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Thanks for the link; I now feel a little more convinced by Euclid's Theorem than last time I looked at it. Though I still don't feel fully convinced by it; I d

by hyp0 12y ago

Thanks for the link; I now feel a little more convinced by Euclid's Theorem than last time I looked at it. Though I still don't feel fully convinced by it; I don't fully see it. It's entirely possible my obstacle is not so much my skepticism as my stupidity :-)

lutusp 12y ago
To see the power of a given proof, try to imagine what would be required to refute it, falsify it. This is by no means the only avenue of attack, but it's instructive. Also, it resembles the approach used by scientists with respect to falsifiable scientific theories (which aren't the same thing as mathematical proofs).
phaemon 12y ago
I'd be interested in hearing why you're not fully convinced. There seem to be 2 parts to the proof. If you have a list of primes: You can generate another number from that list You can always get a prime from that number to add to the list I'm guessing it's the second part that isn't clicking with you, but perhaps I'm wrong. As for 'stupidity', I wouldn't worry about it. The only people I've ever had call me a moron or question my intelligence in any way have always been people who were less intelligent than I am. And that's not because I'm a genius ;-)
- hyp0 12y ago
  I can follow the steps, but not see it. Like turn-by-turn directions, but no map. Perhaps also because I couldn't come up with it on my own - I don't see the family of which it is an instance (partly, this is the magic open-endedness of mathematics, it's not predictable). But I'm seeing more: start with some primes. They needn't be consective or ordered, just some primes. Any old primes will do. eg 2 and 5 are OK (skipping 3). Now multiply them all to get p. Obviously, p is divisible by all the primes we started with, because we just created it by multiplying them. eg 2 * 5 = 10 Note that p will generally be quite a bit bigger than the primes. Typically, you'll have the primes bunched up near the left of the number line, perhaps with some primes skipped between them, then a big gap to p, and continuing to infinity on the right. Now we add one to p. This is just to the right of p on the number line. This p+1 is either prime or it isn't. 1. If it's prime, then there is a prime other than the ones we started with. eg 10 + 1 = 11 2. If it's not prime, it has divisors. This proof claims it must include divisors that are prime but are not among those we started with. <-- THIS IS THE BIT I DON'T GET You can keep doing this, including that new prime (ie either p+1 itself or a prime divisor of it), showing there are infinitely many primes. So, yes, it's the second part.
  - 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

by hyp0 12y ago

lutusp 12y ago

To see the power of a given proof, try to imagine what would be required to refute it, falsify it. This is by no means the only avenue of attack, but it's instructive. Also, it resembles the approach used by scientists with respect to falsifiable scientific theories (which aren't the same thing as mathematical proofs).

phaemon 12y ago

I'd be interested in hearing why you're not fully convinced. There seem to be 2 parts to the proof. If you have a list of primes: You can generate another number from that list You can always get a prime from that number to add to the list I'm guessing it's the second part that isn't clicking with you, but perhaps I'm wrong. As for 'stupidity', I wouldn't worry about it. The only people I've ever had call me a moron or question my intelligence in any way have always been people who were less intelligent than I am. And that's not because I'm a genius ;-)

hyp0 12y ago
I can follow the steps, but not see it. Like turn-by-turn directions, but no map. Perhaps also because I couldn't come up with it on my own - I don't see the family of which it is an instance (partly, this is the magic open-endedness of mathematics, it's not predictable). But I'm seeing more: start with some primes. They needn't be consective or ordered, just some primes. Any old primes will do. eg 2 and 5 are OK (skipping 3). Now multiply them all to get p. Obviously, p is divisible by all the primes we started with, because we just created it by multiplying them. eg 2 * 5 = 10 Note that p will generally be quite a bit bigger than the primes. Typically, you'll have the primes bunched up near the left of the number line, perhaps with some primes skipped between them, then a big gap to p, and continuing to infinity on the right. Now we add one to p. This is just to the right of p on the number line. This p+1 is either prime or it isn't. 1. If it's prime, then there is a prime other than the ones we started with. eg 10 + 1 = 11 2. If it's not prime, it has divisors. This proof claims it must include divisors that are prime but are not among those we started with. <-- THIS IS THE BIT I DON'T GET You can keep doing this, including that new prime (ie either p+1 itself or a prime divisor of it), showing there are infinitely many primes. So, yes, it's the second part.
- 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