$\frac{\phi(m)}{m}$ is dense in $[0,1]$ 
Let $n$ be a natural number, $n \geq 2$, and let $\phi$ be Euler's function; i.e. $\phi(n)$ is the number of positive integers not exceeding $n$ and coprime to $n$. Given any two real numbers $\alpha$ and $\beta$, $0 \leq \alpha < \beta \leq 1$, prove that there exists a natural number $m$ such that $$\alpha<\frac{\phi(m)}{m}<\beta.$$

We know that $$\dfrac{\phi(m)}{m} = \prod_{p \mid m}\left(1-\dfrac{1}{p}\right).$$ How should we choose $m$ so that we can satisfy the above?
 A: Since the sum of $1/p$ is infinite, the product over all primes $p$ of $1 - 1/p$ goes to $0$.   Now  consider the products 
$$ A_{M,N} = \prod_{i=M}^N \left( 1 - \dfrac{1}{p_i}\right)$$
where $p_i$ is the $i$'th prime.  For fixed $M$, these form a sequence decreasing to $0$, with $$1 > A_{M,N+1}/A_{M,N} = 1 - 1/p_{N+1} > 1 - 1/p_M$$
Take $M$ large enough that $1 - 1/p_M > \max(\beta, \alpha/\beta)$, and there will be some 
$N$ so that $\alpha < A_{M,N} < \beta$.  Then take $m = \prod_{i=M}^N p_i$.  
A: Pick a prime $P$ so that $1-\frac{1}{P}>\frac{\alpha}{\beta}$. (There is such a $p$ because $\frac{\alpha}{\beta}<1$.)
Then find the least integer $M\geq P$ so that:
$$\prod_{P\leq p\leq M} \left(1-\frac{1}{p}\right)<\beta.$$
Then, show that the conditions on $M$ and $P$ mean this value is greater than $\alpha.$
You deal with the case $\alpha=0$ by replacing it with $\alpha=\beta/2$.
To ensure $M$ exists, you need to know:
$$\lim_{M\to\infty}\prod_{p<M}\left(1-\frac{1}p\right)=0$$
A: This follows from a much more general theorem.

Theorem: Let $(x_n)$ be a sequence of positive numbers such that $x_n\to 0$ as $n\to\infty$ but $\sum_{n=1}^\infty x_n=\infty$.  Then for any $0\leq a<b\leq \infty$, there is a finite set $A\subset\mathbb{N}$ such that $\sum_{n\in A}x_n\in(a,b)$.

To solve your problem from this theorem, let $x_n=-\log(1-1/p_n)$, where $p_n$ is the $n$th prime.  Then $(x_n)$ satisfies the hypotheses of the theorem: $x_n\to 0$ since $p_n\to \infty$, and $\sum x_n$ diverges since $\log(1+x)\approx x$ for $x$ small and $\sum 1/p_n$ diverges.  Now just apply the theorem with $a=-\log\beta$ and $b=-\log\alpha$ and let $m=\prod_{n\in A} p_n$.
To prove the theorem, choose $N$ such that $x_n<b-a$ for all $n\geq N$.  Let $M\geq N$ be minimal such that $\sum_{n=N}^Mx_n>a$ (such an $M$ exists since $\sum_{n=N}^\infty x_n=\infty$).  Then $$\sum_{n=N}^Mx_n= x_M+\sum_{n=N}^{M-1}x_n<(b-a)+a=b$$ by minimality of $M$.  Thus $A=\{N,N+1,\dots,M\}$ works.
A: Without using too much general theory:
Let $d = \beta - \alpha$. Now let $p_1, p_2, \dots$ be a sequence containing the set of primes for which $\frac{1}{p_i} < d$. Consider the partial products
$$s_n = \prod_{i=1}^n \frac{p_i - 1}{p_i}$$
The sequence $s_n$ is decreasing and each successive difference $s_{n+1} - s_n$ is no more than $\frac{1}{p_{n+1}} < d$. Since the limit of the sequence is zero, one of the partial products must be in the given interval.
If $s_N$ is in the interval, then taking $m = \prod\limits_{i=1}^N p_i$ suffices.
