Prime numbers in the interval [2n, 3n] I have to prove " If $n$ is even then $\displaystyle\prod_{n/2<p \leq 3n/4} p \cdot \prod_{n<p \leq 3n/2}{p} <  {\frac{3n}{2} \choose {n}}.  " $
However,  I can't find at least one prime numbers in $\displaystyle\prod_{n/2<p \leq 3n/4}p  $ when $ n= 2,6,14$. It is product of primes and if I can't find one it doesn't make sense if I will put  $ 0$ since $0$ is not a prime number and the left side of the inequality will be  $0$.My question is "What will I put there and why?  "
 A: One way to think of it is that
$$ \prod_{x:A(x)} x = 1 \cdot x_1 \cdot x_2 \cdot x_3 \cdots x_n $$
where $x_1, x_2, x_3, \ldots x_n$ are all the possible $x$ that have the property $A$.
If there are no $x$ such that $A(x)$, then all that is left on the right-hand side is $1$.

You can also think of it procedurally:
To compute $\prod_{x:A(x)} x$, start by writing $1$ down on a piece of paper. Then, for each $x$ such that $A(x)$, multiply the number on your paper by that $x$ and erase the old number. Then $\prod_{x:A(x)} x$ is the number that's left on your paper when you're done.
In each of these cases, it works fine if there is nothing that satisfies $A$; then the outcome is $1$. But neither of the views will work if there are infinitely many things that satisfy $A$; then we need to go to limits, and the order of those factors may become important.
A: The short answer: the product of no number is $1$.
If $A\subset \Bbb R$ is finite, we expect that
$$\prod_{x\in A}x=\prod_{x\in A}x\prod_{x\in\emptyset} x$$
Then, if $0\notin A$,
$$\prod_{x\in\emptyset} x=1$$
(Honestly, I'm not sure if this can be proved or must be defined).
Known examples are: $3^0=1$. $0!=1$.
