Need help with Associative proof

Question

Let$ G = \{x ∈ R | x \neq -1\} $and * a link to ¨G with $x * y: = x + y + xy.$ Show that $(G, *)$ is a group.

To determine if $ G$ is a group I have to make the associative proof:

Associative?

$(x * y) * z = (x + y +xy)+z -(x + y +xy)*z$

$= x + y + z + xy - xz - yz - xzy$

$= x + y + z + xy -z (x + y + xy)$

$= x + (y + z + xy) -z (x + y + xy)$

I kinda stop here, because I think I made a mistake

Please write $x\neq 1$ and not x!=-1. – Dietrich Burde Nov 28 '18 at 17:56 — Dietrich Burde, Nov 28 '18 at 17:56
You have flipped a sign in the first line of your algebra. – Doug M Nov 28 '18 at 18:06 — Doug M, Nov 28 '18 at 18:06

Cloud JR K · Answer 1 · 2018-11-28T18:10:19.077

1

I don't know from where minus sign from came in your solution . If you're a beginner one easy way is to make seperate equation for LHS and RHS and show they are equal.

$(x*y)*z=(x+y+xy)*z=x+y+xy+z+xz+yz+xyz$ ....(1)

Now consider ,

$x*(y*z)=x*(y+z+yz)= x+y+z+yz +xy+xz+xyz$ ...(2)

from (1) and (2) and using the fact real numbers commute, you can prove associative

edited Nov 28 '18 at 18:10

answered Nov 28 '18 at 18:03

Cloud JR K

2,466

score 1 · Answer 2 · answered Nov 28 '18 at 18:17

It's best not try to force $(x*y)*z = .... = something = .... x*(y*z)$, but to instead do $(x*y)*z = .... = something$ and $x*(y*z) = ... = the same something$.

$(x*y)*z = (x*y) + z + (x*y)z = x + y + xy + z + (x+ y + xy)z = \color{blue}{x + y + z + xy + xz + yz + xyz}$.

And $x*(y*z) = x + (y*z) + x(y*z) = x + y + z+ yz + x(y + z + yz) = \color{blue}{x + y + z + xy + xz + yz + xyz}$.

So $(x*y)*z = \color{blue}{x + y + z + xy + xz + yz + xyz}= x*(y*z)$

Now if you wanted to do it in one line you could with some factoring:

$(x*y)*z = (x*y) + z + (x*y)z = x + y + xy + z + (x+ y + xy)z = x + y + z + xy + xz + yz + xyz= x + y + z+ yz + x(y + z + yz)=x + (y*z) + x(y*z)= x*(y*z)$.

But why kill our brain cells and give us a headache? (Okay, I know. A single equation is more idealist but... let's get practical.)

J.G. · Answer 3 · 2018-11-28T18:54:17.333

The fact that $\Bbb R\backslash\{-1\}$ is a group under $(x+1)(y+1)-1$ is equivalent to the fact that $\Bbb R\backslash\{0\}$ is a group under $xy$; the two groups are isomorphic, since the invertible function $\sigma (x):=x+1$ of inverse $\sigma^{-1}(x):=x-1$ satisfies $x\ast y:=\sigma^{-1}(\sigma(x)\sigma(y))$. In particular, you can prove the associativity in terms of this isomorphism without any messy arithmetic: $$(x\ast y)\ast z=\sigma^{-1}(\sigma(x\ast y)\sigma(z))=\sigma^{-1}(\sigma(x)\sigma(y)\sigma(z))=x\ast (y\ast z).$$

Need help with Associative proof

3 Answers3