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!

  • 2
    You can start with $0 + 0 = 0$, multiply both sides by $a$, and distribute on the left. Then subtract $a\cdot 0$ from both sides. – littleO Oct 16 '15 at 21:11
  • 2
    @littleO You should put that into song form. – PyRulez Oct 17 '15 at 18:24
up vote 10 down vote accepted

Proceed like this

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

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

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

  • Thanks, it is very useful that you listed the properties. It seems to me that we did not use additive closure, multiplicative associativity, or multiplicative identity. – mareoraft Oct 16 '15 at 21:23
  • Therefore, we get this propety in something even more general than Rngs. – mareoraft Oct 16 '15 at 21:24
  • You are welcome. Yes one does not use all properties, and I think you are right in that it is also true in non-associateive algebras. – quid Oct 16 '15 at 21:28

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

$a.0=a.0+0$ (Additive identity)
$=a.0+(a+(-a))$ (Additive inverse)
$=(a.0+a)+(-a)$ (Associativity)
$=(a.0+a.1)+(-a)$ (Multiplicative identity)
$=a(0+1)+(-a)$ (Distributativity)
$=a.1+(-a)$ (Additive identity)
$=a+(-a)$ (Multiplicative identity)
$=0$ (Additive inverse)

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

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

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