First Power of Element in B-Algebra
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Theorem
Let $\left({X, \circ}\right)$ be a $B$-algebra.
Then:
- $\forall x \in X: x^1 = x$
where $x^k$ for $k \in \N$ denotes the $k$th power of the element $x$.
Proof
\(\ds x^1\) | \(=\) | \(\ds x^0 \circ \left({0 \circ x}\right)\) | Definition of B-Algebra Power of Element | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \circ \left({0 \circ x}\right)\) | Definition of B-Algebra Power of Element | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \circ x\) | $0$ in $B$-Algebra is Left Cancellable Element | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) | $0$ in $B$-Algebra is Left Cancellable Element |
Hence the result.
$\blacksquare$