We've been asked to teach ourselves a unit on ordinals for our final exam tomorrow, I grasp how to prove that certain ordinals are distinct but I am having trouble figuring out a proof to show ordinal addition is associative. All the proofs I have found online use methods that we have not covered yet. Would someone be able to guide me through a proof of associativity for ordinal addition?

Here's a general overview of what I know:

Two ordinals are equal if they are order isomorphic.

Ordinal addition for two well ordered sets $a=ord(A, <_A) b=ord(B, <_B)$ then$a+b=ord(AUB, <_+)$ where $x <_+ y$ if either x, y are in A and $x<_A y$, or x, y are in B and $x<_B y$, or x is in A and y is in B

Thank you so much

• In arithmetic, the proof of associativity proceed by induction. In the case of ordinal numbers you need transfinite induction. See P.Suppes, Axiomatic set theory (1960-Dover repr), page 211: Theorem 25. $\alpha + (\beta + \gamma)=(\alpha + \beta) + \gamma$. The proof proceeds by transfinite induction on $\gamma$. Basis: from result of page 206 ($\alpha + 0 = 0 + \alpha = \alpha$). The successor case is straightforward by def of addition. The last case is that $\gamma$ is a limit ordinal and for every $\delta \in \gamma$ : $\alpha + (\beta + \delta)= (\alpha + \beta) + \delta$. – Mauro ALLEGRANZA Apr 18 '14 at 15:01
• @Mauro: With this definition there's no real need for induction. – Asaf Karagila Apr 18 '14 at 15:17

Note that this means that $a+b=c$ if $c$ can be partitioned into an initial segment of order type $a$ and tail segment of order type $b$.
Show that there is a partition of $(a+b)+c$ into three consecutive intervals (initial segment, middle segment, and tail segment) of order types $a,b$ and $c$. And conclude from that that $(a+b)+c=a+(b+c)$.