Can commutativity of multiplication and addition under real numbers be assumed?

[sapnpf6]

Homework Statement

so.. let the operation * be defined as x*y = x + y + xy for every x,y ∈ S,
where S = {x ∈ R : x ≠ -1}. Now i have proven associativity, existence of an identity and inverses, all without commutativity, but i must show that this is an abelian group, so now i have to show commutativity. I know that multiplication is not always commutative, but i am not using matrices, just reals/{-1}.

Homework Equations

the operation * is defined by x*y = x + y + xy.

The Attempt at a Solution

i simply say that first, addition is commutative on the reals, so x+y=y+x, by the commutative law of addition of real numbers, and second that xy=yx, by the commutative law of multiplication of real numbers. this would show that x*y=y*x. Is this allowed or do i have to show commutativity without assuming addition and multiplication are commutative on real numbers? and if i have to show commutative without the assumptions above, can someone point me in the right direction on the proof?

 

[sapnpf6]

in case i did not show enough of an attempt at a solution, here is a more detailed attempt summary...

the operation * is commutative if for every x,y∈{x∈R:x≠ -1} such that x*y = y*x.
But, x*y = x + y + xy, and
y*x = y + x + yx,
= y + x + xy (by commutative law of multiplication of real numbers)
= x + y + xy (by commutative law of addition of real numbers)
So x*y = y*x, and thus, * is commutative.

this is my original solution but i am not entirely sure whether it holds up. it seems like using the law of commutativity of multiplication/addition of real numbers might be too "flimsy" to use in this kind of proof. it would sure be great if i could hear some feedback.

 

[VeeEight]

You are correct in your reasoning. If x*y = x+y + xy, then y*x = x+y +yx, which holds since addition and multiplication are commutative in the reals.

 

[sapnpf6]

so it is not necessary to prove commutativity of addition or multiplication in real numbers here? just making sure...

 

[HallsofIvy]

No, you do not have to "reinvent the wheel" for every problem. You do not have to prove the properties, that have already been proven, for the "usual" operations on the real numbers. if you are given some new operation, defined using the "usual" operations, you can assume all of the properties of the "usual" operations in order to prove (or disprove) those properites for the new operation.