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

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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 $\oplus: S \to S$ be the operation on $S$ defined as:

$\forall \tuple {a, b}, \tuple {c, d} \in S: \tuple {a, b} \oplus \tuple {c, d} := \tuple {a c, a d + b}$


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


Proof

Taking the group axioms in turn:


Group Axiom $\text G 0$: Closure

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

We have that:

\(\ds a\) \(\ge\) \(\ds 0\) Definition of $S$
\(\ds c\) \(\ge\) \(\ds 0\) Definition of $S$
\(\ds \leadsto \ \ \) \(\ds a c\) \(\ge\) \(\ds 0\) Product of Positive Real Numbers is Positive

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 $\struct {S, \oplus}$ is closed.

$\Box$


Group Axiom $\text G 1$: Associativity

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


\(\ds \paren {\tuple {a, b} \oplus \tuple {c, d} } \oplus \tuple {e, f}\) \(=\) \(\ds \tuple {a c, a d + b} \oplus \tuple {e, f}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \tuple {\paren {a c} e, \paren {a c} f + a d + b}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \tuple {a c e, a c f + a d + b}\) simplifying


\(\ds \tuple {a, b} \oplus \paren {\tuple {c, d} \oplus \tuple {e, f} }\) \(=\) \(\ds \tuple {a, b} \oplus \tuple {c e, c f + d}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \tuple {a \paren {c e}, a \paren {c f + d} + b}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \tuple {a c e, a c f + a d + b}\) simplifying


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

$\Box$


Group Axiom $\text G 2$: Existence of Identity Element

We have:

\(\ds \tuple {a, b} \oplus \tuple {1, 0}\) \(=\) \(\ds \tuple {a \times 1, a \times 0 + b}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \tuple {a, b}\)
\(\ds \tuple {1, 0} \oplus \tuple {a, b}\) \(=\) \(\ds \tuple {1 \times a, 1 \times b + 0}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \tuple {a, b}\)


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

$\Box$


Group Axiom $\text G 3$: Existence of Inverse Element

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

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

Hence:

\(\ds a c\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds \dfrac 1 a\)

and:

\(\ds a d + b\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds d\) \(=\) \(\ds -\dfrac b a\)


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

$\Box$


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

$\blacksquare$


Sources

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