0$. So by definition of addition:

$$c^{-1}+a+c=c^{-1}+b+c$$

by associativity and commutativity of addition:

$$(c^{-1}+c)+a=(c^{-1}+c)+b$$

by definition of $c^{-1}$:

$$0+a=0+b$$

by zero property of addition:

$$a=b$$

So we have proven the cancellation law.

Answer 2

I think that you're confused by what the phrase "rules of arithmetic" means. A formal statement of these rules treats $0$ differently, accounting for the fact that you cannot divide by it and so forth. $0$ does not break any rules of arithmetic at all.