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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?

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    $\begingroup$ 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 $\beta-\alpha$. $\endgroup$ – drhab Jun 26 '15 at 11:05
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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 $\gamma=\gamma'$).

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    $\begingroup$ 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? $\endgroup$ – Adi Shavit Oct 17 '17 at 15:08
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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 $\,\alpha+\gamma=\beta$.

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    $\begingroup$ I allready suggested that in a comment. What still lacks is a proof of what "one can show". $\endgroup$ – drhab Jun 26 '15 at 11:10
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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 $$ \alpha - 1 = \cases{\text{undefined (or $0$)} & if $\alpha = 0$\\\alpha & if $\alpha$ is limit ordinal\\ \beta & if $\alpha = \beta + 1$} $$

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  • $\begingroup$ As an ordinal, $\alpha+1=\alpha\cup\{\alpha\}\neq\alpha$ (indeed $\alpha+1>\alpha$). $\endgroup$ – Bernard Jun 26 '15 at 10:43
  • $\begingroup$ 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. $\endgroup$ – drhab Jun 26 '15 at 10:50
  • $\begingroup$ Also, if $\alpha$ is a limit ordinal it is false that $\alpha-1=\alpha$. $\endgroup$ – Bernard Jun 26 '15 at 10:51
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    $\begingroup$ @Bernard we are dealing here with a definion of $\alpha-1$. How could that be false? $\endgroup$ – drhab Jun 26 '15 at 10:53
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    $\begingroup$ @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. $\endgroup$ – Arthur Jun 26 '15 at 11:00

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