# Thread: Algebra, Properties of a Ring

1. ## Algebra, Properties of a Ring

Does anyone know how to prove this:

Suppose that R is a ring. Show that -(a+b) = (-a)+(-b) for a, b in R.

Thanks guys.

2. Originally Posted by suedenation
Does anyone know how to prove this:

Suppose that R is a ring. Show that -(a+b) = (-a)+(-b) for a, b in R.

Thanks guys.
Actually, it isn't unless the ring is commutative. In general the inverse of (a+b) is (-b)+(-a).

Calculate the following:

(a+b)+[(-b)+(-a)]=a+b+(-b)+(-a) by the associative property
=a+[b+(-b)]+(-a) again by associative
=a+0+(-a) by definition of inverse
=a+(-a) by the zero property
=0 by definition of inverse.

Thus (a+b)+[(-b)+(-a)] = 0.
Since the inverse of any element in the ring is unique, (-b)+(-a) = -(a+b).

-Dan

3. Originally Posted by topsquark
Actually, it isn't unless the ring is commutative. In general the inverse of (a+b) is (-b)+(-a).

Calculate the following:

(a+b)+[(-b)+(-a)]=a+b+(-b)+(-a) by the associative property
=a+[b+(-b)]+(-a) again by associative
=a+0+(-a) by definition of inverse
=a+(-a) by the zero property
=0 by definition of inverse.

Thus (a+b)+[(-b)+(-a)] = 0.
Since the inverse of any element in the ring is unique, (-b)+(-a) = -(a+b).

-Dan
It is commutative, a ring has a property that the binary operation must be commutative. However, a commutative ring is one in such multiplication is also.

4. Pffl! That's three posts in a row I've got at least a detail wrong. I'm slipping!

Good catch, ThePerfectHacker.

-Dan