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On the set $\mathbb{R}$, we define the following operation which is called the "pro-sum" of two real numbers :
For all $a, b \in \mathbb{R},\ a \circledast b = (a \times b) + (a+b)$, where $\times$ and $+$ are the classical operations in $\mathbb{R}$.

I was wondering if there was an algebraic property which was respected by the classical sum and not by the "pro-sum"?

I already checked that commutativity, associativity and the identity property (identity element is $0$) are true for the "pro-sum". However I noticed that $-1$ does not have an inverse element, so it means that not every element with this operation is invertible.

Are there references about the "pro-sum" in the literature ?

Thank you in advance !

Maman
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  • It's well-known isomorphic to (R,*) via x+1. This answers everything. I will look for the duplicate(s). Never heard the name "pro-sum". It's really just the product in disguise. – Martin Brandenburg Dec 26 '24 at 00:26
  • See the answer by Andreas Caranti in the linked duplicate. – Martin Brandenburg Dec 26 '24 at 00:30
  • Other related thread: https://math.stackexchange.com/questions/46840/alternative-group-laws-over-the-reals – Martin Brandenburg Dec 26 '24 at 00:35
  • I am forced to answer in a comment because the problem keeps closing before I can make my observation. The operation, is distributive not over addition, but over an augmentrd sum defined by a\oplus b= a+b+1. To wit, by direct algebra webprove that $a\ostar (b\oplus c)=(a\ostar b)\oplus(a\ostar c)$. The identity element of $\oplus$ is $-1$, so $-1$ will behave with $\ostar$ the same way as $0$ would with ordinary multiplication. – Oscar Lanzi Dec 26 '24 at 00:44
  • If anyone can reproduce the circle around the star properly, please reost the above commentcwith this change. We should linitbourselves to standard mathjax characters please. – Oscar Lanzi Dec 26 '24 at 00:46
  • @MartinBrandenburg Perfect ! – Maman Dec 26 '24 at 12:15

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