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

  • Thread starter ArcanaNoir
  • Start date
  • #1
768
4

Homework Statement



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

Homework 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.

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.
 

Answers and Replies

  • #2
I like Serena
Homework Helper
6,577
176
If we want to know if (-x)*y is the additive inverse (x*y), what is the axiom that we should check?
 
  • #3
768
4
I don't know. maybe if I knew what axiom to check I'd have a better idea how to prove it?
 
  • #4
I like Serena
Homework Helper
6,577
176
I don't know. maybe if I knew what axiom to check I'd have a better idea how to prove it?
You wrote that (R,+) is an abelian group.
What is the definition of an inverse in a group?
 
  • #5
768
4
-(a+b)= (-b)+(-a)
right?
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
961
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
I like Serena
Homework Helper
6,577
176
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:
wiki on Ring said:
4. Existence of additive inverse. For each a in R, there exists an element b in R such that a + b = b + a = 0
Btw, the typical shorthand for the additive inverse of [itex]a[/itex] is [itex]-a[/itex].

Does this look familiar?
 
  • #8
768
4
(-x)y+xy=((-x)+x)y=0y=0
ta da?
 
  • #9
I like Serena
Homework Helper
6,577
176
(-x)y+xy=((-x)+x)y=0y=0
ta da?
Yes! :smile:
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
768
4
well I did check that 0y=0, and a+b should equal b+a because its an abelian group. yay :) thanks again!
 
  • #11
I like Serena
Homework Helper
6,577
176
well I did check that 0y=0,
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!)


and a+b should equal b+a because its an abelian group. yay :) thanks again!
Yep!
 

Related Threads on Prove (-x)y=-(xy) for rings

Replies
4
Views
3K
  • Last Post
Replies
16
Views
14K
  • Last Post
2
Replies
33
Views
9K
Replies
3
Views
11K
  • Last Post
Replies
4
Views
924
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
9K
Replies
3
Views
787
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
9
Views
2K
Top