Consider $\mathbb{N}\times\mathbb{Q}$ and $\mathbb{Q}\times\mathbb{N}$ both with the ordering given by $(a,b)\leq(c,d)$ iff ($a<c$ or $a=c$) and $b\leq d$.
Are $\mathbb{N}\times\mathbb{Q}$ and $\mathbb{Q}\times\mathbb{N}$ isomorphic as totally ordered sets?
I think that they aren't so, I need to find a function $f:\mathbb{Q}\times\mathbb{N}\to \mathbb{N}\times\mathbb{Q}$ in order to do that with the use of the following definition:
Definition of isomorphic: Let $(X,≤_X)$ and $(Y,≤_Y)$ be posets. $Y$ is isomorphic to $X$ as a poset if there exists an isomorphism $f:X→Y$ of posets.
By
$(a,b)\leq(c,d)$ iff $a<c$ or $a=c$ and $b\leq d$
I meant the left lexicographic order
They are not isomorphic. Suppose there is such an isomorphism $f \colon \mathbb{N} \times \mathbb{Q} \to \mathbb{Q} \times \mathbb{N}$ of posets. Note that the restriction of an isomorphism to a subposet (hence to any subset) is still an isomorphism (on its image). I assume that $\mathbb{N}$ does not contain $0$, but otherwise you can do exactly the same (the only important thing is that $\mathbb{N}$ has a smallest element).
Consider the poset $f(P) \subseteq \mathbb{Q} \times \mathbb{N}$ consisting of the images of all the elements of the poset $P = \{(1,q) \mid q \in \mathbb{Q}\} \subseteq \mathbb{N} \times \mathbb{Q}$. All the elements of $f(P)$ must be smaller than all the other elements in $\mathbb{Q} \times \mathbb{N}$, since the same is true for $P$. The complement of $f(P)$ thus has a smallest element, because all the elements of the form $(q,n)$ with $q$ negative and big enough (in absolute value) must be elements of $f(P)$. But the complement of $P$ does not have a smallest element, which is a contradiction.
Edit: Another (easier) explanation: in $\mathbb{Q} \times \mathbb{N}$, every element $(q,n)$ has a smallest element bigger than itself, namely $(q,n+1)$. But this does not hold in $\mathbb{N} \times \mathbb{Q}$, hence they cannot be isomorphic.