I'm working through Rudin, and was attempting to prove that since $\mathbb{R}$ satisfies the least-upper-bound property, then each positive element of $\mathbb{R}$ has an $n$-th root. I came to realize, if I can prove between any two numbers there is some $n$-th power of a rational number, then I can prove this theorem a lot easier. Of course, I can see this is true:

Let $0 < x < y$. Then $x^{1/n} < y^{1/n}$, so there is a rational $q$ between them, and then $x < q^n < y$. But is there a slick proof of this fact, without assuming $n$-th roots exist?

  • $\begingroup$ $x\mapsto x^n$ is injective and surjective on $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Feb 25 '18 at 22:23
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    $\begingroup$ I see how the function is injective, but how would I prove it is surjective without implicitly assuming every element has an $n$-th root? $\endgroup$ – Evan Feb 25 '18 at 22:26
  • $\begingroup$ $\mathbb{R}^+$ is unbounded and $a>b>0$ implies $a^n>b^n>0$. Given the injectivity and continuity of $\varphi:x\mapsto x^n$, it follows that $\varphi(\mathbb{R})$ is a dense subset of $\mathbb{R}^+$ containing arbitrarily large intervals, i.e. $\varphi(\mathbb{R}^+)=\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Feb 25 '18 at 22:29
  • $\begingroup$ I'm not seeing how $\phi(\mathbb{R}^+)$ is dense in $\mathbb{R}^+$. And even if I did see that, I don't know why this implies this is all of $\mathbb{R}^+$. Is this explanation out of my reach until I'm further through Rudin? $\endgroup$ – Evan Feb 25 '18 at 22:49
  • $\begingroup$ No, it is pretty simple. $\mathbb{R}^+$ is dense in itself and continuous applications preserve density, hence the range of $\varphi$ has to be dense in itself. $\endgroup$ – Jack D'Aurizio Feb 25 '18 at 22:51

The idea is to to find $p,q \in \Bbb Q^+$ with $p^n\leq x<q^n,$ where $d=(q-p)/p$ is close enough to $0$ that $(1+d)^n<y/x,$ so that $x<q^n=p^n (1+d)^n\leq x (1+d)^n<y.$

(i).If $0<u\leq1$ and $n\in \Bbb N$ then $$(1+u)^n=1+\sum_{j=1}^n\binom {n}{j}u^j\leq$$ $$\leq 1+\sum_{j=1}^n\binom {n}{j}u=1+(2^n-1)u.$$ This crude upper bound will be used in (iii) below.

(ii). For $x>0$ and $m\in \Bbb N$ such that $1/m<x,$ there is a unique $k_m\in \Bbb N$ such that $$(k_m/m)^n\leq x<((1+k_m)/m)^n.$$

Note: There exists $k\in \Bbb N$ with $(k/m)^n>x,$ because if $x<a\in \Bbb N$ and $k=am$ then $(k/m)^n=a^n\geq a>x.$ And $(1/m)^n\leq 1/m<x.$ So $k_m$ is the least $k\in \Bbb N$ such that $((1+k)/m)^n>x.$

(iii). Let $((1+k_m)/m)^n=x(1+d_m).$ We have $d_m>0.$

We would like $d_m<y/x-1$ for some $m$, which would imply $x<((1+k_m)/m)^n<y.$

$$\text {We have } \quad 1+d_m=x^{-1}((1+k_m)/m)^n\leq$$ $$\leq ((k_m/m)^{-n} ((1+k_m)/m)^n=$$ $$=(1+1/k_m)^n\leq$$ $$\leq 1+ (1/k_m)(2^n-1)$$ by (i)..... So we would like $k_m$ to be large enough that $(1/k_m)(2^n-1)<y/x-1.$

(iv). There does not exist $A\in \Bbb N$ such that $1+k_m\leq A$ for every $m\in \Bbb N$ such that $1/m<x.$ Proof: Let $m_0\in \Bbb N$ with $1/m_0<x$ and let $A<m_A\in \Bbb N.$ And let $m=m_0m_A.$ If $1+k_m \leq A$ then $$x<((1+k_m)/m)^n\leq (A/m)^n=(A/m_A)^n(1/m_0)^n<1\cdot (1/m_0)^n\leq 1/m_0<x$$ which is absurd.

So in (iii) we can indeed have $k_m$ as large as desired.

  • $\begingroup$ Hmm, I don't see you're reasoning in the first line of (iii). Is $d_m = ((1+k_m)/m - k_m/m)/(k_m/m)$? As in the formula for $d$ in your outline of the argument at the top? I see that $x$ is bounded in between these two values, but not sure how you got that line. $\endgroup$ – Evan Feb 26 '18 at 22:38
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    $\begingroup$ This line provides a definition of $d_m.$ I expected to be using it more than I actually did, so I wanted an abbreviation. $\endgroup$ – DanielWainfleet Feb 27 '18 at 2:51
  • $\begingroup$ I have re-written that line to clarify it. I appreciate your point. The way I first wrote it would likely send readers backwards, trying to find where $ d_m $was first mentioned $\endgroup$ – DanielWainfleet Feb 27 '18 at 2:57

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