Ordinal addition proof Let $\alpha ,\alpha^{'},\beta$ Be ordinals then prove that  if $\alpha<\alpha^{'}\implies \alpha +\beta \le \alpha^{'}+\beta$ 
My attempt is to use induction on $\beta$, when $\beta=0$ then its trivial that $\alpha < \alpha^{'}$ by the problem we are trying to solve, my problem is how to proceed from here, for when its a successor and a limit ordinal, any help would be gratefully appreciated.
My definition of ordinal addition is as follows $\alpha +\beta= \alpha$ if $\beta=0$  it is equal to $S(\alpha+\gamma)$ if $\beta=S(\gamma)$ here S is referring to the successor, and is equal to $\sup_{\gamma<\beta}(\alpha+\gamma)$ if $\beta$ is a limit ordinal.
 A: I don't think you need induction for this. Note that $\alpha +\beta$ is, by definition, the unique ordinal $\gamma$ that is order isomorphic to $\alpha \sqcup \beta.$ So, if $\alpha<\alpha'$, then there is the obvious order-preserving embedding $\alpha \sqcup \beta\to \alpha' \sqcup \beta$. Now, since no ordinal can be order isomorphic to itself, it must be the case that $\alpha \sqcup \beta\le  \alpha' \sqcup \beta.$
If you want to do it by induction, I think we can argue as follows: 
If $\beta $ is a limit ordinal, we have, by definition,  $\alpha + \beta = \text{sup}(\{\alpha + \delta \; |\; \delta < \beta\})$, and similarly for $\alpha' +\beta$. Then, with $\alpha<\alpha',$ the induction hypothesis says $\alpha + \delta \leq \alpha' + \delta$ for all $\delta < \beta$ and the claim follows on purely set-theoretic grounds. 
Remark: It's easy to show that $\alpha\le \alpha'\Rightarrow \alpha+1\le \alpha'+1$
Now, suppose $\beta= \delta + 1.$ Then, using the the associativity law , the inductive hypothesis, and the remark, we compute $\alpha +\beta = \alpha +(\delta+1)=(\alpha +\delta)+1\le(\alpha'+\delta)+1= \alpha'+(\delta+1)=\alpha'+\beta.$
