# Definition talk:Cantor Normal Form

Don't forget that $a_k$ can be $0$, and thus, we can write any natural number $n$ as simply $\omega^0 n$. --Andrew Salmon (talk) 01:12, 26 August 2012 (UTC)
$\displaystyle \sum_{i \mathop = 1}^n \left({ x^{a_i} b_i }\right) \times x^y = x^{a_1 \mathop + y}$
1. Effectively answered above, is the fact that a finite ordinal can indeed be expressed as $\omega^0 n$, which is CNF. However, burying this nugget on the talk page is insufficient and it needs to be published on its own page, along with the (so far undocumented, AFAIK) fact that $\omega^0 = 1$.
2. Still unresolved is the fact that CNF is defined as $\displaystyle \sum_{i \mathop = 1}^k \omega^{a_i} n_i$, while in Ordinal Multiplication via Cantor Normal Form/Infinite Exponent etc. the summation is $\displaystyle \sum_{i \mathop = 1}^n x^{a_i} b_i$. Either the latter is not CNF, in which case the page needs to be renamed, or it is CNF in which case the original page on CNF needs to be amended to accommodate the fact that $\displaystyle \sum_{i \mathop = 1}^n x^{a_i} b_i$ is also CNF. Sorry to raise my voice but I do think this needs to be resolved.
Oh, and finally, the $k$ in $\displaystyle \sum_{i \mathop = 1}^k \omega^{a_i} n_i$ is undefined in this expression. Yes we "know" it's an ordinal, probably, but it needs to be clarified by means of $\exists k \in \operatorname{On}$ ... or whatever. --prime mover (talk) 09:01, 26 August 2012 (UTC)