$\frac{\phi(m)}{m}$ 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?

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 $m = \prod_{i=M}^N p_i$.  
