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4 votes
1 answer
393 views

Commutative subtraction

It is well known that subtraction is not commutative in general. However, it is commutative in some groups: $\mathbb I$, $\mathbb C_2$, $\mathbb K_4$. I am trying to understand the logic. ...
Alex C's user avatar
  • 1,120
0 votes
1 answer
65 views

Build abelian group containing a set $K$ under an associative, commutative operation $*$ with an identity but the inverses are not always in $K$.

We are given a set K of elements and an operation * . For every element in the set, there exists an inverse element (not necessarily in the set). There are three (additional) rules: 1) K contains e (...
mathjunkie's user avatar
2 votes
2 answers
5k views

Proof of closure and identity element for $a * b = ab - a - b + 2$

For real numbers $x > 1$, which forms the set $G$, it is given that the operation on $a,b$, being $a\ast b$, results in $ab - a - b + 2$ (where $ab$ is the ordinary multiplication of $a$ and $b$). ...
user2768428's user avatar