Ordinal Arithmetic

May 2010
1
0
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
 
Nov 2008
354
185
Paris
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 \(\displaystyle \delta<\gamma\) )