Prove that ordinal multiplication $\alpha \cdot (\beta \cdot \gamma) = (\alpha \cdot \beta) \cdot \gamma$ is associative by using the following facts:

$$\beta \cdot 0=0$$ $$\beta \cdot (\alpha +1)=\beta \cdot \alpha+\beta$$ $$\beta \cdot \alpha=\sup\{ \beta \cdot \gamma \mid \gamma < \alpha \}$$

The second fact is for all $\alpha$ while the third fact is for all limit $\alpha \neq 0$.

I use transfinite induction to prove it. I manage to prove the base case. I stuck at successor case.

My attempt: LHS$=\alpha\cdot (\beta \cdot (\gamma+1))=\alpha(\beta \cdot \gamma+\beta)$

RHS$=(\alpha\cdot \beta)\cdot(\gamma+1)=(\alpha\cdot \beta) \cdot \gamma + (\alpha\cdot \beta)=\alpha \cdot (\beta \cdot \gamma)+(\alpha\cdot \beta)$

I don't know the distributive law, so I cannot equate LHS to RHS. Can anyone help me?

  • $\begingroup$ Maybe you should prove the distributive law $\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$ before you prove the associative law? $\endgroup$ – bof Sep 23 '14 at 8:14
  • $\begingroup$ But the distributive law comes after this question. $\endgroup$ – Idonknow Sep 23 '14 at 8:17
  • $\begingroup$ Is there a penalty for doing them out of order? $\endgroup$ – bof Sep 23 '14 at 8:23
  • $\begingroup$ No, but I think the question can be proven without distributive law since the author arranges them in this way. $\endgroup$ – Idonknow Sep 23 '14 at 8:27
  • $\begingroup$ What book are you using? $\endgroup$ – bof Sep 23 '14 at 8:35

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