# Prove (-x)y=-(xy) for rings

1. Nov 27, 2011

### ArcanaNoir

1. The problem statement, all variables and given/known data

I need to show that (-x)*y=-(x*y) for a ring. unless it's not true.

2. Relevant equations

A ring is a set R and operations +, * such that (R, +) is an Abelian group, * is associative, and a*(b+c)=a*b+a*c and (b+c)*a=b*a+c*a.

3. The attempt at a solution

I don't know what the first step of this proof will be, I'm looking for the *trick*, as it were.

2. Nov 27, 2011

### I like Serena

If we want to know if (-x)*y is the additive inverse (x*y), what is the axiom that we should check?

3. Nov 27, 2011

### ArcanaNoir

I don't know. maybe if I knew what axiom to check I'd have a better idea how to prove it?

4. Nov 27, 2011

### I like Serena

You wrote that (R,+) is an abelian group.
What is the definition of an inverse in a group?

5. Nov 27, 2011

### ArcanaNoir

-(a+b)= (-b)+(-a)
right?

6. Nov 27, 2011

### HallsofIvy

Staff Emeritus
That not a definition, is it? I was under the impression that "-x" was the additive inverse of x: x+ (-x)= 0. So to prove that (-x)y= -(xy), you need to show that (-x)y is the additive inverse of xy: that (-x)y+ (xy)= 0.

7. Nov 27, 2011

### I like Serena

Let's try to find the definition of a ring, and in particular the definition of the additive inverse.

On wikipedia there is an article on a ring.
It lists the requirements (aka axioms) of a ring:
http://en.wikipedia.org/wiki/Ring_math#Formal_definition

In particular we have:
Btw, the typical shorthand for the additive inverse of $a$ is $-a$.

Does this look familiar?

8. Nov 27, 2011

### ArcanaNoir

(-x)y+xy=((-x)+x)y=0y=0
ta da?

9. Nov 27, 2011

### I like Serena

Yes!
That's basically it.

Just a couple of things.

How do you know that 0y=0?

And the definition requires that a+b=b+a=0.
Do both equalities hold?

10. Nov 27, 2011

### ArcanaNoir

well I did check that 0y=0, and a+b should equal b+a because its an abelian group. yay :) thanks again!

11. Nov 27, 2011

### I like Serena

You're making this a bit easy on yourself, aren't you?
How did you check this?
If it is so simple, you should be able to easily reproduce the proof...
(It is not trivial!)

Yep!