Ordinal arithmetic washing line Let $\eta$ be the order type of $\mathbb{Q}$.
I'm trying to calculate $(1+ \eta) \cdot (\eta + 1)$ and $(\eta + 1) \cdot (1+ \eta)$. So for the first one I'm thinking that you just do this
$(1+\eta) \cdot (\eta + 1) = (1+ \eta) \cdot \eta + (\eta + 1) \cdot 1= \eta + \eta + 1= \eta +1$.
Was wondering is this correct. It's hard to think about ordinal arithmetic. I was under the impression you think about it with the washing line method. So left side of the multiplication is the washing line and the right hand side is the clothes. 
It's just $(1+\eta) \cdot (1+ \eta)= (1+\eta) \cdot 2$ by similar reasons. However, I this can't be possible. Or maybe I'm doing something wrong.
 A: Remember that when you multiply order types, you have the left but not the right distributive law. Thus, $(1+\eta)\cdot(\eta+1)=(1+\eta)\cdot\eta+1+\eta$, and $(\eta+1)\cdot(1+\eta)=\eta+1+(\eta+1)\cdot\eta$. For any further simplification, we must see what $(1+\eta)\cdot\eta$ and $(\eta+1)\cdot\eta$ look like.
Suppose that $\langle X,\preceq\rangle$ is a linear order of type $(1+\eta)\cdot\eta$. Then by the definition of $(1+\eta)\cdot\eta$ we can decompose $X$ into pairwise disjoint sets $X_q$ for $q\in\Bbb Q$ in such a way that


*

*if $p,q\in\Bbb Q$ with $p<q$, $x\in X_p$, and $y\in X_q$, then $x\prec y$, and  

*each $X_q$ is ordered in type $1+\eta$ by $\preceq$ and hence is order-isomorphic to $\Bbb Q\cap[0,1)$ with its usual order. 


Note that each $X_q$ is densely ordered by $\preceq$: if $x,y\in X_q$ with $x\prec y$, there is a $z\in X_q$ such that $x\prec z\prec y$.
Clearly $X$ is countably infinite. Now let $x\in X$ be arbitrary; $x\in X_q$ for some $q\in\Bbb Q$. Choose any $y\in X_{q-1}$ and $z\in X_{q+1}$ and note that $y\prec x\prec z$, so that $x$ is neither a first or last element of $\langle X,\preceq\rangle$. Since $x\in X$ was arbitrary, $\langle X,\preceq\rangle$ has no first or last element. 
Finally, let $x,y\in X$ with $x\prec y$; there are $p,q\in\Bbb Q$ such that $x\in X_p$ and $y\in X_q$. Clearly $p\le q$. If $p<q$, let $r$ be any rational number between $p$ and $q$; then for each $z\in X_r$ we have $x\prec z\prec y$. If $p=q$, use the earlier observation that $X_p$ is densely ordered to conclude that there is a $z\in X_p$ such that $x\prec z\prec y$. It follows that $\preceq$ densely orders $X$.
Thus, $\langle X,\preceq\rangle$ is a dense linear order without endpoints and is therefore order-isomorphic to $\Bbb Q$; that is, it has order type $\eta$. 
We can now conclude that $(1+\eta)\cdot(\eta+1)=(1+\eta)\cdot\eta+1+\eta=\eta+1+\eta$. Now the set $\Bbb Q^-$ of negative rationals has order type $\eta$, as does the set $\Bbb Q^+$ of positive rationals, so $\Bbb Q=\Bbb Q^-\cup\{0\}\cup\Bbb Q^+$ has order type $\eta+1+\eta$, and $\eta+1+\eta=\eta$. (Or you could argue as before that $\eta+1+\eta$ is a countable dense order type without endpoints.) The final conclusion is that $$(1+\eta)\cdot(\eta+1)=\eta\;.$$
That was pretty detailed, and the other one is quite similar, so I’ll leave it to you.
