# Question in rings

• Oct 19th 2010, 04:04 AM
raed
Question in rings
Hi All:
I need the solution of the following question:

Let (R,+,.) satisfy all conditions of a ring except the condition of a+b=b+a for all a ,b in R ( abelian with +); If 1 in R (R is arig with unity) prove that a+b=b+a for all a,b in R.

Best wishes
• Oct 19th 2010, 05:03 AM
tonio
Quote:

Originally Posted by raed
Hi All:
I need the solution of the following question:

Let (R,+,.) satisfy all conditions of a ring except the condition of a+b=b+a for all a ,b in R ( abelian with +); If 1 in R (R is arig with unity) prove that a+b=b+a for all a,b in R.

Best wishes

Existence of additive inverse: $(a+b)+(-(a+b))=0$

But also $a+b+(-b-a)=a+(b+(-b))+(-a)=a+0+(-a)=a+(-a)=0$ , so by uniqueness of inverse

$-b-a = -(a+b)=-a-b$ .

Justify each step above and end the argument.

Tonio
• Oct 19th 2010, 05:30 AM
raed
Quote:

Originally Posted by tonio
$a+b+(-b-a)=a+(b+(-b))+(-a)=a+0+(-a)=a+(-a)=0$ , so by uniqueness of inverse

$-b-a = -(a+b)=-a-b$ .

Tonio

Thank for your reply but this ring is not a belian under additive is it true that $-(a+b)=-a-b=-b-a$
• Oct 19th 2010, 07:56 AM
tonio
Quote:

Originally Posted by raed
Thank for your reply but this ring is not a belian under additive is it true that $-(a+b)=-a-b=-b-a$

I didn't say it was: that's what you must prove! But you said that "ring" fulfills all the other ring axioms, so there's distributivity:

$-(a+b)=(-1)(a+b)=(-1)a+(-1)b=-a-b$ . That this equals $-b-a$ follows from uniqueness of the additive inverse.

Tonio
• Oct 19th 2010, 11:35 PM
raed
Quote:

Originally Posted by tonio
I didn't say it was: that's what you must prove! But you said that "ring" fulfills all the other ring axioms, so there's distributivity:

$-(a+b)=(-1)(a+b)=(-1)a+(-1)b=-a-b$ . That this equals $-b-a$ follows from uniqueness of the additive inverse.