I've been encountered with the following question:
Let $\alpha>1$ be a real number and let $n \in \mathbb{N}$ be natural number.
Now we claim that there exists $x$ ("large enough") s.t. for all $y> x$ : |{$p\in \mathbb{P}: y<p<\alpha y$}| > n , where $\mathbb{P} $ is the set of all primes.
in other words, the problem is to show that for every such $\alpha$ and $n$, for any "large enough" number ($z$) there are more than n primes in the interval $[z,\alpha z]$.
My try (the unimportant part): I thought i could use the prime number theorem to show that $\pi(\alpha z)-\pi(z) > n$ for large enough $z$ (maybe by $\lim_{z\to \infty} \pi(\alpha z)/ \pi(z) > 1$). I am not sure if this is correct, and if it's answer the need for $">"$ rather then $"\geq"$ (if it doesn't make sense you may skip this part).
thanks in advance
Your approach is right indeed. The prime number theorem says that: $$ \lim_{x\to\infty}\frac{\pi(x)\log x}{x}=1. $$ This can be written as: $$ \lim_{y\to\infty}\frac{\pi(\alpha y)\log(\alpha y)}{\alpha y}=1. $$ Combining the two we have: $$ \lim_{y\to\infty}\frac{\pi(\alpha y)\log(\alpha y)}{\alpha y}\frac{y}{\pi( y)\log( y)}=1 \implies \lim_{y\to\infty}\frac{\pi(\alpha y)}{\pi(y)}=\alpha>1. $$
This proves that there are arbitrarily many prime numbers inside the interval $[y,\alpha y]$.
If you want a specific bound, you can inspect a bit the proof of Chebyshev's theorems, for instance here as well as books on analytic number theory. There you can use likes of the following inequalities for $n\geq 2$ $$ \pi(2n)\geq \frac{n\log 2}{\log(2n)} \text{ and } \pi(2n)-\pi(n) \leq \frac{2n\log 2}{\log n} $$ to explicitly find the $x$ in the problem by lower bounding $\pi(\alpha y)-\pi(y)$.