# Ordinal Arithmetic

#### Jerbear

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

#### clic-clac

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 $$\displaystyle \delta<\gamma$$ )