There is a simple, concrete way to do this. Consider the set $\mathbb{R}(x)$ of all rational functions of a real variable,
$$
f(x) = \frac{p(x)}{q(x)},
$$
where $p(x)$ and $q(x)$ are polynomials in $x$. [I'm not going to worry about the distinction between 'polynomials' and 'polynomial functions', nor shall I worry about the fact that the denominator of $f(x)$ may be zero for a finite number of values of $x$ - these worries can be taken care of.]
$\mathbb{R}(x)$ includes the set of all constant functions $f(x) = t$, where $t$ is any real number, in particular the zero function $f(x) = 0$. Rational functions can be added, subtracted, multiplied, and divided (except that of course you can't divide by the zero function); these operations obey the usual algebraic laws; and their results coincide with the results of the usual operations in the case of the constant functions. We say that $\mathbb{R}(x)$ is a field, which extends the real field $\mathbb{R}$. [All of this can be made quite formal and watertight.] In practice, we make no distinction between elements of $\mathbb{R}$ and the corresponding 'constant' elements of $\mathbb{R}(x)$. [I don't think that this practice is ever actually made formal - but again let's just not worry about it now!]
Although $x$ here is a 'variable' and not a 'number' [and it is important to make this distinction when one defines polynomials properly - especially when dealing with fields other than $\mathbb{R}$], there is a clear sense in which $x$ can be regarded as an infinitesimally small 'number'. (There is also a similar but different treatment which makes $x$ an infinitely large 'number'.) The point is that (i) the field $\mathbb{R}(x)$ can be totally ordered, in a way that extends the usual ordering of $\mathbb{R}$, and (ii) in this ordering of $\mathbb{R}(x)$, we have $0 < x < t$ for every positive real number $t$.
The definition of the total ordering of the field $\mathbb{R}(x)$ is straightforward. [Caution: I haven't written out the details before, so it is a good and necessary exercise to check that I haven't slipped up! Also, I don't have an authoritative reference to hand at the moment. The Web page A non Archimedean ordered field is helpfully clear, but unfortunately it is also erroneous - it gets confused, at the end, between the two constructions in which $x$ may be either infinitesimal or infinite.] Say that a non-zero rational function,
$$
f(x) = \frac{p(x)}{q(x)} = \frac{a_nx^n + a_{n+1}x^{n+1} + a_{n+2}x^{n+2} + \cdots}{b_mx^m + b_{m+1}x^{m+1} + b_{m+2}x^{m+2} + \cdots},
$$
where $a_n \ne 0$ and $b_m \ne 0$, is positive if $a_n$ and $b_m$ have the same sign.
(Intuitively, if $x$ is small, then all powers of $x$ but the lowest are negligible, so $f(x) \bumpeq a_nb_m^{-1}x^{n-m}$. If $x$ is positive, it follows that $f(x)$ is positive if and only if $a_n$ and $b_m$ have the same sign.)
This definition is independent of the particular expression of $f(x)$ as a ratio of polynomials, because if also
$$
f(x) = \frac{r(x)}{s(x)} = \frac{c_kx^k + c_{k+1}x^{k+1} + c_{k+2}x^{k+2} + \cdots}{d_lx^l + d_{l+1}x^{l+1} + d_{l+2}x^{l+2} + \cdots},
$$
where $c_k \ne 0$ and $d_l \ne 0$, then $p(x)s(x) = q(x)r(x)$, whence $a_nd_l = b_mc_k$, from which it is clear that $a_n$ and $b_m$ have the same sign if and only if $c_k$ and $d_l$ have the same sign.
For rational functions $f(x)$ and $g(x)$, we now write $f(x) > g(x)$ if and only if $f(x) - g(x)$ is positive. [The symbols $\geqslant$, $<$ and$\leqslant$ may then defined in terms of $>$, for $\mathbb{R}(x)$, in the usual way.]
In $p(x)/q(x)$, because $b_m \ne 0$, we can divide $p(x)$ and
$q(x)$ by $b_m$, and thus can assume without loss of generality
that the trailing coefficient of $q(x)$ is $1$. Then
$p(x)/q(x)$ is positive if and only if $a_n > 0$.
Proposition. If $f(x)$ and $g(x)$ are positive, then so are $f(x) + g(x)$ and $f(x)g(x)$.
Proof. Let $f(x) = p(x)/q(x)$, and $g(x) = r(x)/s(x)$,
where the polynomials $p(x)$, $q(x)$, $r(x)$ and $s(x)$ are as
above, and $b_m = d_l = 1$. Suppose $f(x)$ and $g(x)$ are positive,
i.e. $a_n > 0$ and $c_k > 0$.
$$
f(x)g(x) = \frac{p(x)r(x)}{q(x)s(x)}.
$$
The trailing coefficient of
$p(x)r(x)$ is $a_nc_k > 0$, and the trailing coefficient of
$q(x)s(x)$ is $b_md_l = 1$, therefore $f(x)g(x)$ is positive.
$$
f(x) + g(x) = \frac{p(x)s(x) + q(x)r(x)}{q(x)s(x)}.
$$
Again, the
trailing coefficient of $q(x)s(x)$ is $b_md_l = 1$. The trailing
coefficient of $p(x)s(x)$ is $a_nd_l = a_n$, and the trailing
coefficient of $q(x)r(x)$ is $b_mc_k = c_k$. Therefore, the trailing
coefficient of $p(x)s(x) + q(x)r(x)$ is $a_n$, $c_k$, or
$a_n + c_k$, according as $n+l < m+k$, $n+l > m+k$, or
$n+l = m+k$, respectively. In all three cases, the coefficient is
positive, therefore $f(x) + g(x)$ is positive. $\square$
["$\square$", or a similar symbol, is often used to mark the end of a proof.]
This proposition enables us to operate just as confidently with the order properties of $\mathbb{R}(x)$ as with its arithmetic (or perhaps I should say, algebraic) properties. Elements of $\mathbb{R}(x)$ thus behave far more like 'numbers' than one would at first expect. $\mathbb{R}(x)$ certainly extends the familiar structure of $\mathbb{R}$ in terms of its total ordering, as well as in terms of addition, subtraction, multiplication, and division - even if one chooses not to regard the non-constant elements of $\mathbb{R}(x)$ as 'numbers'.
As promised, some of these new 'numbers', including our old friend $x$, are infinitesimally small: if $t$ is any positive real number, no matter how small, then
$$
t - x = \frac{t - x}{1} > 0,
$$
simply because $t$ and $1$ have the same sign. Similarly, the rational function $\frac{1}{x}$ is infinitely large: for all real $t$,
$$
\frac{1}{x} - t = \frac{1 - tx}{x} > 0,
$$
because the coefficients of the lowest powers of $x$ in the numerator and denominator are both $1$.
