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!
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6You 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
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4@littleO You should put that into song form. – Christopher King Oct 17 '15 at 18:24
3 Answers
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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 $ab +a(-b)= a(b + (-b))= a0= 0$.
quid
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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
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Therefore, we get this propety in something even more general than Rngs. – mareoraft Oct 16 '15 at 21:24
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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
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@Filippo Post a new question on this website asking "What is the 'additive closure' property in the context of rings?" and somebody should help you. – mareoraft Feb 08 '22 at 22:16
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@mareoraft As far as I know this simply means that the sum of two elements of the ring is an element of the ring, I would just appreciate a confirmation that this is what you mean because the comment really surprised me. – Filippo Feb 09 '22 at 06:39
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1@Filippo Yes, that is what is what I mean. Typically we first define addition as a binary operation (which includes closure in its definition), then we define a group, then we define a ring. So if you want to be really careful, you may have to go back and investigate these definitions. In my second comment I was using the term rng to refer to a non-unital ring (https://en.wikipedia.org/wiki/Rng_%28algebra%29). Have a great day! – mareoraft Feb 09 '22 at 15:33
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@mareoraft Thank you for the explanation and the link and have a great day, too :) – Filippo Feb 09 '22 at 18:33
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$a\cdot0=a\cdot0+0$ (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.
