How can you prove this using only the ring axioms?

[dndod1]

Homework Statement

Using only the ring axioms, prove that in a general ring (R, +,X)
aX (x-z) = (aXx)- (aXz) where all a,x,z are elements of R

Homework Equations

Group axiom 3: G3= There is an inverse for each element g^-1 *g =e

Ring axiom 3: R3= Two distributive laws connect the additive and multiplicatie structures.
For any x,y,z xX(y+z) = (xXy)+ (xXz)
and (x+y) X z= (xXz) + (yXz)

The Attempt at a Solution

My attempt. I thought that this would actually be straight forward; that I would just need to put -z as the addition of its inverse. I expected the rest to just fall into place.

Here's what I did:

aX (x-z) = (aXx)- (aXz)


Left hand side aX (x-z)
= aX(x + z^-1) from G3
= (aXx) + (aXz^-1) from R3
= (aXx)+ (aX -z) from G3
= (aXx)- (aXz), as required


I'm not sure whether I am allowed to just write the last line or whether I have left out some all important step!

Thank you very much in anticipation of your assistance.

 

[Dick]

You are asking why aX(-z)=(-aXz)? Well, aX(-z)+aXz=aX(-z+z) from your distributive axiom. What does that tell you?

 

[dndod1]

Thank you Dick. I got there! Thank you for your extremely quick reply. I really appreciate your help. Many thanks.

 

[Hurkyl]

For the record, when the group operation is x+y, the inverse of x is usually written as -x.

 

[dndod1]

Thank you! I shall alter my notation. Much appreciated!

 

[Dick]

For the record, when the group operation is x+y, the inverse of x is usually written as -x.

True, it is kind of confusing to be mixing the '*' notation for the group operation with the ring notation of '+' for the group operation, but dndod1 seemed to be dealing with it ok.