# Ordinal Numbers

## Solved Problems

Click or tap a problem to see the solution.

### 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.

### Example 2

Show that the following rule does not hold for all ordinals: ${\text{If } \alpha + \gamma = \beta + \gamma,\;}\kern0pt{\text{then } \alpha = \beta .}$

### Example 3

Show that the following rule does not hold for all ordinals: ${\left( {\alpha \cdot \beta } \right)^2} = {\alpha ^2} \cdot {\beta ^2}.$

### Example 4

Show that the following rule does not hold for all ordinals:

$$\text{If } \gamma \gt 0 \text{ and } \alpha \cdot \gamma = \beta \cdot \gamma,\;$$ $$\text{then } \alpha = \beta .$$

### Example 5

Simplify the ordinal expression $\left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right).$

### 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

$2 \cdot \omega = \left( {1 + 1} \right) \cdot \omega = \omega \ne 1 \cdot \omega + 1 \cdot \omega = \omega + \omega.$

### 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

$\alpha + \gamma = 1 + \omega = \omega,\;\;\beta + \gamma = 2 + \omega = \omega,$

so $$\alpha + \gamma = \beta + \gamma = \omega,$$ though

$\alpha = 1 \ne \beta = 2.$

### 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

$\left( {\omega \cdot 2} \right)^2 \ne {\omega^2} \cdot {2^2}.$

Ordinal multiplication is associative. Since $$n \cdot \omega = \omega,$$ the left-hand side ($$LHS$$) of this expression is written as

${\left( {\omega \cdot 2} \right)^2} = {\left( {\omega \cdot 2} \right)^{1 + 1}} = \left( {\omega \cdot 2} \right) \cdot \omega \cdot 2 = \omega \cdot \left( {2 \cdot \omega} \right) \cdot 2 = \omega \cdot \omega \cdot 2 = {\omega^2} \cdot 2.$

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

$\alpha \cdot \gamma = 2 \cdot \omega = \omega,\;\;\beta \cdot \gamma = 3 \cdot \omega = \omega.$

So $$\alpha \cdot \gamma = \beta \cdot \gamma = \omega,$$ but

$\alpha = 2 \ne \beta = 3.$

### 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 $$2\omega = \omega:$$

$\alpha = \left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right) = \left( {1 + \omega} \right)\left( {3 + \omega + {\omega^2}} \right).$

Now recall that for limit ordinals, $$n + \omega = \omega.$$ Therefore,

$\alpha = \left( {1 + \omega} \right)\left( {3 + \omega + {\omega^2}} \right) = \omega\left( {\omega + {\omega^2}} \right).$

Using left distributivity and the exponentiation property $${\omega^{\beta + \gamma }} = {\omega^\beta } \cdot {\omega^\gamma },$$ we get

$\alpha = \omega\left( {\omega + {\omega^2}} \right) = \omega \cdot \omega + \omega \cdot {\omega^2} = {\omega^2} + {\omega^3}.$