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By letting $R$ be any ring I have to show these following equations hold in $R$:

a) $x\cdot 0=0$

b) $0\cdot x=0$

c) $-(-x)=x$

d) $(-x)\cdot(-y)=x\cdot y$

I'm not really looking for the answer just for a hint to get me started. I guess I'm confused on how to show these properties hold within $R$.

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2 Answers 2

Hint : $-1$ is the additive inverse of $1$ and $-1$ is the multiplicative inverse of $-1$.

$$ 1 -1 = 0$$ $$(-1)\times (-1) = 1$$

Then use distributivity.

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So proving all the above this is what I've got for my proof: –  Gamecocks99 Feb 4 '13 at 11:39
    
We know the (1-1)=0 ==> x*0=0 ==> x*(1-1)=0 ==> x-x=0 ==> 0=0 –  Gamecocks99 Feb 4 '13 at 11:40
    
Sorry I pressed enter by accident. But continuing 0*x is the same as the first method but the only reason you have to prove this is because multiplication is some cases is not commutative. For part c, since (-1)*(-1)=1 ==> -(-x)=x ==> (-1)*(-1)*x=x ==> x=x. Finally for the last part, since (-1)(-1)=1 ==> (-x)*(-y)=x*y ==> (-1)*(x)*(-1)*(y)=x*y ==> (-1*-1)*(x*y)=x*y ==> x*y=x*y. Therefore all the equations hold in R. –  Gamecocks99 Feb 4 '13 at 11:44
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First of all, what if the ring doesn't have a $1$? Like, the ring of all even integers? Second, if the ring does have a $1$, you have to prove $(-1)(-1)=1$. Third, proving $(x)(0)=0$ implies $0=0$ doesn't prove $(x)(0)=0$. I think there is still a lot of work to do. –  Gerry Myerson Feb 4 '13 at 11:51
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@DamienL One needs to be precise at this level. Does the minus symbol in $\,1 - 1\,$ denote a binary ring operation, or is it the unary inverse operation in the additive group? One of the main points of the exercise is to rigorously clarify how the two are related, so one cannot gloss over that. –  Math Gems Feb 4 '13 at 14:55
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  1. $x \cdot 0 = 0$: Observe that $x \cdot 0 = x \cdot (0 + 0) = x \cdot 0 + x \cdot 0$ Hence we have $x \cdot 0 = x \cdot 0 + x \cdot 0$. If we subtract $x \cdot 0$ from both sides, we get $0 = x \cdot 0$.
  2. $0 \cdot x = 0$: Similar reasoning.
  3. $-(-x) = x$: Observe that since $R$ is a ring, it is an additive group, so elements have additive inverses. Thus for all $x \in R$, $x + (-x) = 0$. Subtract $(-x)$ from both sides and get $x = -(-x)$.
  4. $(-x)(-y) = xy$: Before we show this, observe that $a(-b) = (-a)b = -ab$. Now observe that $0 = x \cdot 0 = x(y + (-y)) = xy + x \cdot(-y)$. Subtracting $x\cdot (-y)$ from both sides yields $xy = -x(-y) = (-x)(-y)$.

Why does $a(-b) = (-a)b = -ab$? Well observe that $0 = a \cdot 0 = a(b + (-b)) = ab + a(-b)$ which implies $a(-b) = -ab$. Similarly $0 = 0 \cdot b = (a + (-a))b = ab + (-a)b$ which implies $(-a)b = -ab$. Hence $a(-b) = (-a)b = -ab$.

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