Holomorph

Ouch, I can't believe how careless I was when I wrote that, sorry ... I couldn't find how to edit that post so I'll just put the corrections here. Of course, it should actually be y-x instead of x-y in the whole post and in the last phrase of the second paragraph, p is actually a and q is actually b (this was because i wrote them as p and q on paper but then changed the notation when posting it here ) Also, in the bracketed phrase about the archimedian property it x can be any real number, not necessarily positive, only y needs to be positive (of course, these are not the same x and y as in the rest of the proof...). Hopefully it should make more sense now, sorry again... Edit: Strange, I can edit this post but not the other ...

It suffices to prove that one can find a rational (strictly) between any 2 distinct real numbers because then you can easily make an infinite sequence of rationals between the 2 (for example, if [math]x < y[/math] then we can find a rational [math]r_1[/math] between the 2 and reiterate the process with [math]x[/math] and [math]r_1[/math] or [math]r_1[/math] and [math]y[/math] to give a sequence [math](r_n)_n[/math] of rationals between [math]x[/math] and [math]y[/math]). Here's a proof of this (we take [math]x < y[/math] and name [math]r[/math] the rational we want to find): suppose, without loss of generality, that [math]x[/math] and [math]y[/math] are positive (for if [math]x[/math] and [math]y[/math] are of different signs, we can pick the rational [math]r = 0[/math] and the case where they are both negative is analogous). Since [math]x-y > 0[/math], we can use the archimedian property of the real numbers (which is: given any positive real number [math]x[/math] and [math]y > 0[/math], one can find a positive integer [math]n[/math] such that [math]ny > x[/math]) to find a positive integer [math]b[/math] such that [math]b (x-y) > 1[/math], i.e. [math]x-y > \frac{1}{b}[/math]. Now let [math]a[/math] be the smallest integer such that [math]\frac{a}{b} > x[/math], then [math]r = \frac{a}{b}[/math] works since [math]r>x[/math] and also [math]r < y[/math] since otherwise we would have [math]r \geq y = x + (y - x) > \frac{1}{q} + x[/math] and hence [math]x < \frac{p-1}{q}[/math], contradicting the minimality of [math]p[/math]. Incidentally this is true not only for the reals but for any ordered field with the archimedian property (such a field always contains a sub-field isomorphic to [math]\mathbb{Q}[/math]).