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The set $\mathbb{Q1}$ of all rational numbers other than 1 will form a group under the operation $a*b=a+b-ab$

If we consider another set $\mathbb{Q}$ of all rational numbers.Will the set of rational numbers $\mathbb{Q}$ form a group under the same operation? Looks like This set satisfies the properties of group under the above operation !

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  • $\begingroup$ I don't think I am correct. That's why I posted it. There must be something wrong. $\endgroup$ – Kangkan Oct 5 '17 at 4:36
  • $\begingroup$ I can not really find a proper question. $\endgroup$ – Cornman Oct 5 '17 at 4:38
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    $\begingroup$ What's the inverse of $1$? $\endgroup$ – Lord Shark the Unknown Oct 5 '17 at 4:40
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    $\begingroup$ Incidentally, note that $(1-a) * (1-b) = (1-ab)$ $\endgroup$ – Hurkyl Oct 5 '17 at 6:23
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What is the inverse of $1$ under this group operation? There isn't one: $1 * b = 1+b-b=1 $ for any $b$.

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Well. Now I figured it. For any two element a & its inverse b in Q1 we will get $a*b=0$ as 0 is its identity element. This will give $a+b-ab=0$ and $b=a/(a-1)$. So if the set consists 1 then inverse will not exist.

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This operation is simply multiplication on $\mathbb{Q}$ under the transformation $\phi:x\mapsto 1-x$. Indeed: $\phi(a)\phi(b)=(1-a)(1-b)=1-(a+b-ab)=\phi(a*b)$. Since $1=\phi(0)$, asking whether we can include $1$ in this group is precisely the same as asking whether we can include $0$ in the multiplicative group of the rationals. Clearly, we cannot.

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