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Math Help - Ordinal Arithmetic

  1. #1
    Newbie
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    May 2010
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    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 = βγ + δ and δ < β.

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

    Thanks x
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  2. #2
    Senior Member
    Joined
    Nov 2008
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    Hi

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