# Prove that $\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$ is dense in $[0,1]$

suppose that $$\phi(n)$$ is Euler function. prove that, $$\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$$ is dense in $$[0,1]$$

(if $$A_n=\{1 \leq m \leq n \mid m \in \Bbb N ; \gcd(n,m)=1\}$$ then $$\phi(n)=|A_n|$$)

I think : If $$p$$ is prime then $$\phi(p)=p-1$$. for any $$\varepsilon > 0$$ there is a prime number $$p$$ such that $$1-\varepsilon \leq \frac{p-1}{p} \leq 1$$. for other elements i don't know what can i do.

This hinges on the fact that $$\sum_{p\text{ prime}}\frac1p=+\infty$$ which in turn implies that $$\prod_{p\text{ prime}}\frac{p-1}p=0\,.$$ Once you know this fact, the proof is easy. Let $p_n$ be the $n$-th prime. Let $\epsilon>0$. There is some prime number $q=p_{N}$ such that $\frac{q-1}{q}\geq 1-\epsilon$. Then consider the sequence of numbers $$k_n=\prod_{i=0}^n\frac{p_{N+i}-1}{p_{N+i}}=\frac{\phi\left(\prod_{i=0}^np_{N+i}\right)}{\prod_{i=0}^np_{N+i}}$$ Then $k_n$ tends to zero, and the difference between consecutive terms is $\leq \epsilon$. Hence any real number $\in[0,1]$ can be approximated by some $\frac{\phi(n)}n$ up to arbitrary $\epsilon>0$.

• I understand that there is a sequence in the set $\{ \dfrac{\phi(n)}n \}$ converging to $0$ , but I cannot understand how does this imply that for any $c \in [0,1]$ we can find a sequence from the mentioned set converging to $c$ ? Please help
– user123733
Mar 31, 2015 at 13:50
• @user123733 The key point is that the $1-\epsilon\leq k_0\leq 1$, and for all $i$, $0<k_{i}-k_{i+1}\leq \epsilon$ and $k_i\to 0$. This means that any $c$ is $\epsilon$-close to some $\phi(n)/n$ Mar 31, 2015 at 13:55
• But that you already said in you r answer ; I want a construction from definition of limit of sequence that we can get a member from the set in $(a,b)$ with $0<a<b<1$
– user123733
Mar 31, 2015 at 13:59
• Just take $\epsilon=\frac{b-a}2$. The reasoning above shows that there exists some $i$ with $k_n\in(a,b)$. Mar 31, 2015 at 14:05
• take this $\epsilon$ where ? Please don't be mad , 'cause I really can't understand ; the trouble I'm having is something like I know $\{ \dfrac 1{n+3}\}$ converges to $0$ , but this sequence is not dense in $[0,1]$ ...
– user123733
Mar 31, 2015 at 14:12