Ordinal Numbers
Solved Problems
Example 1.
Show that the right distributive law \[\left( {\alpha + \beta } \right) \cdot \gamma = a \cdot \gamma + \beta \cdot \gamma \] does not hold for all ordinals.
Solution.
We know that \(2 \cdot \omega = \omega.\) Then we have
Example 2.
Show that the following rule does not hold for all ordinals: \[{\text{If } \alpha + \gamma = \beta + \gamma,\;}\kern0pt{\text{then } \alpha = \beta .}\]
Solution.
Let \(\alpha = 1,\) \(\beta = 2,\) \(\gamma = \omega.\) We see that
so \(\alpha + \gamma = \beta + \gamma = \omega,\) though
Example 3.
Show that the following rule does not hold for all ordinals: \[{\left( {\alpha \cdot \beta } \right)^2} = {\alpha ^2} \cdot {\beta ^2}.\]
Solution.
Let \(\alpha = \omega\) and \(\beta = 2.\) Check that
Ordinal multiplication is associative. Since \(n \cdot \omega = \omega,\) the left-hand side (\(LHS\)) of this expression is written as
We see that \(LHS \ne RHS.\)
Example 4.
Show that the following rule does not hold for all ordinals:
\(\text{If } \gamma \gt 0 \text{ and } \alpha \cdot \gamma = \beta \cdot \gamma,\;\) \(\text{then } \alpha = \beta .\)
Solution.
Let \(\alpha = 2,\) \(\beta = 3,\) and \(\gamma = \omega.\) For these ordinal numbers, we have
So \(\alpha \cdot \gamma = \beta \cdot \gamma = \omega,\) but
Example 5.
Simplify the ordinal expression \[\left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right).\]
Solution.
We denote this expression by \(\alpha.\) Given that \(n \cdot \omega = \omega,\) we can substitute \(4\omega = \omega\) and \(2\omega = \omega:\)
Now recall that for limit ordinals, \(n + \omega = \omega.\) Therefore,
Using left distributivity and the exponentiation property \({\omega^{\beta + \gamma }} = {\omega^\beta } \cdot {\omega^\gamma },\) we get