I have a task to show that $G:= \mathbb Q\setminus \{-1\}$ is a group to some binary operation *

(x is the "times" symbol) with $A*B:=A\times B + A + B$

I could show that it's associative, that the neutral Element is $0$, but I'm stuck finding the inverse Element...

Maybe I started in the wrong place and don't see the wood for the trees, but I just found a Element for $1$ with $A=1$ and $B=-1/2$

so that $A*B=[1 \times (-1/2)]+1 - 1/2 = 0$

but because of the one you can't derive a general rule from that

but is there a general inverse Element??

  • $\begingroup$ Hint: $G\to\mathbb Q\setminus\{0\}$ $a\mapsto x+1$ is a group homomorphism. $\endgroup$ – Hagen von Eitzen Oct 19 '14 at 19:46
  • $\begingroup$ You have to define some things here: what is that Q, what then does $\;Q\{-1\}\;$ mean, what is a "linkage operation" ...? $\endgroup$ – Timbuc Oct 19 '14 at 19:49
  • $\begingroup$ ok i see....it escaped the \ ....sorry $\endgroup$ – R-Carver Oct 19 '14 at 19:53
  • $\begingroup$ The English term you're looking for to mean "linkage operation" is "binary operation." I don't know if I can help with the rest of your question, though. $\endgroup$ – apnorton Oct 19 '14 at 20:01
  • $\begingroup$ thank you...ill remember that :-) $\endgroup$ – R-Carver Oct 19 '14 at 20:12

If the neutral element is $0$ and you want to find the inverse of $a \in G$, that means you want to find $B$ such that $0 = a*b$. This implies:

$0 = a*_{\scriptsize G} b = a\times_{\scriptsize \mathbb Q} b+_{\scriptsize \mathbb Q}a+_{\scriptsize \mathbb Q}b = (a+_{\scriptsize \mathbb Q}1)\times (b+_{\scriptsize \mathbb Q}1) -_{\scriptsize \mathbb Q} 1$

$\iff 1 = (a+_{\scriptsize \mathbb Q}1)\times_{\scriptsize \mathbb Q}(b+_{\scriptsize\mathbb Q}1)$

$\iff 1/_{\scriptsize \mathbb Q}(a+_{\scriptsize \mathbb Q}1) -_{\scriptsize \mathbb Q} 1 = b$

$ \qquad \qquad \qquad \qquad \qquad \:\:= -_{\scriptsize \mathbb Q}a/_{\scriptsize \mathbb Q}(a+_{\scriptsize\mathbb Q}1)$

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  • $\begingroup$ Corrected it, I made a mistake. $\endgroup$ – flawr Oct 19 '14 at 20:07
  • $\begingroup$ It seems to me you are multiplying by the inverse of $(a+1)$, but isn't part of the problem to determine the inverse of an arbitrary element? $\endgroup$ – Clayton Oct 19 '14 at 20:14
  • $\begingroup$ @Clayton Your confusion arises from mixing up the two multiplication operations. $\endgroup$ – user714630 Oct 19 '14 at 20:15
  • $\begingroup$ thank you very much flawr $\endgroup$ – R-Carver Oct 19 '14 at 20:19
  • $\begingroup$ @Clayton I now denoted each operation with the corresponding algebraic structure. $\endgroup$ – flawr Oct 19 '14 at 20:26

Hint: What happens when you expand $(a+1)(b+1)$?

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  • $\begingroup$ thanks...afterwards it's so simple $\endgroup$ – R-Carver Oct 19 '14 at 20:23

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