# "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?

• I am sure of this: if $\alpha<\beta$ then there is a unique $\gamma$ with $\alpha+\gamma=\beta$. I encountered it as an exercise in a good book about sets. Unfortunately I am not familiar enough with the stuff to provide a proof of it. I suspect that $\gamma$ is the type of the well-ordening on $\beta-\alpha$. Jun 26 '15 at 11:05

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 $\gamma=\gamma'$).

• Why is it called "Subtraction on the left" or "left-subtraction" (and respectively with right) and how would they look when using the $\alpha-\beta=\gamma$ subtraction operator in each case? Oct 17 '17 at 15:08
• @AdiShavit It simply means the result of the subtraction is added on the left side i.e. $\alpha-\beta=\gamma\iff\alpha=\gamma+\beta$ Apr 8 '20 at 1:52
• @SimplyBeautifulArt your definition of subtraction on the left does not match the one given in the answer, which uses $\alpha = \beta + \gamma$ instead. Apr 19 at 10:33

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 $\,\alpha+\gamma=\beta$.

• I allready suggested that in a comment. What still lacks is a proof of what "one can show". Jun 26 '15 at 11:10

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 $$\alpha - 1 = \cases{\text{undefined (or 0)} & if \alpha = 0\\\alpha & if \alpha is limit ordinal\\ \beta & if \alpha = \beta + 1}$$

• As an ordinal, $\alpha+1=\alpha\cup\{\alpha\}\neq\alpha$ (indeed $\alpha+1>\alpha$). Jun 26 '15 at 10:43
• Actually you proved that addition is not commutative. If $\alpha<\beta$ and $\gamma$ is the type of the wellordening on $\beta-\alpha$ don't we have $\alpha+\gamma=\beta$ then? I am not familiar enough with this stuff to make a statement of that. Jun 26 '15 at 10:50
• Also, if $\alpha$ is a limit ordinal it is false that $\alpha-1=\alpha$. Jun 26 '15 at 10:51
• @Bernard we are dealing here with a definion of $\alpha-1$. How could that be false? Jun 26 '15 at 10:53
• @Bernard In that case, what do you suggest $\omega - 1$ is so that $(\omega - 1) + 1 = \omega$? I know you would like that to be true, but it just can't be done for limit ordinals. The definition in my answer is the second-best thing. Jun 26 '15 at 11:00

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).