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Given group $x*y = x+y+xy, x\in \mathbb{R}: x \ne -1,$ where $xy$ denotes the usual product of two real numbers, find $2^{-1}*6*5^{-1}.$ Hence, solve the equation $2*x*5=6.$

First need find the identity element. Let there be a real number $m,$ and let its inverse be $m^{-1}.$

So, have$m*m^{-1} = m + m^{-1} + e =e \implies m = - m^{-1}.$ Hence, identity element is $0.$

So, $m*m^{-1} = m + m^{-1} + mm^{-1} = e \implies m^{-1}(1+m) = e-m$ $ \implies m^{-1} = \frac{e-m}{1+m}.$
This means that for $m^{-1}$ to exist, need discard $-1$ from the group.
Hence, need find the identity element, by some other equation.
Say, $\forall m\in G, m*e = m + e +me = m \implies e(1+m) = 0 \implies e=0,$ as $m\ne -1,$ as proved above.
Also, want to add that anyway a single value of $m$ cannot be a solution to finding the inverse for any element $m,$ in the group.

The reason for $-1$ not being allowed in the group, is given here.

Hence, $2^{-1} =-\frac{2}{3}, 5^{-1} = -\frac{5}{6}.$ Hence, $2^{-1}*6*5^{-1} = (-\frac{2}{3})*6*(-\frac{5}{6}),$ and by the associativity property of the group, have $(-\frac{2}{3})*(6*(-\frac{5}{6})) = ((-\frac{2}{3})*6)*(-\frac{5}{6}).$

$(-\frac{2}{3})*(6*(-\frac{5}{6})) = -\frac{2}{3}*(6-\frac{5}{6}-5) = -\frac{2}{3}*(1-\frac{5}{6}) $
$= -\frac{2}{3}*\frac{1}{6} = -\frac{2}{3}+\frac{1}{6}-\frac{2}{3}\frac{1}{6}= -\frac{2}{3}+\frac{1}{6}-\frac{1}{9}$
$= \frac{-12+3-2}{18}=-\frac{11}{18}$


$((-\frac{2}{3})*6)*(-\frac{5}{6}) = (-\frac{2}{3}+6-4)*(-\frac{5}{6}) = (-\frac{2}{3}+2)*(-\frac{5}{6})$
$= (\frac{4}{3})*(-\frac{5}{6}) = \frac{4}{3}-\frac{5}{6}-\frac{10}{9} = \frac{24-15-20}{18} = -\frac{11}{18}.$


Need solve the equation $2*x*5 = 6,$ for which I assume that as the term $xy$ follows the usual rule of multiplication for real numbers, so can write it also as $yx.$

Using $(2*x)*5 = (2+x+2x)*5$ get:
$=> (2+3x)*5 = 6 => 2+ 3x+5+10+ 15x = 6 => 17+ 18x = 6$
This yields $17 +18x = 6 => x = -\frac{11}{18}.$
Testing by substituting this value in the given equation, get:
$(2*-\frac{11}{18})*5 = 6$
LHS:
(i):$(2-\frac{11}{18}-\frac{11}9)*5 = \frac16*5 = \frac16+5+\frac56= 6$
(ii): $2*(-\frac{11}{18}*5) = 2*(-\frac{11}{18}+5-\frac{55}18) = 2*(\frac43) = 2+\frac43-\frac83= 6$
RHS: 6
Match.

Also, by computing using $2*(x+5+x5) =6 => 2*(x+5+5x) =6;$ get
$(2+x+2x+5+10+5x+10x) = 6 => 17 + 18x =6 => 18x = -11 => x = \frac{-11}{18}.$
//same as above, i.e. associativity applies.

Edited above, as per comments, and answer by Kan't

jiten
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    I think you have a confusion in one of your preliminary calculations: you write $m * m^{-1} = m + m^{-1} +e,$ where this $e$ term presumably comes from the $mm^{-1}$ term in the expression for $mm^{-1}$. But in the expression $x y = x+y +xy,$ the $xy$ term is just the regular multiplication of $x$ and $y$. – ckefa Jan 21 '25 at 13:55
  • @ckefa Should I delete this question then, and present anew in the light of this. – jiten Jan 21 '25 at 13:59
  • @ckefa Then, have $m*m^{-1} = m+m^{-1} +mm^{-1} =e \implies m^{-1}(1+m) = e-m \implies m^{-1} = \frac{e-m}{1+m}.$ That helps only to prove why $m\ne 1,$ but how to find the value of $e?$ – jiten Jan 21 '25 at 14:06
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    I think similar questions have been answered on this site before: (like https://math.stackexchange.com/questions/1971224/let-g-mathbb-r-together-with-the-operation-xy-x-y-xy-forall-x-y?noredirect=1&lq=1), so I would recommend against reposting the question. As for calculating $e$ and computing inverses, consider the factorization $x +y + xy = (1+x)(1+y) - 1.$ If you get stuck there you can check the previous times this question has been answered. – ckefa Jan 21 '25 at 14:19
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    @ckefa Request to show why only one of the four calculations above, ascribes to $x+y+xy = (1+x)(1+y)-1.$ – jiten Jan 21 '25 at 14:25
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    To make this group operation seem less arbitrary, let $f(x)=x+1$ and show that this is a group isomorphism (to nonzero reals under ordinary multiplication). This also gives a recipe to compute the desired products more simply. – Semiclassical Jan 21 '25 at 14:31
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    You can see that $x0=x$ directly so we know $e=0$. However $-22=-2+2+(-2)2=-4 \neq 0$. So $2^{-1} \neq -2$. This is because you're assumed $mm^{-1}=e$. You now have $\frac{e-m}{1+m}$ so use $e=0$ for the general inverse. – CyclotomicField Jan 22 '25 at 00:21
  • @CyclotomicField Still if could derive $e=0,$ from $m^{-1}= \frac{e-m}{1+m}.$ Am not comfortable with deriving $e=0$ from $x0=x.$ A better way would be to use the expression $x + e + xe = x \implies e(1+x)=0;$ which can mean $e=0,$ or $x=-1.$ But, $x=-1$ is not possible. Still if could derive directly from $m^{-1}= \frac{e-m}{1+m},$ then better. – jiten Jan 22 '25 at 01:18
  • @CyclotomicField Please find the edited OP. – jiten Jan 22 '25 at 01:37
  • Also, I am the downvoter here. This is a badly thought-out question. The reason being is that there are too many backtracks and corrections here. It is on the person asking to have their question be clear when it is posted. You have been here long enough to know this! – Mike Jan 22 '25 at 05:22
  • @Mike If had known the corrections, then why would have come here, in the first place. I come here to ask for guidance when fail to see light. – jiten Jan 22 '25 at 05:35

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Let $G:=(\mathbb R\setminus\{1\},*)$. Now, first: find out what is the identity element, say $e$. By its very definition: $\forall x\in G:x*e=x\iff$ $\forall x\in G:x+e+xe=x\iff$ $\forall x\in G:e(1+x)=0\iff$ $e=0$. Next, find out what is the inverse of each element $x$, say $x^{-1}$. Again, by its very definition: $x*x^{-1}=0\iff$ $x+x^{-1}+xx^{-1}=0\iff$ $(1+x)x^{-1}=-x\iff$ $x^{-1}=-\frac{x}{1+x}$.

Now that you know what $x^{-1}$ is, apply "$*$"'s recipe to $(2^{-1}*6)*5^{-1}$.

Btw, $2*x*5=6\iff$ $(2*x)*5=6\iff$ $(2+x+2x)*5=6\iff$ $(2+3x)*5=6\iff$ $2+3x+5+(2+3x)5=6\iff$ $7+3x+10+15x=6\iff$ $18x=-11\iff$ $x=-\frac{11}{18}$. But your "Hence" is not clear to me, as this is unrelated to finding out the previous $(2^{-1}*6)*5^{-1}$.

Kan't
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