I'm currently going through "An Introduction to the Theory of Numbers" by Hardy and Wright and at one point, they discuss why the distance from one prime to the next must have a long chunk of composites in between. I'm trying to understand the reasoning but I'm having some difficulty. Here is what they say:

Suppose that $2, 3, \ldots, p$ are the primes upto $p$. Then all the numbers up to $p$ are divisible by one of these primes, and therefore, if $2\cdot3\cdot5 \cdots p = q$ all of the $p-1$ numbers i.e. $q + 2, q+3, q+4 \ldots , q+p$ are composite.

I'm having particular difficulty understanding the $q + 2, q+3, q+4, \ldots , q+p$ part. If $q$ is equal to the product of primes from $2$ to $p$ then how can how can we have all the $p-1$ numbers be equal to $q+2, q+3, q+4, \ldots q + p$?

Can someone please clarify this? or if not, just explain what they are trying to say here.

Thanks a bunch!

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    $\begingroup$ q+2 is divisible by 2, q+3 is divisible by 3, q+4 is divisible by 2 again, q+5 is divisible by 5,.... $\endgroup$ – Will Jagy Jan 15 '14 at 2:42
  • $\begingroup$ @WillJagy I understand what you have said there but the main confusion I'm having is how can you have $p-1$ numbers be equal to $q+2, q+3, \text{etc}$ $\endgroup$ – Jeel Shah Jan 15 '14 at 2:44
  • $\begingroup$ What does "$p-1$ numbers be equal to $q+2,q+3$ etc." mean? $\endgroup$ – anon Jan 15 '14 at 2:45
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    $\begingroup$ Huh? ${}{}{}{}$ $\endgroup$ – anon Jan 15 '14 at 2:50
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    $\begingroup$ Yes: the numbers $q+2,\cdots,q+p$ (of which there are $p-1$) are composite. That is what the passage says. By the way, here is the original text. (I cannot figure out how could believe the text is saying $q+2$ is less than $p-1$ from this!) $\endgroup$ – anon Jan 15 '14 at 3:09

Let's translate the case $p=7$:

Suppose that $2,3,5,7$ are the primes up to $7$. Then all the numbers up to $7$ are divisible by one of these primes, and therefore, if $2\cdot3\cdot5\cdot7=210$ all of the six numbers $212, 213, 214, 215, 216, 217$ are composite.

Hopefully, the above made sense. The authors are talking about the numbers $212, 213, 214, 215, 216,$ and $217$. There are six numbers listed, because $7-1=6$.

Grammatically, the clause "$p-1$" is being used to modify the plural noun "numbers". The authors are not equating, or comparing, any other variables with $p-1$.

  • $\begingroup$ This makes a $\large{A\,LOT}$ more sense. Thanks a bunch! $\endgroup$ – Jeel Shah Jan 15 '14 at 3:07

They are saying that

$q+2$ is divisible by $2.$

$q+3$ is divisible by $3.$

$q + 4$ is divisible by $2.$

And so on.

  • $\begingroup$ @anon I didn't read the whole post, I was just doing a slightly naïve question just in case. Reading it, I delete my comment. $\endgroup$ – Pedro Tamaroff Jan 15 '14 at 3:13

Let $n > 0$ then $ n! + 2, .... n! + n$ are all non-prime. You can create a nonprime streak of arbitrary length using this process.


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