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Is proving that there is always a prime number in the interval $(46n,47n]$ important?

Thanks in advance!

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    If $n=2$ this is false. Or if $n=4$ or $n=7$. – 2017-02-03
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    yes, you're right :). Let's say this is true for natural numbers that bigger than another natural number k. Is this an important statement? – 2017-02-03
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    Maybe if the proof was interesting. We already know from the Prime Number Theorem that, given any positive $\epsilon$, then for sufficiently large $n$ there is always a prime between $n$ and $(1+\epsilon)n$. See, e.g., [this](https://en.wikipedia.org/wiki/Bertrand's_postulate). Taking $\epsilon=\frac 1{46}$ and $n=46m$ for large $m$ gives your claim. – 2017-02-03
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    *Important* for who? If it is an exam question, it might be important for you. Anyway, Chebyshev's theorem (weak version of the PNT) states that $\pi(n)\geq \frac{n\log(2)}{\log(n)}$ for any $n$ big enough, and your statement is a consequence of it. – 2017-02-03

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A result due to Ingham that states there is always a prime in the interval $n^3, (n+1)^3$ (for big enough $n$) means that your result would be true for $n\ge 58383$.

In reality even the biggest prime gaps for numbers in that range (about $2.7$ million) are a factor of $1000$ smaller than the limit needed to guarantee your interval is occupied. A simple search shows that the last gap of suitable size occurs between $1327$ and $1361$, but this does not correspond to any range produced by integer $n$. Similarly the gaps $887\to 907$, $523 \to 541$, $509\to 521$, $467 \to 479$, $421 \to 431$, $409\to 419$, $359 \to 367$, $337\to 347$ do not fall in line with integer $n$ and we need to take $n=7$ to find the last prime-free space from $322$ to $329$.