# A proof of $n*0=0$?

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".

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What do you mean by "$0$" if you don't assume $0$ follows all the rules of arithmetic? How to you expect to prove anything about it? Are you asking how to convince a confused layperson? –  Zev Chonoles May 23 '13 at 20:33
@Ovi What is the question? Are you asking for a proof that $n\cdot 0$? I don't see it on the question's body. How do you define $\cdot$ and $0$? –  Git Gud May 23 '13 at 20:33
I'm saying that how come we don't take 0*n to be undefined since in order to prove that 0*n=0, we need to assume that the properties of arithmetic (such as the distributive property used by Andy Bromberg below)hold for 0. The reason why I think we would have reason to doubt that the properties of arithmetic hold for 0 is that 0 gives us problems with one of the properties, namely $a/b=c$ iff $b*c=a.$ –  Ovi May 23 '13 at 20:45
@Ovi note that in the field axioms (mathworld.wolfram.com/FieldAxioms.html), the multiplicative inverse property is held only for $a\neq0$. However, we don't even need that property to show that $a\cdot0=0$. –  Andy Bromberg May 23 '13 at 20:51
@Ovi: Except $$a/b=c\text{ iff }b\cdot c=a$$ is not an axiom/rule of arithmetic (if anything it is a definition of the symbol $a/b$, and at any rate needs to include the assumption that $b\neq 0$). The notation $a/b$ has to *mean something before you start making statements about it, and it does not mean anything when $b=0$. –  Zev Chonoles May 23 '13 at 20:51

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 $c^{-1}$:

$$0+a=0+b$$

$$a=b$$

So we have proven the cancellation law.

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nice. to expand just a little more, one could say that a reasonnable definition of 0 is that it is the neutral element of the addition law, i.e. $0+n=n$ for all $n$. then from this your answer derives. –  Glougloubarbaki May 23 '13 at 20:38
@Glougloubarbaki, yes, I used the field axioms in the proof. The zero property of addition is one of those axioms and so can be taken as true. –  Andy Bromberg May 23 '13 at 20:41
yes, I just assumed that the OP is not familiar with the notion of field (or "abstract" algebraic structure) so I just mentionned that one can write "0+0=0" by definition of what 0 is. –  Glougloubarbaki May 23 '13 at 21:03
@Glougloubarbaki Gotcha! Thanks for clarifying, then. –  Andy Bromberg May 23 '13 at 21:06

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.

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