Is this proof of $0 * a = 0$ correct? I need to prove that $0*a = a*0 = 0$ using only axiomatic properties for real numbers(It's a exercise from Apostol's book.
My try:
We can write $0$ as $(a - a)$. Hence we can rewrite the equation as follows:
$0*a = a*0 = a*(a-a) = 0$
By distributive law we have: $aa -aa = 0$, where $aa - aa$ evaluates to $0$ because $(-aa)$ is the negative of $aa$
Note: It's given that $a(b - c) = ab - ac.$ (Theorem 1.4)
 A: Given the Theorem 1.4, this proof seems fine. It uses only the results available to you and the axioms.
Unfortunately, there is something bugging me in general about the proof. One has to be careful with establishing an arithmetic with minus signs. Little things can trip you up. For example, it's natural to look at $(-a)b$ and think it just means $-(ab)$, but conceptually, they're different.
Yes, the theorem does seem to cover this, but it still bothers me, because establishing Theorem 1.4 often relies on establishing $0a = 0$ first. Here's the way I'm used to the proof going:
$$0a = (0 + 0)a = 0a + 0a,$$
which implies $0a = 0$ by cancellation (add the inverse of $0a$ to both sides, and simplify). Then, to establish that $(-a)b = -(ab) = a(-b)$, we observe that
$$(-a)b + ab = ((-a) + a)b = 0b = 0,$$
and similarly $ab + (-a)b = 0$. Therefore $(-a)b$ fits the definition of the inverse of $ab$. Similarly, $a(-b) = -(ab)$ as well. Finally, this yields Theorem 1.4:
$$ab - ac = ab + (-(ac)) = ab + a(-c) = a(b + (-c)) = a(b - c).$$
I don't really know how you'd prove Theorem 1.4 without relying on $0a = 0$.
