Density of n-th Power of Rationals in $\mathbb{Q}$ 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?
 A: 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.
