# Exponent Not Equal to Zero

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## Theorem

Let $x$ and $y$ be ordinals.

Let $x \ne 0$.

Then:

- $x^y \ne 0$

## Proof

The proof shall proceed by Transfinite Induction on $y$.

### Basis for the Induction

- $x^0 = 1$ by the definition of ordinal exponentiation.

Therefore, $x^0 \ne 0$.

This proves the basis for the induction.

### Induction Step

The inductive hypothesis supposes that $x^y \ne 0$.

\(\displaystyle x^{y^+}\) | \(=\) | \(\displaystyle x^y \times x\) | definition of Definition:Ordinal Exponentiation | ||||||||||

\(\displaystyle x^y\) | \(\ne\) | \(\displaystyle 0\) | Inductive Hypothesis | ||||||||||

\(\displaystyle x\) | \(\ne\) | \(\displaystyle 0\) | Hypothesis | ||||||||||

\(\displaystyle x^y \times x\) | \(\ne\) | \(\displaystyle 0\) | by Ordinals have No Zero Divisors |

This proves the induction step.

### Limit Case

The inductive hypothesis says that:

- $\forall z \in y: x^z \ne 0$

\(\displaystyle \forall z \in y: \ \ \) | \(\displaystyle x^z\) | \(\ne\) | \(\displaystyle 0\) | Inductive Hypothesis | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(\in\) | \(\displaystyle x^z\) | by Ordinal Membership is Trichotomy | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(\in\) | \(\displaystyle \bigcup_{z \in y} x^z\) | by the fact that limit ordinals are nonempty | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle 0\) | \(\in\) | \(\displaystyle x^y\) | definition of ordinal exponentiation | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x^y\) | \(\ne\) | \(\displaystyle 0\) | definition of empty set |

This proves the limit case.

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 8.32$