# Prime numbers....is there a pattern? **URL:** \ **Category:** Factual Questions\ **Created:** [June 12, 2003, 1:40pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317 "2003-06-12T13:40:45Z")\ **Posts on this page:** 7\ **Page:** 3
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [June 15, 2003, 5:08pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317/41 "2003-06-15T17:08:23Z")
Oops, punched in the wrong values in the calculator. **Hari Seldon** ’s count of number of primes below 2[sup]1024[/sup] is right, not mine. It’s still a huge enough number. ---
**Author:** ![Wendell\_Wagner](https://avatars.discourse-cdn.com/v4/letter/w/8491ac/32.png) [@Wendell\_Wagner](https://boards.straightdope.com/u/Wendell_Wagner)\ **Post date:** [June 15, 2003, 7:57pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317/42 "2003-06-15T19:57:06Z")
Hari Seldon writes: \> The number of primes below n = 2^1024 is roughly n/710, \> which is something like 2^1021. (2^1024)/710 doesn’t equal 2^1021. Let’s start from the beginning. The number of primes below 2^1024 is approximately (2^1024)/ln(2^1024), which does equal approximately (2^1024)/710. But that equals about 2^1014.53. ---
**Author:** ![Moe](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/moe/32/2972_2.png) [@Moe](https://boards.straightdope.com/u/Moe)\ **Post date:** [June 15, 2003, 10:35pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317/43 "2003-06-15T22:35:06Z")
> [@](#): > > \*Originally posted by ultrafilter \* > \*\*No, there are actually as many primes as there are non-primes. > \*\* Just out of curiousity, is this proven? How can we be sure that the number of primes is infinite? ---
**Author:** ![Jabba](https://avatars.discourse-cdn.com/v4/letter/j/b5e925/32.png) [@Jabba](https://boards.straightdope.com/u/Jabba)\ **Post date:** [June 15, 2003, 10:45pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317/44 "2003-06-15T22:45:04Z")
That was actually proven by Euclid, and is another classic proof by contradiction. Suppose there are in fact finitely many primes, and list them as 2, 3, 5, …, P ( so P is the largest prime). Form the number Q = 2.3.5…P + 1 Now every integer greater than 2 is either prime or has a prime divisor. In particular, this is true of Q. But: Q is not divisible by 2 ( it leaves a remainder of 1). Similarly, Q leaves a remainder of 1 when divided by 3. In fact, it is clear that Q leaves a remainder of 1 when divided by each of 2, 3, 5, …, P. This is our contradiction, for we supposed that 2, 3, 5, …, P were all the primes, and we have shown that none of them divides Q. Therefore our assumption that there are only finitely many primes must be false. ---
**Author:** ![Moe](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/moe/32/2972_2.png) [@Moe](https://boards.straightdope.com/u/Moe)\ **Post date:** [June 15, 2003, 11:06pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317/45 "2003-06-15T23:06:37Z")
Maybe this is just beyond me but I don’t quite get it. What do you mean by: “Form the number Q = 2.3.5…P + 1” Is Q a set (including all primes and P + 1), or a single number (P + 1)? “But: Q is not divisible by 2 ( it leaves a remainder of 1)” Why is this? Actually I guess my confusion is still over what Q is. Obviously it can’t be P + 1 since that is divisible 2, no? ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [June 15, 2003, 11:11pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317/46 "2003-06-15T23:11:07Z")
Q is the number **dre2xl** was talking about yesterday. 🙂 ---
**Author:** ![Jabba](https://avatars.discourse-cdn.com/v4/letter/j/b5e925/32.png) [@Jabba](https://boards.straightdope.com/u/Jabba)\ **Post date:** [June 15, 2003, 11:11pm UTC](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317/47 "2003-06-15T23:11:48Z")
To form Q, take the product of the numbers on the list 2, 3, 5, …, P ( i.e. multiply them together) and then add 1 to the result. The 1 that is added at the end is the remainder in each of the divisions mentioned later in the proof. [Previous page](https://boards.straightdope.com/t/prime-numbers-is-there-a-pattern/181317.md?page=2)