"Subtracting" Ordinals - possible?

Question

I was wondering whether it is possible to "subtract" ordinals, or in other words - does there exist an ordinal $\gamma$ for every pair of infinite ordinals $\alpha$ and $\beta$ such that $\alpha+\gamma=\beta$ or $\gamma+\alpha=\beta$...

Any ideas?

Answer 1

Subtraction on the right is not always possible (i.e., there exist ordinals $\alpha,\beta$ such that $\alpha\geq\beta$ and that, for no ordinal $\gamma$, we have $\gamma+\beta=\alpha$). An example is when $\alpha=\omega$ and $\beta=1$. The uniqueness doesn't hold either (e.g., $1+\omega=\omega=0+\omega$).

Subtraction on the left is, however, always possible (i.e., for every ordinals $\alpha,\beta$ such that $\alpha\geq\beta$, there exists a unique ordinal $\gamma$ such that $\beta+\gamma=\alpha$). This can be proven by transfinite induction (use Bernard's hint to show the existence of $\gamma$; for the uniqueness part, show by induction on the ordinal $y$ that, if $x,z$ are ordinals such that $x<y$, then $z+x<z+y$, which then implies that ordinal addition is left-cancellative, namely, for an ordinal $\gamma'$, if $\beta+\gamma=\beta+\gamma'$, then $\gamma=\gamma'$).

Answer 2

No, there isn't. Say $\alpha + 1= \beta$. Then for $\alpha + \gamma$ to be equal to $\beta$, $\gamma$ must be $1$. But if $\alpha$ is infinite, then $1+\alpha = \alpha \neq \beta$.

That being said, we do have the notion of $\alpha - 1$, which is usually defined as $$ \alpha - 1 = \cases{\text{undefined (or $0$)} & if $\alpha = 0$\\\alpha & if $\alpha$ is limit ordinal\\ \beta & if $\alpha = \beta + 1$} $$

Answer 3

A left subtraction is always possible: let $\gamma$ be the unique ordinal isomorphic to the well-ordered set $$\bigl\{\xi\mid \alpha\le\xi<\beta\bigr\}.$$ One can show $\,\alpha+\gamma=\beta$.