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> Reading this story in Hofstadner's GEB destroyed my ability to accept mathematical proofs as "proven". That's too bad, because the anecdote doesn't challenge

by lutusp 12y ago

> Reading this story in Hofstadner's GEB destroyed my ability to accept mathematical proofs as "proven". That's too bad, because the anecdote doesn't challenge the basis for mathematical proofs or logical reasoning, in fact it requires it as a precondition for the anecdote to move forward. Remember that Gödel's incompleteness theorems don't argue that there are no true statements, only that some of them cannot be proven true. > So, instead of proof, I have to fall back on intuition ... You might be better off reviewing the structure of logic and mathematical proof. Start here: http://en.wikipedia.org/wiki/Euclid's_theorem http://en.wikipedia.org/wiki/Euclid's_theorem My reasoning is that, if there's one proof sufficiently transparent to win acceptance from a skeptic of logic, then there might be two ... ad infinitum.

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.
tjgq 12y ago
In my humble opinion, proofs by contradiction are not the best examples of mathematical reasoning to be presented to the uninitiated. It has been my experience that people untrained in mathematics find it difficult to (intuitively) accept them as valid.
- phaemon 12y ago
  Euclid's proof isn't a proof by contradiction. As the wikipedia page says: >"Euclid is often erroneously reported to have proved this result by contradiction" It simply says that if you are constructing a list of primes, you can always add one more to the list, therefore there are infinitely many.
  - tjgq 12y ago
    The overall structure of the proof is not by contradiction, but one of the steps is. The Wikipedia article calls this out, right after the sentence you quoted.
    - lutusp 12y ago
      Also, even though it's correct that Euclid didn't pose it as a proof by contradiction, it can certainly be posed that way, and I often use that form when presenting it to nonmathematicians.
      - tjgq 12y ago
        Not everyone thinks the same way. There are certainly lay persons out there who do not find proofs by contradiction jarring when they come across them for the first time. I remember I was one of them. But the impression I get from my (possibly biased) sample is that most non-trained people intuitively see proof by contradiction (or any form of nonconstructive proof, really) as a way of "cheating", because it asserts something does or does not exist without actually producing a (counter)example. YMMV, of course.

by lutusp 12y ago

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.

tjgq 12y ago

In my humble opinion, proofs by contradiction are not the best examples of mathematical reasoning to be presented to the uninitiated. It has been my experience that people untrained in mathematics find it difficult to (intuitively) accept them as valid.

phaemon 12y ago
Euclid's proof isn't a proof by contradiction. As the wikipedia page says: >"Euclid is often erroneously reported to have proved this result by contradiction" It simply says that if you are constructing a list of primes, you can always add one more to the list, therefore there are infinitely many.
- tjgq 12y ago
  The overall structure of the proof is not by contradiction, but one of the steps is. The Wikipedia article calls this out, right after the sentence you quoted.
  - lutusp 12y ago
    Also, even though it's correct that Euclid didn't pose it as a proof by contradiction, it can certainly be posed that way, and I often use that form when presenting it to nonmathematicians.
    - tjgq 12y ago
      Not everyone thinks the same way. There are certainly lay persons out there who do not find proofs by contradiction jarring when they come across them for the first time. I remember I was one of them. But the impression I get from my (possibly biased) sample is that most non-trained people intuitively see proof by contradiction (or any form of nonconstructive proof, really) as a way of "cheating", because it asserts something does or does not exist without actually producing a (counter)example. YMMV, of course.