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Prove that formula $m \circ n = 2mn+2m+2n+1$ determines binary operation on the set $\mathbb{R} \setminus \{-1\}$ and $(\mathbb{R} \setminus \{-1\}, \circ)$ is a group.

I am pretty sure how to prove that given formula forms group, however don't really understand what is needed to show that this formula determines binary operation? Any hints are appreciated. I've been thinking about binary operation definition, which in terms of our course states that binary operation on set A is such an mapping, that $A \times A \rightarrow A$, but how it fits here?

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Note that $f(x):=2(x+1)$ give a bijection $\Bbb R\setminus\{-1\}\to \Bbb R\setminus\{0\}$. Then show that $f(x\circ y)=f(x)\cdot f(y)$. Conclude.

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The operation $\circ$ is defined as a function from $\Bbb R\times\Bbb R$ to $\Bbb R$. It’s clearly a binary operation, so to show that it’s a binary operation on $\Bbb R\setminus\{-1\}$, you have to show that its restriction to $(\Bbb R\setminus\{-1\})\times(\Bbb R\setminus\{-1\})$ is a function from $(\Bbb R\setminus\{-1\})\times(\Bbb R\setminus\{-1\})$ to $\Bbb R\setminus\{-1\}$. To put it in more algebraic terms, you have to show that the set $\Bbb R\setminus\{-1\}$ is closed under the operation, i.e., that if $m,n\in\Bbb R\setminus\{-1\}$, then $m\circ n\in\Bbb R\setminus\{-1\}$.

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