1. ## Ordinal Arithmetic

Hi,
Wikipedia says, regarding ordinals:

Left division with remainder : for all α and β, if β > 0, then there are unique γ and δ such that α = β·γ + δ and δ < β.

Why is this true? Can anyone help construct an induction for it or something?

Thanks x

2. Hi

You may know that, if $\displaystyle \beta>0,$ right multiplication ( $\displaystyle y\mapsto\beta.y$ ) is strictly increasing. Therefore, the class of ordinals $\displaystyle \{x\ ;\ \beta.x>\alpha\}$ is non empty and has a minimum, which must be a successor (or else, let's call this minimum $\displaystyle \lambda,$ since right multiplication is continuous, we would have $\displaystyle \beta.\lambda=\sup\{\beta.\delta\ ;\ \delta<\lambda\}$ and since $\displaystyle \alpha<\beta.\lambda,$ this means there is a $\displaystyle \delta<\lambda$ such that $\displaystyle \alpha<\beta.\delta,$ absurd) so let's name this minimum $\displaystyle \gamma+1$ and consider $\displaystyle \gamma.$
It is such that $\displaystyle \beta.\gamma\leq\alpha$ and is maximal for this property.
Use now that right addition is also a strictly increasing continuous function, therefore there is a minimal ordinal $\displaystyle x$ such that $\displaystyle \beta.\gamma+x>\alpha,$ you find it again to be a successor, let's say $\displaystyle x=\delta+1,$ and conclude. ( Also, if $\displaystyle \delta\leq\gamma,$ since right addition is an (strictly) increasing function, you get $\displaystyle \beta.\gamma+\delta\geq\beta.\gamma+\gamma=\beta.( \gamma+1)>\alpha$ by hypothesis on $\displaystyle \gamma,$ contradiction; therefore $\displaystyle \delta<\gamma$ )