B-Algebra Power Law
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Theorem
Let $\struct {X, \circ}$ be a $B$-algebra.
Let $n, m \in \N$ such that $n \ge m$.
Then:
- $\forall x \in X: x^n \circ x^m = x^{n - m}$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.
$B$-Algebra Power Law with Zero
- $\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n - m}$
Proof
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\forall m \in \N_{> 0}, m \le n: \forall x \in X: x^n \circ x^m = x^{n - m}$
Basis for the Induction
$\map P 1$ is true, as this just says:
- $x \circ x = 0$
which follows from the definition of the zeroth power in $B$-algebra.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\forall m \in \N_{> 0}, m \le k: \forall x \in X: x^k \circ x^m = x^{k - m}$
Then we need to show:
- $\forall m \in \N_{> 0}, m \le k + 1: \forall x \in X: x^{k + 1} \circ x^m = x^{k + 1 - m}$
Induction Step
This is our induction step:
First we show that:
- $\forall x \in X: x^{k + 1} \circ x = x^k$
Thus:
\(\ds x^{k + 1} \circ x\) | \(=\) | \(\ds \paren {x^k \circ \paren {0 \circ x} } \circ x\) | Definition of $B$-Algebra Power of Element | |||||||||||
\(\ds \) | \(=\) | \(\ds x^k \circ \paren {x \circ \paren {0 \circ \paren {0 \circ x} } }\) | $B$-Algebra Axiom $(\text A 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x^k \circ \paren {\paren {x \circ x} \circ 0}\) | $B$-Algebra Axiom $(\text A 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x^k \circ \paren {x \circ x}\) | $B$-Algebra Axiom $(\text A 2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x^k \circ 0\) | $B$-Algebra Axiom $(\text A 1)$ | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds x^{k + 1} \circ x\) | \(=\) | \(\ds x^k\) | $B$-Algebra Axiom $(\text A 2)$ |
By induction, it follows that:
- $\forall n \in \N_{>0}: \forall x \in X: x^n \circ x = x^{n - 1}$
Now let $1 \le m \le k$.
We have:
\(\ds x^{k + 1} \circ x^{m + 1}\) | \(=\) | \(\ds x^{k + 1} \circ \paren {x^m \circ \paren {0 \circ x} }\) | Definition of $B$-Algebra Power of Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x^{k + 1} \circ x} \circ x^m\) | $B$-Algebra Axiom $(\text A 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x^k \circ x^m\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x^{k - m}\) | Induction Hypothesis |
By induction, it follows that:
- $\forall x \in X: \forall n, m \in \N: n \ge m \implies x^n \circ x^m = x^{n - m}$
$\blacksquare$
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