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On the set $I=(0,1)=\{x\in\Bbb R\mid 0<x<1\}$, the operation $*$ is defined by: $$\forall s,t \in I, s*t= \frac{1}{1+\frac{(1-s)}{s} \frac{(1-t)}{t}}$$ Operations are usual ones with $\Bbb R$. We have been told to assume that $*$ is associative.

I believe the identity is $e=\frac{1}{2}$. But I cannot not find the inverse. I'm assuming it doesn't have have one. Thus it would not be a group.

Also, I need to show that $I$ is closed under $*$. I believe that I can claim closure. If $s,t \in\Bbb R$ s.t. $0<s,t<1$ then $a*b$ is closed because the fraction will always have values between $(0,1)$.

Yes? No? Prettier ways to say it?

Shaun
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Blip2020
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    Follows Immediately by transport of structure as explained in the dupes. Here we have $,h(x) = (1-x)/x = 1/x-1,\ h^{-1}(x) = 1/(1+x)\ \ $ – Bill Dubuque Apr 05 '21 at 17:10

2 Answers2

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First note that the group operation is commutative, by symmetry of its definition.

You can rewrite:

$$s*t = \frac{1}{1+\frac{1-s-t+st}{st}} = \frac{1}{2+\frac{1-s-t}{st}}$$

From which: $$s*\frac{1}{2} = \frac{1}{2+\frac{\frac{1}{2}-s}{\frac{1}{2}s}} = \frac{1}{2+\frac{1-2s}{s}}=\frac{1}{\frac{1}{s}} = s$$

So $e=\frac{1}{2}$ is indeed the identity element.

Now, given any element $s \in S$, you want to find $t$ such that $s*t = \frac{1}{2}$. In other words, you have to solve the equation in $t$:

$$\frac{1}{2+\frac{1-s-t}{st}} = \frac{1}{2}$$ $$\frac{st}{2st+1-s-t} = \frac{1}{2}$$ $$2st = 2st + 1 - s -t$$ $$t = 1-s$$

Just to verify, let's compute: $$s*(1-s) = \frac{1}{2+\frac{1-s-(1-s)}{s(1-s)}} = \frac{1}{2+\frac{0}{s(1-s)}} = \frac{1}{2}$$

For closure, you need to show that $s*t < 1$ (it is obviously positive). $$\frac{1}{2+\frac{1-s-t}{st}} <^? 1$$ $$1 <^? 2+\frac{1-s-t}{st}$$ $$-st <^? 1-s-t$$ $$-(1-s)(1-t)<^? 0$$

which holds, since $1-s$ and $1-t$ are elements of $H$.

We are done!

AnalysisStudent0414
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You can note that $$ \frac{1-s}{s}=\frac{1}{s}-1 $$ and that the map $f\colon I\to(0,\infty)$, $s\mapsto s^{-1}-1$ is bijective, with $$ f^{-1}(x)=\frac{1}{1+x} $$ Now consider that $s*t=1/(1+f(s)f(t))$ and so $$ s*t=f^{-1}(f(s)f(t)) $$ which means that $$ f(s*t)=f(s)f(t) $$ Thus the operation is just the transfer to $I$ of the multiplication on $(0,\infty)$ through $f^{-1}$. Since $(0,\infty)$ is a group, so is $I$.

What's the neutral element? $f^{-1}(1)=1/2$. What's the inverse of $s$? It's $$ f^{-1}((f(s))^{-1})=\frac{1}{1+(f(s))^{-1}}=1-s $$

egreg
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