If $\alpha,\xi,\rho$ are ordinal numbers with $\alpha=\xi+\rho$, then $\xi$ is called a **segment** of $\alpha$, and $\rho$ is called a **remainder** of $\alpha$. If $\alpha$ is such that there exists no decomposition $\alpha=\xi+\rho$ with $\xi,\rho$ both strictly less than $\alpha$, then $\alpha$ is called a **prime component**. That is, a prime component is an ordinal which cannot be written as a sum of two strictly lesser ordinals.

Proposition 1.Each ordinal number has only finitely many distinct remainders.

Proof.Let $\alpha=\xi_0+\rho_0=\xi_1+\rho_1$ with $\rho_0<\rho_1$, so that $\rho_0,\rho_1$ are distinct remainders of $\alpha$. Then we have that $\rho_1=\rho_0+\delta$ for some ordinal $\delta\ne 0$, which means that $\xi_0+\rho_0=\xi_1+\rho_0+\delta$, implying that $\xi_0+\rho_0 > \xi_1+\rho_0$ and therefore $\xi_0 > \xi_1$. Hence, we have that $\rho_0<\rho_1\implies\xi_0>\xi_1$.Now suppose that there exist infinitely many distinct remainders of $\alpha$. Then there must either exist some infinite ascending sequence of remainders $\rho_0<\rho_1<\cdots$ or some infinite descending sequence of remainders $\cdots<\rho_1<\rho_0$. Clearly the latter is impossible, since there is no infinite descending sequence of ordinals. On the other hand, if $\rho_0<\rho_1<\cdots$ is an infinite ascending sequence of remainders, then the corresponding segments form an infinite descending sequence $\cdots<\xi_1<\xi_0$ which is also impossible! Hence, $\alpha$ cannot have infinitely many distinct remainders at all. $\blacksquare$

Proposition 2.The smallest nonzero remainder of a nonzero ordinal must be a prime component.

Proof.Let $\alpha=\xi+\rho$ with $\rho\ne 0$ the smallest nonzero remainder of $\alpha$. If $\rho$ can be decomposed in the form $\rho=\xi'+\rho'$ with $\xi',\rho'$ both smaller than $\xi'$, then we would have $\alpha=(\xi+\xi')+\rho'$, meaning that $\rho'<\rho$ is a smaller remainder of $\alpha$, and further that $\rho'$ is nonzero since $\xi'\ne \rho$ by assumption. This would contradict the fact that $\rho$ was supposed to be the smallest nonzero remainder of $\alpha$, so we have that there cannot exist a decomposition of the form $\rho=\xi'+\rho'$, and therefore $\rho$ must be a prime component as claimed. $\blacksquare$

Proposition 3.Every ordinal number has a unique decomposition as a sum of monotone decreasing prime components. Furthermore, this decomposition uses only finitely many prime components.

Proof.If $\alpha$ is a prime component, then clearly the only way to decompose it as a finite sum of monotone decreasing prime components is the trivial decomposition which consists only of $\alpha$ itself. If $\alpha$ is not a prime component, then we may iteratively construct such a decomposition by letting $\rho$ be the smallest nonzero remainder of $\alpha$ (which is necessarily a prime component by the above proposition), so that $\alpha=\xi+\rho$, and then letting $\alpha'=\xi$ and repeating this process with $\alpha'=\alpha-\rho$. This process must eventually terminate with some $\alpha^{(n)}$ being a prime component, for otherwise $\alpha$ would have infinitely many distinct remainders.On the other hand, the remainder $\rho$ is uniquely determined at each step. For if is a decomposition of $\alpha$ as a finite sum of monotone decreasing prime components, then $\rho_0$

mustequal the smallest nonzero remainder of $\alpha$. For because $\rho$ is the smallest nonzero remainder of $\alpha$, we must have that $\rho_0 \geq \rho$, and since $\rho$ must also be a remainder of $\rho_0$, we have that $\rho_0=\xi+\rho$ for some $\xi$. But this contradicts the primality of $\rho_0$ unless either $\xi=\rho_0$ or $\rho=\rho_0$. The former is impossible, since this would imply $\rho_0=\rho_0+\rho$. Thus, we must have that $\rho_0=\rho$.Hence, since the decomposition of $\alpha$ into a finite sum of monotone decreasing prime components can be computed in finitely many steps, with the prime components $\rho_i$ uniquely determined at each step, we have that the decomposition is unique. $\blacksquare$