"Subtracting" Ordinals - possible? 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?
 A: 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 $\,\alpha+\gamma=\beta$.
A: 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
$$
\alpha - 1 = \cases{\text{undefined (or $0$)} & if $\alpha = 0$\\\alpha & if $\alpha$ is limit ordinal\\ \beta & if $\alpha = \beta + 1$}
$$
A: 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 $\gamma=\gamma'$).
A: In addition to the other answers, note that you can construct the Grothendieck group of the ordinals under Hessenberg arithmetic, (which is commutative), just as one constructs the integers from the natural numbers under the latter’s arithmetic. This allows you to define subtraction in the usual way. See this question for an example.
Alternatively, you can define a truncated subtraction, or monus, on the ordinals themselves, provided you use the Hessenberg sum (which again is commutative).
