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 $\oplus: S \to S$ 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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.7 \ \text {(a)}$