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EDIT I can see the divisor that must exist cannot be one of the given primes: taking just one of them, multiplied by the product of the rest, the next number i

by hyp0 12y ago

EDIT I can see the divisor that must exist cannot be one of the given primes: taking just one of them, multiplied by the product of the rest, the next number it divides after p must be one extra addition of it, which will be greater than our number p+1. Therefore, it isn't a divisor. The same argument excludes all the other initial primes. So this means: it has a divisor not in the initial primes (actually, I think it must have two). But why should it be prime? I think a given divisor does not need to be prime; but it must not be divisible by an initial prime. I guess this means that either it itself is prime, or it has divisors which in turn are either prime or have divisors etc. None of these divisors are an initial prime, because then they would also be divisors of p+1, which we have established they are not. So I guess that's the proof... but I don't feel sure of it. There are too many steps, and I'm not 100% sure of them, and can't see the whole. Perhaps I've not covered some possibility in some step - how could I be sure I've covered them all? Maybe as it becomes more familiar, I will come to see it.

lutusp 12y ago
Remember the role of axioms, which another poster has explained in a different way. The issue in question (not itself an axiom but one that requires acceptance of axioms) is whether each composite (i.e. non-prime) is uniquely composed of primes. To prove this for yourself, try assembling a composite number out of non-prime factors. Then, to make sure of your result, decompose your factors into the primes from which they were composed. Finally, restate your factorization by replacing your factors with the primes that compose them. Example: the composite number 32 is normally factored as 2^5, i.e. four multiplications of the prime number 2. Let's say I want to falsify the idea that all positive integers are either prime or uniquely composed of primes, so I instead compose 32 using the nonprime factors 8 and 4. Then I factor 8 and 4, and discover that their prime factors are also factors of 32 -- 8 = 2^3 4 = 2^2 32 = 2^5 -- so I have proven the original thesis: all positive integers are either themselves prime or are uniquely composed of primes. The idea I am trying to convey is that the original claim doesn't mean one cannot assemble a composite out of non-primes, only that the composite number is also representable by a unique prime factorization. More depth here: http://en.wikipedia.org/wiki/Prime_factor http://en.wikipedia.org/wiki/Prime_factor
phaemon 12y ago
> "So this means: it has a divisor not in the initial primes (actually, I think it must have two). But why should it be prime?" Ok, this is the heart of the issue. If the new number is a prime, all is well and good, but if the number isn't prime, why should it's divisors be? The simple answer is: they don't have to be. If you have divisors that aren't prime, then keep dividing till you hit some that are. The definition of a prime number is one that can only be divided by itself, so for any non-prime number, you must be able to keep finding factors until they're all prime! Let's take an example. Our list of primes is {5,7} which are nice small numbers to use. By following the rule of "multiply and add 1" we get: 5 * 7 + 1 = 36. Ok, so let's break 36 down. We get: 36 = 2 * 18 Right, well, 2 isn't on our list, but let's face it: 2 isn't a real prime. None of the other primes like it. It's even. Nor is 18 on our list, but that's not prime (and that was your objection before), so let's break 18 down. 36 = 2 * 2 * 9 Well, that's a bit better. We have another unpopular 2, but we also got a 9, and even though 9 isn't prime, it's probably primier than 2 is. Continue on: 36 = 2 * 2 * 3 * 3 There we go. Now we actually have a proper prime number, "3", that we can add to our list. And you see (I hope) that none of these numbers could possibly be on our original list, because all the numbers already on that list give a remainder of "1" when we divide "36". Yet we must, inevitably, hit a prime number because we just keep dividing till we do!
- lutusp 12y ago
  > Right, well, 2 isn't on our list, but let's face it: 2 isn't a real prime. I can't tell whether you're taking this position or ridiculing it, but if 2 isn't accepted as a prime number, this would falsify the Fundamental Theorem of Arithmetic for all even numbers. http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmet... Quote: "In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1[3] either is prime itself or is the product of prime numbers ..." If your purpose was satire, then perhaps this post will inform other readers who may not detect your satirical intent.
  - phaemon 12y ago
lutusp 12y ago
> I think a given divisor does not need to be prime; but it must not be divisible by an initial prime. But that's how prime is defined -- indivisible by any other numbers except 1. If you statement is true -- that a given number "must not be divisible by an initial prime", that means the number is itself prime. Positive integers fall into precisely two categories: 1. Not divisible by any smaller numbers except 1. 2. Divisible by one or more smaller numbers. Those in category (1) are prime. Those in category (2) are composite. There is no third possibility.
- hyp0 12y ago
  I think you're interpreting "initial primes" as all the primes up to p+1; but here, I defined it to be just the particular primes we happened to start with. There can be gaps. > I think a given divisor does not need to be prime; but it must not be divisible by an initial prime. I will disprove by counter-example that not being divisible by an initial prime implies it is prime: initial primes: 5 and 7 p = 5 * 7 = 35 p+1 = 36 36 has a few divisors. Lets take 18 as a "given divisor". 18 is not divisible by any of the "initial primes", 5 and 7. Yet 18 itself is not prime. (While it is divisible by the primes 2 and 3, but they aren't "initial primes".)

by hyp0 12y ago

lutusp 12y ago

Remember the role of axioms, which another poster has explained in a different way. The issue in question (not itself an axiom but one that requires acceptance of axioms) is whether each composite (i.e. non-prime) is uniquely composed of primes. To prove this for yourself, try assembling a composite number out of non-prime factors. Then, to make sure of your result, decompose your factors into the primes from which they were composed. Finally, restate your factorization by replacing your factors with the primes that compose them. Example: the composite number 32 is normally factored as 2^5, i.e. four multiplications of the prime number 2. Let's say I want to falsify the idea that all positive integers are either prime or uniquely composed of primes, so I instead compose 32 using the nonprime factors 8 and 4. Then I factor 8 and 4, and discover that their prime factors are also factors of 32 -- 8 = 2^3 4 = 2^2 32 = 2^5 -- so I have proven the original thesis: all positive integers are either themselves prime or are uniquely composed of primes. The idea I am trying to convey is that the original claim doesn't mean one cannot assemble a composite out of non-primes, only that the composite number is also representable by a unique prime factorization. More depth here: http://en.wikipedia.org/wiki/Prime_factor http://en.wikipedia.org/wiki/Prime_factor

phaemon 12y ago

> "So this means: it has a divisor not in the initial primes (actually, I think it must have two). But why should it be prime?" Ok, this is the heart of the issue. If the new number is a prime, all is well and good, but if the number isn't prime, why should it's divisors be? The simple answer is: they don't have to be. If you have divisors that aren't prime, then keep dividing till you hit some that are. The definition of a prime number is one that can only be divided by itself, so for any non-prime number, you must be able to keep finding factors until they're all prime! Let's take an example. Our list of primes is {5,7} which are nice small numbers to use. By following the rule of "multiply and add 1" we get: 5 * 7 + 1 = 36. Ok, so let's break 36 down. We get: 36 = 2 * 18 Right, well, 2 isn't on our list, but let's face it: 2 isn't a real prime. None of the other primes like it. It's even. Nor is 18 on our list, but that's not prime (and that was your objection before), so let's break 18 down. 36 = 2 * 2 * 9 Well, that's a bit better. We have another unpopular 2, but we also got a 9, and even though 9 isn't prime, it's probably primier than 2 is. Continue on: 36 = 2 * 2 * 3 * 3 There we go. Now we actually have a proper prime number, "3", that we can add to our list. And you see (I hope) that none of these numbers could possibly be on our original list, because all the numbers already on that list give a remainder of "1" when we divide "36". Yet we must, inevitably, hit a prime number because we just keep dividing till we do!

lutusp 12y ago
> Right, well, 2 isn't on our list, but let's face it: 2 isn't a real prime. I can't tell whether you're taking this position or ridiculing it, but if 2 isn't accepted as a prime number, this would falsify the Fundamental Theorem of Arithmetic for all even numbers. http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmet... Quote: "In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1[3] either is prime itself or is the product of prime numbers ..." If your purpose was satire, then perhaps this post will inform other readers who may not detect your satirical intent.
- phaemon 12y ago

lutusp 12y ago

> I think a given divisor does not need to be prime; but it must not be divisible by an initial prime. But that's how prime is defined -- indivisible by any other numbers except 1. If you statement is true -- that a given number "must not be divisible by an initial prime", that means the number is itself prime. Positive integers fall into precisely two categories: 1. Not divisible by any smaller numbers except 1. 2. Divisible by one or more smaller numbers. Those in category (1) are prime. Those in category (2) are composite. There is no third possibility.

hyp0 12y ago
I think you're interpreting "initial primes" as all the primes up to p+1; but here, I defined it to be just the particular primes we happened to start with. There can be gaps. > I think a given divisor does not need to be prime; but it must not be divisible by an initial prime. I will disprove by counter-example that not being divisible by an initial prime implies it is prime: initial primes: 5 and 7 p = 5 * 7 = 35 p+1 = 36 36 has a few divisors. Lets take 18 as a "given divisor". 18 is not divisible by any of the "initial primes", 5 and 7. Yet 18 itself is not prime. (While it is divisible by the primes 2 and 3, but they aren't "initial primes".)