Proof that a group is a group

Question

Hi all I'm stuck on a homework question. The question is as follows:

"For $a,b\in\mathbb R$ we define $a∗b:=a+b+ab\in\mathbb R$. Furthermore let $G =\mathbb R\setminus\{-1\}$.

Show that $G$ together with the binary operation $G × G → G, (a, b) → a ∗ b$, is a group"

I know I must show that it's associative, there exists a neutral element and there exists an inverse. So far I've managed to show it's associative and I think the neutral element is letting $b=0$ but I don't know what the inverse element would be so that $a*b=0$. Any help would be very much appreciated.

Your question's title is funny... – DonAntonio Feb 08 '17 at 10:57 — DonAntonio, Feb 08 '17 at 10:57
All of the deals excluding -1 – Thomas Feb 08 '17 at 10:57 — Thomas, Feb 08 '17 at 10:57
All of the reals excluding -1 – Thomas Feb 08 '17 at 10:58 — Thomas, Feb 08 '17 at 10:58

DonAntonio · Accepted Answer · 2017-02-08T11:08:41.547

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Hint: remember that $\;a\neq-1\;$:

$$0=a*a':=a+a'+aa'=a+(1+a)a'$$

edited Feb 08 '17 at 11:08

answered Feb 08 '17 at 10:59

DonAntonio

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More exactly, $a\ne -1$. – Bernard Feb 08 '17 at 11:02
Of course how'd I miss that – Thomas Feb 08 '17 at 11:06
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@BernardGood catch on that typo, thanks. Edited. – DonAntonio Feb 08 '17 at 11:09

score 1 · Answer 2 · answered Feb 08 '17 at 11:05

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Finding the inverse to a given $a\in G$ corresponds to finding the solution of the equation $a+x+ax=0$ with $a\in G$ being a parameter and $x \in G$ the unknown, so you get:

$a+x+ax=0$

$x(1+a)=-a$

$x= \frac{-a}{1+a}$.

answered Feb 08 '17 at 11:05

pepa.dvorak

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score 1 · Answer 3 · answered Feb 08 '17 at 11:06

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Hint:

Don't forget you have to show this is an internal law.

Rewriting $a*b$ as $(a+1)(b+1)-1$, finding a inverse amounts to solving $$(a+1)(b+1)=1.$$

answered Feb 08 '17 at 11:06

Bernard

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Proof that a group is a group

3 Answers3