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i need some tips on how to start searching for inverses on a group structure, I seem to have a difficulty with actually finding an inverse when the operation is specified for example $$(Q\backslash \{-1\},*), \ \ a*b=a+b+ab$$ the first thing i think is the algebraic inverse or rather $\frac{-p}{q}$ but in cases like this that the operation is specified and "new" i am lost on how to find it, i know that $\frac{-p}{q}$ in this case wouldn't work for the exclusion of -1.

Are there any tips on how to proceed when proving inverse?

I know that in this example that 0 is the identity element but i don't know how to show this in more abstract terms: $$a*a^{-1}=\frac{p}{q}+\frac{r}{s}+\frac{pr}{qs}=0$$ i've tried putting them togheter like this $$\frac{ps+rq+pr}{qs}=0$$ or $$\frac{ps+rq}{qs}=-\frac{pr}{qs}$$

but i still am lost

If this question is inappropiate please let me know so i can delete it

Thanks in advance!

ancient mathematician
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Ramiro genta
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  • If this question is inappropiate please let me know so i can delete it --- Seems fine to me (but I don't know why you would think to tag this with algebraic-number-theory; didn't you look at the tag's description and some questions with that tag?), but maybe it would help to indicate why, when you set up the equations that the inverse is supposed to satisfy, what is holding you back: First find the identity by finding an element $e$ such that for each element $a$ we have $ae = ea = a,$ then solve for the inverse of an element $a$ by solving for $x$ in $ax=xa=e.$ – Dave L. Renfro Mar 15 '21 at 16:05
  • For more practice on this type of problem, see my answer to Groups of real numbers. – Dave L. Renfro Mar 15 '21 at 16:08
  • You say the identity is $0$. So $a^{-1}$ is that element $b$ for which $a+b+ab=0$. Can you solve that for $b$? – ancient mathematician Mar 15 '21 at 16:32
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    Probable duplicate of https://math.stackexchange.com/questions/513045/need-to-prove-that-s-cdot-defined-by-the-binary-operation-a-cdot-b-aba – lhf Mar 15 '21 at 16:55

1 Answers1

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I don't know what $Q/[-1]$ means, so I'll just call it $G$.

Step 1: find the identity element $-$ look for an element $e\in G$ such that $a*e = e*a = a$ for all $a$.

Step 2: given $a\in G$, look for $b\in G$ such that $a*b=e$ (for the right inverse) or $b*a=e$ (for the left inverse). Your group is commutative, so these are the same.

That's it!

TonyK
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  • Sorry, i don't know how to write it but $Q/[-1]$ is the set of rational numbers with the exception of -1. I tried doing it and found that the inverse is the number 0, however the problem still comes when i try to find an inverse, because i know it should equal 0 and can't find examples in a general way, as to write them $\frac{r}{s}. \forall r,s \in\mathbb{R}$ – Ramiro genta Mar 15 '21 at 16:14
  • Yes i'm sorry i meant that 0 is the identity$a*e=\frac{p}{q}+0+\frac{p}{q}0= \frac{p}{q} =a$ – Ramiro genta Mar 15 '21 at 16:32