B-Algebra Power Law with Zero
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Theorem
Let $\struct {X, \circ}$ be a $B$-algebra.
Let $n, m \in \N_{>0}$ such that $n > m$.
Then:
- $\forall x \in X: n, m \in \N_{>0} \implies x^m \circ x^n = 0 \circ x^{n - m}$
where $x^k$ for $k \in \N_{>0}$ denotes the $k$th power of the element $x$.
Proof
First we show that:
- $\forall x \in X: x \circ x^2 = 0 \circ x$
\(\ds x \circ x^2\) | \(=\) | \(\ds x \circ \paren {x^1 \circ \paren {0 \circ x} }\) | Definition of Power ($B$-Algebra) | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {x \circ \paren {0 \circ x} }\) | First Power of Element in $B$-Algebra | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ x} \circ x\) | $B$-Algebra Axiom $(\text A 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \circ x\) | $B$-Algebra Axiom $(\text A 1)$ |
$\Box$
Now we show that:
- $\forall x \in X: m \in \N_{>0}: x \circ x^m = 0 \circ x^{m - 1}$
\(\ds x \circ x^m\) | \(=\) | \(\ds x \circ \paren {x^{m - 1} \circ \paren {0 \circ x} }\) | Definition of Power ($B$-Algebra) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ x} \circ x^{m - 1}\) | $B$-Algebra Axiom $(\text A 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \circ x^{m - 1}\) | $B$-Algebra Axiom $(\text A 1)$ |
$\Box$
Now we prove the original proposition using a proof by induction.
The base case for $n = 1, m = 2$ was established in the first lemma.
Let us assume $x^m \circ x^n = 0 \circ x^{n - m}$ for some $n, m \in \N_{>0}$.
Then:
\(\ds x^{m + 1} \circ x^n\) | \(=\) | \(\ds \paren {x^m \circ \paren {0 \circ x} } \circ x^n\) | Definition of Power ($B$-Algebra) | |||||||||||
\(\ds \) | \(=\) | \(\ds x^m \circ \paren {x^n \circ x}\) | Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x^m \circ x^{n - 1}\) | $B$-Algebra Power Law | |||||||||||
\(\ds \) | \(=\) | \(\ds x^0 \circ x^{n - m - 1}\) | Induction using the previous lemma | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \circ x^{n - m - 1}\) | Definition of Zeroth Power of Element |
Hence the result.
$\blacksquare$