An inequality containing prime numbers. Suppose $a,b \in \mathbb{N} $, and suppose $\gcd(a,b) = 1$ also suppose
$$ \frac{a}{b} \leq p \leq \exp\left(\frac{a^2}{b^2} \right) - 1 $$
for all prime $p$. What do you have to say about this inequality? Can this lead me to some sort of contradiction? 
 A: It is a known fact that there are infinitely many distinct prime numbers, and you can use it to get a contradiction.
Your inequality implies that every prime number lies in the interval
$$\left[\;\frac{a}{b}, \;\;\;\;\; \exp\left(\frac{a^2}{b^2}\right) - 1 \;\right] $$
There is only a finite number of integers in that interval, hence it follows that there exists only a finite number of prime integers, a contradiction.
A: I will not use the term of the  infinite  set of primes,but rather the dymanic idea of Euclid.(which means that the set of primes is infinite of course-but i think that,where we can avoid the term infinite,it's best to avoid it-here though it doesn't make any problem)
Euclid proved that given $n$ primes $p_1<p_2<...<p_n$ there is a prime $p_{n+1}$ greater than the others.
Using now the Foundamental Theorem of Number Theory, we have that (number of primes less or equal to $x$) $π(x)\sim \frac {x}{logx}$.
Consider now the Chebychev's approximation $0,92129\frac {x}{logx}<π(x)<1,2555\frac {x}{logx}$. 
So, you can approximately count the number of primes in the area $[2,\frac {a}{b}]$ and in the area $[2,\exp\big(\frac{a^2}{b^2} \big) - 1]$. Then you can find the number of primes in the area $[\frac {a}{b},\exp\big(\frac{a^2}{b^2} \big) - 1]$. Due to Euclid's propostition, there will be a prime,out of this area. So contradiction as you wanted.
A: Let $q=\frac ab\in\mathbb Q$, then the inequality goes like
$$
 q\leq p\leq e^{(q^2)}-1
$$
for all primes $p$. For $p=2$ you get
$$
 q\leq 2
$$
and therefore
$$
 e^{(q^2)}-1 \leq e^4 - 1 < 55
$$
Then, if such $q\in\mathbb Q$ existed you would have
$$
 2 \leq p < 55
$$
for all primes $p$, which obviously is not true (take $p=59$).
