Prove that $\mathbb{Q}+ \mathbb{R}$ is not order isomorphic to $\mathbb{R}+ \mathbb{Q}$: is my proof correct? I must prove that $\mathbb{Q}+ \mathbb{R}$ is not order isomorphic to $\mathbb{R}+ \mathbb{Q}$. Both sets have linear order. 
I'd like to know whether my proof is correct or not.
$\triangleleft$
For convenience, let us denote the  $\mathbb{Q}+ \mathbb{R}$ as  $\mathbb{Q_1}+ \mathbb{R_1}$  and $\mathbb{R}+ \mathbb{Q}$ as $\mathbb{R_2}+ \mathbb{Q_2}$.

Important edit:
The $+$ does not stand for Minkowski sum.
$\mathbb{R_2}+ \mathbb{Q_2} = \{\dots, q_{r1}, q_{r2}, q_{ir1}, \dots; \dots, \dot q_{r1}, \dot q_{r2}, \dots\}$, where $q_{rn}$ is a rational number from $\mathbb{R_2}$ and $\dot q_{rn}$ is its copy from $\mathbb{Q_2}$. By $q_{irn}$, irrational numbers from $\mathbb{R_2}$ are denoted.
It's analogous for $\mathbb{Q_1}+ \mathbb{R_1}$, except that the copies of rational numbers are now in $\mathbb{R_1}$.

Linear order is: every element of $\mathbb{Q_1}$ is greater than any element of $\mathbb{R_1}$, and every element of $\mathbb{R_2}$ is greater than every element of $\mathbb{Q_2}$. Within the $\mathbb{Q_1}$, $\mathbb{R_1}$, $\mathbb{R_2}$, and $\mathbb{Q_2}$, the order is natural.
Assume the contrary:  $\mathbb{Q_1}+ \mathbb{R_1}$ $\approx$ $\mathbb{R_2}+ \mathbb{Q_2}$. Then, every element of $\mathbb{Q_1}$ maps to an interval $(a_2, +\infty)$ of $\mathbb{R_2}$. (Naturally, the set of rational numbers cannot be mapped to the set of real numbers: they have different cardinalities, $\aleph_0$ and $\mathfrak{c}$). 
Now, let's take $a_2$. It is mapped to some element $a_1$ in $\mathbb{R_1}$. In $\mathbb{R_1}$ there exists an element $b_1$ that is greater than $a_1$ and should be mapped to some element in $\mathbb{R_2}$ that is greater than $a_2$. However, we can't do this, since $(a_2, +\infty)$ is already completely 'occupied'. Contradiction. 
$\triangleright$
I also have another, slightly different version of the proof: after the paragraph beginning with 'Assume the contrary...' ends, we could say instead that, since order isomorphism exists, then some interval $(x, +\infty)$ $\subset$ $\mathbb{R_1}$ should be mapped to some part of $\left(r, a_2\right] \subset$ $\mathbb{R_2}$ . In other words, set $(x, +\infty)$ with no greatest element is order isomorphic to a set with the greatest element $a_2$. Impossible.
It would also be great to know if there are any alternative ways of proving this.
Thank you in advance. 
 A: Remark. I'm using the standard definition of $A+B$ where elements of $A$ are smaller than the elements of $B$.
Your proof doesn't make much sense, unfortunately. Elements of $\mathbb{Q}_1$ don't map to intervals, they map to elements. Since the rest of your proof hinges on that, your proof isn't a proof.
Here is a simple proof. In $\mathbb{Q} + \mathbb{R}$, there exists an element $x$ such that 1. there is an uncountable number of elements bigger than $x$; 2. there exists elements $a<b$ smaller than $x$ and such that there is only a countable number of elements between $a$ and $b$. Namely, you can take $x = 0 \in \mathbb{R}$ and $(a,b) = (0,1) \in \mathbb{Q} \times \mathbb{Q}$.
Such an element $x$ does not exist in $\mathbb{R} + \mathbb{Q}$. Indeed, suppose there were one. Then since there is an uncountable number of elements bigger than $x$, then $x$ must belong to the copy of $\mathbb{R}$. But then if you take any pair of elements smaller than $x$, they must belong to $\mathbb{R}$ too (by definition of the order sum), and between two different elements of $\mathbb{R}$, there always is an uncountable number of elements.
This proves that there cannot be an order-isomorphism $f : \mathbb{Q} + \mathbb{R} \to \mathbb{R} + \mathbb{Q}$. Otherwise you could find an $x$: it would just be, for example, $f(0)$ where $0 \in \mathbb{R}$.
