1. ## Prove a ring

Define a new operation of addition in Z by x(+)y = x + y -1 with a new multiplication in Z by x(*)y = x+y-xy. Verify that Z forms a ring with respect to these operations.

In order to show it is a ring, I must show that R forms an abelian group with respect to addition, R is closed with respect to an associative multiplication, and two distributive laws hold in R.

How should I start proving those 3 things.

for abelian group: x + y -1 = y + x -1 and x+y-xy = y+x-yx is this right?

for associative : (x+y) - xy = x + ( y - xy ) is this right?

for distributive law: z(x+y-1) = (x+y-1)z and z(x+y-xy) = (x+y-xy)z is this right?

2. Originally Posted by rainyice
Define a new operation of addition in Z by x(+)y = x + y -1 with a new multiplication in Z by x(*)y = x+y-xy. Verify that Z forms a ring with respect to these operations.

In order to show it is a ring, I must show that R forms an abelian group with respect to addition, R is closed with respect to an associative multiplication, and two distributive laws hold in R.

How should I start proving those 3 things.

for abelian group: x + y -1 = y + x -1 and x+y-xy = y+x-yx is this right?

for associative : (x+y) - xy = x + ( y - xy ) is this right?

for distributive law: z(x+y-1) = (x+y-1)z and z(x+y-xy) = (x+y-xy)z is this right?
To prove that this is an abelian group under + you must also prove that it is a group! You need to find the additive identity, prove that your group is closed (although it is as you cannot add or multiply two integers and not get an integer!) and prove it is associative.

Currently, you have only shown that addition is commutative (you also proved that multiplication is commutative which do not need to prove). The rest isn't proving anything...

To prove that addition (I have denoted this by $\circ$ as this is NOT the normal addition) is associative, you need to prove that

$(x\circ y)\circ z = x \circ (y\circ z)$ where $x \circ y = x+y-1$.

Does that make sense? Similarly, multiplication is different in this ring so you should use a different symbol for it.