Prove the following properties of a ring R:
(a) For all a,b,c in R we have $a(b-c) = ab -ac$
(b) If R has multiplicative identity 1, then $(-1)(-1) = 1$.
(c) If R has multiplicative identity 1, then for all a in the ring we have $(-1)a = -a$
Attempt: I will use for the following axioms for all a,b,c in R
i) $(ab)c = a(bc)$
ii) $a(-b) = -ab$
iii) $a + (-a) = 0$
Proof(a): Suppose a,b,c are in R. Then $a(b-c) = a(b + (-c)) = ab + a(-c) = ab + (-ac) = ab -ac$
Proof (b): 1 is in R, then $(-1)(-1) + (-1) = (-1)[(-1) + (1)] = (-1)(0) = 0$
Proof (c) : Suppose 1 is in R, then $(-1)a = (-1)a + (1)(a) - a = 0 - a = -a$
Please can anyone please verify this are all correct, and if not, can anyone please give some feedback. I would really appreciate it. Thank you.
First proof looks good to me.
The second proof shows that $(-1)(-1) + (-1) = 0$. This is close to the final conclusion you wish to draw; push it one step further.
The third proof is good, though you elided over a bit in the middle. After the second inequality, you might write:
$$ = (-1 + 1)a - a = 0a - a = 0 - a = -a.$$
Even in the added steps above, it assumes familiarity with the fact that $0a = 0$.
(If this has not been established, use the standard trick of $0a = (0+0)a = 0a + 0a$ and cancel.)