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I'm trying to follow this proof for two elements in some ring, but am struggling to justify the final two steps in the proof. I'm only to use either the properties of a ring or the fact that $(-a)b = -ab$. Here's what I have so far.

First, using this proof and letting $a = x$ and $b = -y$, we have \begin{align*} (-x)(-y) = (-a)b = -ab = -\left(x\left(-y\right) \right). \end{align*} Since multiplication in a ring is commutative, we have \begin{align*} -\left(x\left(-y\right) \right) = -\left(\left(-y\right)x \right). \end{align*} Applying the aforementioned property with $a = y$ and $b = x$, we have \begin{align*} -\left(\left(-y\right)x \right) = - \left( (-a)b\right) = - \left(-ab\right) = - \left(-yx \right). \end{align*} From here, the solution I'm trying to follow simply concludes that $-(-yx) = yx = xy$. The last step surely follows from commutativity, but I'm struggling why $-(-yx) = yx$, aside from a simple appeal to it being "obvious." In particular, the given ring is arbitrary, and we aren't given the fact that the ring has the property of multiplicative inverses. It seems to me that this step requires us to assume that $-1$ is its own inverse, which we could certainly prove, but doesn't that require that the ring is closed under inverses?

This seems somewhat trivial, but I'd grealty appreciate any insights on this, as I'm trying to follow the proof in full.

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  • $\begingroup$ $-(-yx)$ is a ring element $w$ such that $w + (-yx) = 0$ $\endgroup$
    – gd1035
    Commented Jul 26, 2018 at 18:53
  • $\begingroup$ Ring has two operations $+$ and $\times$. The $-r$ refers to the additive inverse of $r$, so $-(-r)$ is the (additive) inverse of the additive inverse of $r$, which is $r$ (based on group theory). $\endgroup$
    – Anurag A
    Commented Jul 26, 2018 at 18:53
  • $\begingroup$ "Multiplication in a ring is commutative"? $\endgroup$
    – saulspatz
    Commented Jul 26, 2018 at 18:55
  • $\begingroup$ The proof will be similar to this proof for field, the only thing you need is replace the "multiplicative is commutative" part by another lemma "$\forall y, y0 = 0$". $\endgroup$ Commented Jul 26, 2018 at 19:09

4 Answers 4

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It holds in any group (with additive notation) that $-(-a)=a$ -- and therefore especially in the additive group of a ring.

This is fast to see once you know that inverses in a group are unique: By definition $-(-a)$ is a solution to $x+(-a)=0$, but $a$ is a solution (by definition of $-a$), and because the solution is unique, $-(-a)$ must be the same as $a$.

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To show that $-1$ is its own inverse, observe that $0 = -1 \cdot 0 = -1(1 + -1) = -1 + (-1)^2$, so that $1 = (-1)^2$.

Now to show that $-yx = (-1)yx$, consider that $0 = 0(yx) = (1 + -1)(yx) = yx + (-1)yx$.

So $-(-yx) = (-1)((-1)yx) = (-1)^2yx = 1yx = yx$.

Fortunately none of this requires closure under inverses for the ring. Closure under inverses means that every nonzero element of the ring has a multiplicative inverse; a ring can have an element that has an inverse without that saying anything either way about whether the ring is closed under inverses. In fact, in every ring $1$ is its own inverse.

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You can use the distributive law twice

$$0=(a+(-a))\cdot b=a\cdot b +(-a)\cdot b$$ whence $$-((-a)\cdot b)=a\cdot b$$ then $$0=(-a)\cdot (b+(-b))=(-a)\cdot b+(-a)\cdot(-b)$$

whence $$-((-a)\cdot b)=(-a)\cdot(-b)$$

and it follows that $$a\cdot b=(-a)\cdot(-b)$$

Note that this does not assume that multiplication is commutative.

Since $-a$ is the additive inverse of $a$, and you are looking for a result involving multiplication, it is inevitable that there will be an actual or implied appeal to the distributive law, since this is the basic property which links addition and multiplication.

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  • $\begingroup$ It looks like he has already proved that $(-a)b=-ab$, though. So further explicit appeals to distributivity should not be necessary. $\endgroup$ Commented Jul 26, 2018 at 19:16
  • $\begingroup$ @HenningMakholm he says he is able to use that result or the basic properties of the ring. I thought this method with giving, though it is longer than necessary. Is it possible to avoid two appeals - explicit or implicit - to the distributive law (one for each side)? $\endgroup$ Commented Jul 26, 2018 at 19:41
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Just for fun, if you want it in one long chain of equalities:

$$(-a)(-b)=(-a)(-b)+0=(-a)(-b)+\Big(\big(-(ab)\big)+ab\Big)=$$ $$=\Big((-a)(-b)+\big(-(ab)\big)\Big)+ab=\big((-a)(-b)+(-a)b\big)+ab=$$ $$=(-a)\big((-b)+b\big)+ab=(-a)0+ab=0+ab=ab.$$

Try to identify at each step the axiom or property applied (I assumed as proven $(-a)b=-(ab)$ and $0a=a0=0$).

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