Remark 2

Every mapping $g$ of $I$ into $\mathbf{R}_+$ which satisfies the same conditions as $f$ is of the form $x \to \lambda f(x)$ where $\lambda > 0$. For if $\lambda = g(a) > 0$, the relations $z^p \leq x \leq z^{p+1}, z^q \leq a \leq z^{q+1}$ imply, by hypothesis,

whence and therefore, passing to the limit with respect to $\tilde{x}$, we have $g(x) = \lambda f(x)$.

Let us seek conditions under which $f(I)$ is an interval of $\mathbf{R}_+$. Clearly the following two conditions are necessary:

(GR_{IIIa}) The set of elements $> \omega$ in $I$ is not empty and has no smallest element, and given any two elements $x, y$, of $I$ such that $x < y$, there exists $z \in I$ such that $xz = y$ ("subtraction" of magnitudes).

(GR_{IVa}) Every increasing sequence of elements of $I$, which is bounded above by an element of $I$ has a least upper bound in $I$.

We shall show that these conditions are also sufficient, and moreover that they allow us to dispense with axiom (GR_{IV}) (Archimedes' axiom). To be precise, we shall prove the following proposition: