Alternative proof of $a\times0= 0$

Question

I was trying to find a proof of $a\times0 = 0$ by myself (assuming commutativity, associativity, distributivity, etc) and I came up with $$ a+0=a(1) \implies 1 = \frac{a+0}{a} = \frac aa + \frac 0a = 1 + \frac 0a \implies \frac 0a = 0 \implies a(0)=0$$ Is there anything wrong with this proof, or does it assume anything that shouldn't be assumed?

Answer 1

There's nothing wrong with your proof per se, but it looks like you do use the distributive law, which is the standard way, but you disguise it by first dividing through by $a$. The standard proof generally looks like: $$ a0 = a(0+0) = a0 + a0 $$ and then use additive cancellation to conclude that $a0=0$. This proof is valid in any ring, and furthermore, any proof that $a0=0$ or $0a=0$ in an arbitrary ring must use the distributive law since it is the only ring axiom that links the additive and multiplicative structures of the ring, and $0$ is defined as the additive identity and we're proving a multiplicative property.

Now, in specific number systems, the proof may look different. In the natural numbers, for instance, multiplication is usually defined inductively as: $$ a0 = 0 $$ and $$ a(n+1) = an + a $$ and so there, $a0=0$ because it is defined that way. Other number systems usually define arithmetic in ways that reduce to natural number arithmetic, so it all really comes back to definitions.