Embedding of $\mathbb{Q}$ in $\mathbb{R}$ respects $\leq$. I have to verify that the embedding of $\mathbb{Q}$ in $\mathbb{R}$ respects $\leq$. That is, verify for $a,b\in \mathbb{Q}$ that if $a\leq b$ as rationals, then $a\leq b$ as real numbers.
I have this idea. We can embed $a$ and $b$ in $\mathbb{R}$ by associating them with the Dedekind cuts $(-\infty,a]|(a,\infty)$ and $(-\infty,b]|(b,\infty)$.
$a\leq b$ in $\mathbb{R}$ $\Leftrightarrow (-\infty,a]\subseteq (-\infty,b] \Leftrightarrow \in \mathbb{Q}$
Which holds.
 A: First of all lets be precise. If you define $\mathbb{R}$ as Dedekind cuts of $\mathbb{Q}$ then $\mathbb{Q}$ embeds into $\mathbb{R}$ via $q\mapsto\big((-\infty, q), [q,\infty)\big)$. The total ordering on the Dedekind cuts is defined by the inclusion on the left component, i.e $(A,B)\leq (A', B')$ iff $A\subseteq A'$.
So let $p, q\in\mathbb{Q}$ and $p\leq q$ in $\mathbb{Q}$. Then obviously $(-\infty,p)\subseteq(-\infty, q)$ and thus $p\leq q$ in $\mathbb{R}$.
A: EDIT (October 25, 2017). This is an updated version, according to the comments below, of an answer using Cauchy sequences, and written before the OP's editing of the question.
We can define the set $\mathbb{Q}^+$ of non-negative rational numbers as $0$ and all the rational numbers whose numerator and denominator have the same sign. Then the definition of $a \leq b$ as rationals can be given by saying there exists $x \in \mathbb{Q}^+$ such that $b = a+x$ (it is an easy exercise to check that this is equivalent to the standard definition $u/v \leq w/z$ if $uz  \leq vw$). 
Now identify $x$ with the corresponding real number via the natural embedding $\mathbb{Q} \hookrightarrow \mathbb{R}$ that associates to $x \in \mathbb{Q}$ the (constant) Cauchy sequence $\{x\}$. This sequence is eventually non negative, hence $0 \leq x$ in $\mathbb{R}$ and so $a \leq a+x$ in $\mathbb{R}$.
But this precisely means that $a \leq b$ as real numbers, and you are done.
A: I would use the following definitions for order:
for $\mathbb{Q}$: $\frac{a}{b}\leq\frac{c}{d}$ if and only if $ad\leq bc$
for $\mathbb{R}$: $a\leq b$ if and only if $\exists c(a+c^2=b)$
as far as I know these are these are fairly standard definitions relying on only multiplication, addition and ordering of integers.
But as has been said it, the definitions that you use will shape your answer.
