Ordinal Addition/Examples/Ordinal Addition by Two
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Theorem
Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Let $2$ denote the successor of the ordinal $1$.
Then:
- $x + 2 = x^{++}$
where $+$ denotes ordinal addition.
Proof
\(\ds x + 2\) | \(=\) | \(\ds x + 1^+\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + 1}^+\) | Definition of Ordinal Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds x^{++}\) | Ordinal Addition by One |
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 3$ Some ordinals