Group/Examples/ac, ad+b on Positive Reals by Reals

Example of Group
Let $\R$ denote the set of real numbers.

Let $\R_{\ge 0}$ denote the set of positive real numbers.

Let $S = \R_{\ge 0} \times \R$ denote the Cartesian product of $\R_{\ge 0}$ and $\R$.

Let $\circ: S \to S$ be the operation on $S$ defined as:
 * $\forall \tuple {a, b}, \tuple {c, d} \in S: \tuple {a, b} \circ \tuple {c, d} := \tuple {a c, a d + b}$

Then the algebraic structure $\struct {S, \circ}$ is a group.

Proof
Taking the group axioms in turn:

Let $\tuple {a, b}, \tuple {c, d} \in S$ be arbitrary.

We have that:

Then from Real Numbers form Field we haveL
 * $a d + b \in \R$

Thus $\tuple {a c, a d + b} \in \R_{\ge 0} \times \R$ and so $\struct {S, \circ}$ is closed.

Let $\tuple {a, b}, \tuple {c, d}, \tuple {e, f} \in S$ be arbitrary.

Thus $\paren {\tuple {a, b} \circ \tuple {c, d} } \circ \tuple {e, f} = \tuple {a, b} \circ \paren {\tuple {c, d} \circ \tuple {e, f} }$ and so $\struct {S, \circ}$ is associative.

We have:

Thus $\tuple {1, 0}$ is the identity element of $\struct {S, \circ}$.

We have that $\tuple {1, 0}$ is the identity element of $\struct {S, \circ}$.

Hence we need to find $\tuple {c, d} \in S$ such that $\tuple {a, b} \circ \tuple {c, d} = \tuple {1, 0}$.

Hence:

and:

Thus every element $\tuple {a, b}$ of $\struct {S, \circ}$ has an inverse $\tuple {\dfrac 1 a, -\dfrac b a}$.

All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.