Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 8 / a

Subtraction fails associativity: $(0\ast 0)\ast 1=-1$ but $0\ast (0\ast 1)=1$. Thus it violates the group axioms and is not a group. It has a right identity $0$, but no element can be a two-sided identity element.

The operation is not defined for every pair in the proposed set: choosing $a=1,b=-1$ makes its denominator zero. Thus it is not even a binary operation on $\mathbb{C}\setminus \{0\}$, and is not a group. In particular, removing zero from the domain does not prevent the sum in the denominator from vanishing.

The substitution $F(a)=e^a$, with $F(-\infty )=0$, is a bijection onto $[0,\infty )$ and obeysThus the log-sum-exp function transfers nonnegative addition to the given set. It is closed, associative and commutative, with identity element $-\infty$. But a finite $a$ has $e^a>0$, and adding a nonnegative number cannot give zero, so no finite $a$ has an inverse element. Therefore it is a commutative monoid but not a group.

The sum of two odd integers plus one is odd, so the operation is closed. Both associations of a triple give $a+b+c+2$, proving associativity. Its identity element is $-1$, and the inverse element of $a$ is $-a-2$, also odd, since $a+(-a-2)+1=-1$. Thus it is an abelian group.

More explicitly, shifted addition on odd integers is ordinary integer addition in different coordinates:This bijection is a group isomorphism to $(\mathbb{Z},+)$.