Math question

[Justforfun000]

I heard a quote like this somewhere..."23 is the perfect odd number, divisible by nothing - hence perfect in its ratio to one."

Can someone explain that to me?

[Melchior]

It simply says that it is a prime number, as far as I can figure.

[Batman]

That's about all I can say about it too, I'm afraid. And prime numbers aren't exactly rare.
No clue why 23 would be extra special, sorry.

[The Spartan]

Yeah all that says is that it's a prime number. And there are an infinite number of those.

He's probably thinking of this:23

[Surlethe]

I heard a quote like this somewhere..."23 is the perfect odd number, divisible by nothing - hence perfect in its ratio to one."

That simply says that 23 is prime. However, there is in fact such a thing as a perfect number: a positive integer is perfect iff it is equal to the sum of all of its proper divisors (i.e., those divisors which are positive and not equal to the number itself). For example, 6 = 1+2+3, 28 = 1+2+4+7+14, etc. Needless to say, no prime number can be perfect (can you prove this?), so 23 is not a real perfect number.

And prime numbers aren't exactly rare.

That depends on what you mean by "rare". They aren't rare in the sense that there are a countably infinite number of them, true; however, in the sense that they are scattered very sparsely through the whole numbers, they are rare: given any whole number n, you can find a sequence n in length of composite numbers.

[Justforfun000]

Interesting. I'm assuming it must have been meant in the prime sense. Maybe it was a comparison of other numbers and not meant that 23 itself was extra special. Thanks guys.

[Kuroneko]

And prime numbers aren't exactly rare.

That depends on what you mean by "rare". They aren't rare in the sense that there are a countably infinite number of them, true; however, in the sense that they are scattered very sparsely through the whole numbers, they are rare: given any whole number n, you can find a sequence n in length of composite numbers.

In yet another sense, Sum_{p prime}[ 1/p ] is divergent, so there are the primes are a "large" subset of the integers--for example, larger than the squares, since Sum[ 1/n² ] is finite. Thus, "on average", there must be more than one prime between any consecutive squares. (Unfortunately, proving that there is at least one for any pair of consecutive squares is an unsolved problem.)

---

And now, for something completely different...

If we let sets of the form nZ+m = {z in Z: z≡m mod n}, n nonzero, be a topological basis, we find that those basis sets are actually clopen (because we can take Z and exclude all nZ+m', m' not congruent to m mod n). Then, since the only integers with no prime divisors are ±1, Union_{p prime}[ pZ ] = Z\{±1}. Z\{±1} is not closed because no finite set is open (nZ+m is trivially infinite for every n,m). But since every finite union of closed sets is closed, we cannot have finitely many primes. Hence the primes form an infinite set.

Only using Fermat's Last Theorem to prove 2^{1/n} is irrational for n>2 is sweeter.

Q: Is the set of primes dense in the integers under this topology?

[Surlethe]

And now, for something completely different...

If we let sets of the form nZ+m = {z in Z: z≡m mod n}, n nonzero, be a topological basis, we find that those basis sets are actually clopen (because we can take Z and exclude all nZ+m', m' not congruent to m mod n). Then, since the only integers with no prime divisors are ±1, Union_{p prime}[ pZ ] = Z\{±1}. Z\{±1} is not closed because no finite set is open (nZ+m is trivially infinite for every n,m). But since every finite union of closed sets is closed, we cannot have finitely many primes. Hence the primes form an infinite set.

That is a thing of beauty.

Only using Fermat's Last Theorem to prove 2^{1/n} is irrational for n>2 is sweeter.

Indeed.

Q: Is the set of primes dense in the integers under this topology?

I haven't the time for a proper exploration of this question right this second because it's an hour and a half past my bedtime, but it seems to boil down to the question of given a pair of integers (n,m) does there exist a prime p such that p≡m mod n? This seems like it might involve Fermat's Little Theorem or the Chinese Remainder Theorem.

[Kuroneko]

I haven't the time for a proper exploration of this question right this second because it's an hour and a half past my bedtime, but it seems to boil down to the question of given a pair of integers (n,m) does there exist a prime p such that p≡m mod n?

Yes.

This seems like it might involve Fermat's Little Theorem or the Chinese Remainder Theorem.

You're over-thinking it, surely. (Egads, I should've thought up a better question. But we can rephrase the above with another condition to make it more interesting...)

[Justforfun000]

I love Kuroneko's posts. I almost feel live a divine intelligence has deigned to drop a pearl of infinite wisdom on the rest of us poor mortals. The man has more brains then a Borg colony.

[The Grim Squeaker]

I love Kuroneko's posts. I almost feel live a divine intelligence has deigned to drop a pearl of infinite wisdom on the rest of us poor mortals. The man has more brains then a Borg colony.

You mean one? .

This seems like it might involve Fermat's Little Theorem or the Chinese Remainder Theorem.

What's Fermat's "Little" theorem? I'm only aware of his (Solved) last theorem .

On a sidenote, anyone who actually tries to read what these two mathemagicians conjure between themselves, try reading some of Simon Singh's books, such as the aforementioned "Fermat's last Theorem" to get an easier primer to the game of maths first. (Although it still won't help with understanding the words of the K&S )

[SpacedTeddyBear]

I love Kuroneko's posts. I almost feel live a divine intelligence has deigned to drop a pearl of infinite wisdom on the rest of us poor mortals. The man has more brains then a Borg colony.

When I got to the word "clopen" I had to stop, reboot my brain, and proceeded to google "clopen".

Hell the only thing I noticed, is that 23 is the 9th prime number, and that 3^2 is 9.

[Surlethe]

I haven't the time for a proper exploration of this question right this second because it's an hour and a half past my bedtime, but it seems to boil down to the question of given a pair of integers (n,m) does there exist a prime p such that p≡m mod n?

Yes.

Unless you want to include negative primes such as -2, (4,6) should work as a counterexample: 6Z+4 are all even, and the smallest positive element of 6Z+4 is 4, which is not prime. So this would then be an open set with no primes in it, indicating that the primes are not dense in the space.

If you want to include negative primes, I think 8Z+4 = 4Z works.

... in fact, nZ, n composite, has no primes in it; therefore, the primes are not dense in the space.

This seems like it might involve Fermat's Little Theorem or the Chinese Remainder Theorem.

You're over-thinking it, surely. (Egads, I should've thought up a better question. But we can rephrase the above with another condition to make it more interesting...)

Ooh, what would it be?