consider all the rationals except 1 with the binary operation a*b=a+b-ab for any a,b elements in Q\ {1}.

show that this is a group.

why is 1 excluded

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- Oct 23rd 2009, 03:06 PMbettywhitgroup theory abstract algebra
consider all the rationals except 1 with the binary operation a*b=a+b-ab for any a,b elements in Q\ {1}.

show that this is a group.

why is 1 excluded - Oct 23rd 2009, 03:39 PMproscientia
Verify that $\displaystyle \left\langle\mathbb Q\setminus\{1\},\star\right\rangle$ satisfies all the group axioms. (Do you know what the group axioms are?) Hint for identity: $\displaystyle a\star0=0\star a=\ldots\,?$

This group is isomorphic to $\displaystyle \left\langle\mathbb Q\setminus\{0\},\times\right\rangle$ (the group of nonzero rationals under ordinary multiplication) via the mapping $\displaystyle a\mapsto\frac1{1-a}$ for $\displaystyle a\in\mathbb Q\setminus\{1\}.$ - Oct 25th 2009, 03:04 PMbettywhit
- Oct 25th 2009, 03:23 PMtonio
- Oct 26th 2009, 04:41 PMproscientia
Something is wrong here.

$\displaystyle a\star1\ =\ a+1-a\cdot1=a+1-\color{red}a\color{black}=\color{red}1$

You just**cannot**have two identities, even in a semigroup! Including $\displaystyle 1$ would make $\displaystyle \left\langle\mathbb Q,\star\right\rangle$ a semigroup rather than a group (the element $\displaystyle 1$ will not have an inverse). If a semigroup has an identity, then that identity is always unique. - Oct 26th 2009, 04:54 PMsfspitfire23
If 1 was included, the inverse would also be undefined,