$\mathbb{Z}\rtimes\mathbb{Z}$ is left-orderable but not right-orderable. I was working in the following problem:
Prove that $\mathbb{Z}\rtimes\mathbb{Z}=\langle x,y\mid x^{-1}yx=y^{-1}\rangle$ is left-orderable but not right-orderable.
Where a group $G$ is right-orderable if there exists a total order $<$ such that $\forall~x,y,z\in G~x<y\Rightarrow xz<yz$, and it is left-orderable if there exists a total order $<$ such that $\forall~x,y,z\in G~x<y\Rightarrow zx<zy$.
Proving that $\mathbb{Z}\rtimes\mathbb{Z}$ is not right-orderable is easy:

*

*$1<y\Rightarrow y^{-1}<1\Rightarrow x^{-1}yx<1\Rightarrow yx<x\Rightarrow y<1$, which is a contradiction.


*$1>y\Rightarrow y^{-1}>1\Rightarrow x^{-1}yx>1\Rightarrow yx>x\Rightarrow y>1$, which is a contradiction.
However I can't prove that $\mathbb{Z}\rtimes\mathbb{Z}$ is left-orderable. I know that there are some results that solve this as an immediate consequence, but I want to find the explicit order (if possible) or to know what is the easiest result (the one which needs less previous results) that solves this.
Thanks for your help.
 A: Notice that $yx=xy^{-1}$ implies that every element of $\mathbb{Z}\rtimes\mathbb{Z}$ is of the form $x^\alpha y^\beta$ with $\alpha,\beta\in\mathbb{Z}$.
Hence we can define the order $x^{\alpha_1} y^{\beta_1}>x^{\alpha_2} y^{\beta_2}$ if $\alpha_1>\alpha_2$ or $\alpha_1=\alpha_2$ and $\beta_1>\beta_2$, let us see that this order makes $\mathbb{Z}\rtimes\mathbb{Z}$ a left-ordered group. Let $a,b,c\in \mathbb{Z}\rtimes\mathbb{Z}$, then $a=x^{a_1}y^{a_2}$, $b=x^{b_1}y^{b_2}$ and $c=x^{c_1}y^{c_2}$ for some $a_1,a_2,b_1,b_2,c_1,c_2\in\mathbb{Z}$.
We have to prove that $a<b$ implies $ca<cb$. Notice that $a<b\Leftrightarrow x^{a_1}y^{a_2}<x^{b_1}y^{b_2}\Leftrightarrow a_1<b_1$ or $a_1=b_2$ and $a_2<b_2$. On the other hand $ca=x^{c_1}y^{c_2}x^{a_1}y^{a_2}=x^{c_1}x^{a_1}y^{(-1)^{a_1}c_2}y^{a_2}=x^{a_1+c_1}y^{a_2+(-1)^{a_1}c_2}$ and  $cb=x^{b_1+c_1}y^{b_2+(-1)^{b_1}c_2}$. Then, if $a_1<b_1$, we have $a_1+c_1<b_1+c_1$ which implies $ca<cb$ and if $a_1=b_2$ and $a_2<b_2$ then $a_1+c_1=b_1+c_1$ and $a_2+(-1)^{a_1}c_2<b_2+(-1)^{b_1}c_2$, which implies $ca<cb$.
