Proof: let $A$ a ring, then $(-a) \cdot (-b) = a \cdot b $ $\forall a,b \in A$

Question

I must prove this property:

Property: let $A$ a be ring, then $(-a) \cdot (-b) = a \cdot b $, $\forall a,b \in A$.

Proof: let $a \in A$ and $b \in A$, by hypothesis $A$ is a ring then $a \cdot 0=0$ and $b + (-b) = (-b) + b= 0$ and $(-b) \in A $, therefore $a \cdot (b + (-b))=0$, but by hypothesis $\cdot$ is distributive then $a \cdot (b + (-b))=a \cdot b + a \cdot (-b) =0$, therefore $-(a \cdot (-b))=a \cdot b$, but in a ring is true that $-(c \cdot d)=(-c) \cdot d$, $\forall c,d \in A$, therefore we have $(-a) \cdot (-b)=-(a \cdot (-b))=a \cdot b$.

It is correct?

Thanks in advance!!

Pedro · Accepted Answer · 2013-04-15T02:04:59.407

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In light of MathGem's answer, let's look at your statement again. You said, "in a ring it is true $-(a\cdot b)=(-a)\cdot b$." What you're saying is that $$\tag 1 a\cdot b+a\cdot (-b)=0$$

which is true.

Thus, you in turn claim that in a ring

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

which is true. Then, it follows

$$\tag 2 (-a)\cdot (-b)+a\cdot (-b)=0$$

But since inverses are unique $(1),(2)$ imply $a\cdot b=(-a)\cdot(-b)$.

There is a nice proof, that goes as follows:

Look at the expression $$a\cdot b+a\cdot (-b)+(-a)\cdot(-b)$$ and use the distributive laws in two different ways. One would be the above is$$a\cdot b+a\cdot (-b)+(-a)\cdot(-b)= \\a\cdot (b+(-b))+(-a)\cdot(-b)=\\a\cdot 0+(-a)\cdot (-b)=\\ 0+(-a)\cdot (-b)=\\(-a)\cdot (-b)$$

What is the other?

edited Apr 15 '13 at 02:04

answered Apr 15 '13 at 00:53

Pedro

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okok but my process is correct?? Your proof is interesting!!! Thanks soo much!! – mle Apr 15 '13 at 00:59
@GarnakOlegovitc Sorry, yes. Your proof looks OK. – Pedro Apr 15 '13 at 00:59
$a \cdot b + a \cdot (-b) + (-a) \cdot (-b) = a \cdot b + (a + (-a)) \cdot (-b) = a \cdot b + 0 \cdot (-b) = a \cdot b $ Correct? – mle Apr 15 '13 at 01:04
@GarnakOlegovitc Correct. All in all, $(-a)\cdot(-b)=a\cdot b$. – Pedro Apr 15 '13 at 01:04
I don't think Math Gems is Bill, Peter. The registration date on the account predates Bill's account being suspended. Although their 'styles' are quite similar... – Potato Apr 15 '13 at 02:06
@Potato I've seen him link to Bill's post quite a few times. – Pedro Apr 15 '13 at 02:08
@PeterTamaroff It's suspicious, I agree. But this would mean Bill had an alternate account long before he got hit with the yearlong suspension, and I have no idea why he would do that. However, it's very interesting that their answer styles are virtually identical... – Potato Apr 15 '13 at 02:10
@PeterTamaroff I have wondered about this for a while now, to be honest. – Potato Apr 15 '13 at 02:11
@Potato Here he links to a search result of Bill's answers. – Pedro Apr 15 '13 at 02:16
Here again. – Pedro Apr 15 '13 at 02:21
@PeterTamaroff It's almost certainly him, then. Bizarre. – Potato Apr 15 '13 at 02:21
@Potato I'm glad he is not gone, but is Gone! – Pedro Apr 15 '13 at 02:22
They're both ring theorists, too. It is him. – Potato Apr 15 '13 at 02:22

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

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There is an error. The statement "but in a ring is true..." begs the principle. Instead, let $\rm\:b = -c\:$ in what you just proved. More explicitly, you have proved $\rm\:x(-\color{#c00}y) = -(xy)\:$ for all $\rm\:x,y.\:$ Thus, applying this twice we deduce $\rm\ (-a)(-\color{#c00}b) = -((-a)b) = -(b(-\color{#c00}a)) = -(-(ba)) = ba = ab.$

A more conceptual way to view the proof is that $\rm\:(-a)(-b)\:$ and $\rm\:ab\:$ are both inverses of $\rm\:(-a)b,\:$ hence they are equal by uniqueness of inverses.

Equivalently, evaluate $\ \rm\overline{(-a)(-b)\! \,+\,} \overline{ \underline {(-a)b}} \underline{\,+\,ab_{\phantom{_{1}}}}\:$in two ways, over or underlined first.

See also prior posts on this Law of Signs in Rings.

edited Apr 13 '17 at 12:21

Community

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answered Apr 15 '13 at 01:47

Math Gems

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Why do $a,b$ get swapped in your first paragraph? – Pedro Apr 15 '13 at 01:59
After writing out the OPs statement, and hoping I interpret it alright, it doesn't look wrong. – Pedro Apr 15 '13 at 02:06
@Peter So one may apply the law just proved, which pulls the negation out of the second argument of a product. Alternatively, one could extend the law by using commutativity, i.e. extend it to $\rm: −(xy)=x(−y)=(−y)x,:$ so one can pull the negation out of either argument. – Math Gems Apr 15 '13 at 02:15
@Peter Without any proof of the claim "but in a ring it is true that P(c,d)" the proof is either incomplete or circular. That the OP thinks this statement does not require proof indicates a possible conceptual error, since that statement is equivalent to the statement he just proved. – Math Gems Apr 15 '13 at 02:17
True. But as I see it, he observes $a\cdot b+(-a)\cdot b=0$ and $c\cdot d+c\cdot (-d)=0$, then lets $c=-a$, $d=-b$. – Pedro Apr 15 '13 at 02:19
@Peter The "but in a ring..." seems to indicate otherwise. How do you explain its purpose in your view of the proof? – Math Gems Apr 15 '13 at 02:22
When he says "but in a ring..." he's asserting that $c\cdot d+c\cdot (-d)=0$, as I see it. That's all. I think, however, we should let him explain what he really means, instead filling this thread with comments that might lead nowhere. – Pedro Apr 15 '13 at 02:25
@MathGems, if you want I can proof "$-(c \cdot d)=(-c) \cdot d, \forall c,d \in A$"!!! ;) – mle Apr 15 '13 at 16:26
@Garnak It is puzzling why you thought that you did not need to prove/justify that statement, esp. since you already did that for a closely related statement. Why did you think that? – Math Gems Apr 15 '13 at 16:46
@Gemme Math, sorry.. i do not read all comments... i think that to proof this property... !!! Now, I understand!! ;) – mle Apr 15 '13 at 17:54

Proof: let $A$ a ring, then $(-a) \cdot (-b) = a \cdot b $ $\forall a,b \in A$

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