In a ring, how do we prove that a * 0 = 0?

Question

In a ring, I was trying to prove that for all $a$, $a0 = 0$.

But I found that this depended on a lemma, that is, for all $a$ and $b$, $a(-b) = -ab = (-a)b$.

I am wondering how to prove these directly from the definition of a ring.

Many thanks!

Answer 1

Proceed like this

• $a0 = a(0+0)$, property of $0$.
• $a0 = a0 + a0$, property of distributivity.
• Thus $a0+ (-a0) = (a0 + a0) +(-a0)$, using existence of additive inverse.
• $a0+ (-a0) = a0 + (a0 + (-a0))$ by associativity.
• $0 = a0 + 0$ by properties of additive inverse.
• Finally $0 = a0$ by property of $0$.

Your lemma is also true, you can now prove it easily:

Just note that $ab +a(-b)= a(b + (-b))= a0= 0$.

Answer 2

$a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$

Answer 3

$a\cdot0=a\cdot0+0$ (Additive identity)
$=a\cdot0+(a+(-a))$ (Additive inverse)
$=(a\cdot0+a)+(-a)$ (Associativity)
$=(a\cdot0+a\cdot1)+(-a)$ (Multiplicative identity)
$=a\cdot(0+1)+(-a)$ (Distributivity)
$=a\cdot1+(-a)$ (Additive identity)
$=a+(-a)$ (Multiplicative identity)
$=0$ (Additive inverse)

Then, we can prove that $-a=(-1)\cdot a$:
If $a+(-1)\cdot a=0$, then $(-1)\cdot a=-a$ (there's a unique additive inverse element)
$a+(-1)\cdot a=a\cdot1+a\cdot(-1)=a\cdot(1+(-1))=a\cdot0=0$ (we used the first theorem)

So, the first theorem is necessary to prove the second one, but not conversely.