4 ms·

A little disappointing that it all boiled down to the lack of a way to handle a negative coefficient, but fun problem nonetheless.

jerf 14y ago
Perhaps so, but it interesting to observe that a lot of very, very smart people stared at those axioms for a long time without noticing that, and that the corresponding proof of that fact was non-trivial. I find it amazing how simple axioms can be, and yet how they are such tricky and subtle beasts. Seemingly obvious axioms can stand for decades until massive, fundamental failures are found in them, and the replacement are only slightly less simple axioms that upon visual inspection reveal none of their heritage. Truly amazing little buggers.
- mturmon 14y ago
  I remember in a mathematical probability class, the professor writing down the three basic axioms. The first two are very simple...the third is just additivity: P(A1 U A2 U ...) = P(A1) + P(A2) + ... for disjoint Ai's. He explained the intent, and asked us if we thought the axiom was reasonable. He seemed very disappointed when everyone nodded and smiled that it seemed okay, unobjectionable, even obvious. He re-emphasized that this was an infinite union. He said something to the effect of, you seem too complacent about this one, because the union is allowed to be infinite, and you can play all kinds of games with that. It was only later that some of us came across the huge literature that alters these axioms (some by weakening the additivity axiom above), to see what more general systems come out.
  - btilly 14y ago
    The third should be just additivity over countable sets of disjoint Ai's. If you leave out countable, then I can make each Ai be a single real number. Under any continuous probability distribution, they have individual probability 0. Yet all together we want them to have probability 1.
    - mturmon 14y ago
      Indeed. I used the ... notation rather than an abstract index to indicate a countable index set. If you expand out all the details, it's tedious. For instance, each Ai also has to be measurable.
    - Someone 14y ago
      Countable does not make a difference. If you pick a random integer, each has probability zero, but you will pick one of them.
      - btilly 14y ago
        Not a counterexample. There exists no continuous probability distribution over the integers, and therefore there is no way to "pick a random integer, each with probability zero".
      - mturmon 14y ago
        No, it needs to be countable unions only. And to your specific example: There is no probability measure on the integers that assigns zero probability to each integer.
        Someone 14y ago
        Yep, that was a lapse of mind. I was thinking of divergent double sums vs divergent double integrals.
jpwagner 14y ago
Remember that this is over the positive integers. I think if you were interested enough to give it deeper thought, you would not be so disappointed.
brazzy 14y ago
Seems to me quite interesting that you need methods involving negative polynomial coefficients to prove statements that use only positive numbers. Actually, it seems to be a concrete example for Gödel's incompleteness theorems: a very simple system can be used to make statements whose veracity can be proven only by using concepts outside the system.
- carstimon 14y ago
  Something similar: If you want to write down a formula for solving a cubic: ax^3+bx^2+cx+d = 0, for give a,b,c,d, you can't avoid using complex numbers. This is true even if you only care about cubics which have real solutions. To clarify that last sentence, contrast it with the quadratic formula, which has no mention of the imaginary unit i. It's true that a quadratic with real coefficients might have only complex solutions, but you might just interpret that as having no solutions: once the formula tells you "take the square root of a negative", you know it has no real solutions. Maybe this could be worked into something similar to the high school algebra problem, but with a list of axioms about real numbers.
- visarga 14y ago
  That's what I thought too: no fixed set of axioms can prove all the possible true affirmations. There is always going to be one that is outside of their scope. That makes mathematics more interesting, almost like a living organism, full of surprises.
tlb 14y ago
It's interesting that you need negative numbers in order to prove certain statements containing only positive numbers. It's analogous to needing complex numbers to prove certain things about reals.

jerf 14y ago

Perhaps so, but it interesting to observe that a lot of very, very smart people stared at those axioms for a long time without noticing that, and that the corresponding proof of that fact was non-trivial. I find it amazing how simple axioms can be, and yet how they are such tricky and subtle beasts. Seemingly obvious axioms can stand for decades until massive, fundamental failures are found in them, and the replacement are only slightly less simple axioms that upon visual inspection reveal none of their heritage. Truly amazing little buggers.

mturmon 14y ago
I remember in a mathematical probability class, the professor writing down the three basic axioms. The first two are very simple...the third is just additivity: P(A1 U A2 U ...) = P(A1) + P(A2) + ... for disjoint Ai's. He explained the intent, and asked us if we thought the axiom was reasonable. He seemed very disappointed when everyone nodded and smiled that it seemed okay, unobjectionable, even obvious. He re-emphasized that this was an infinite union. He said something to the effect of, you seem too complacent about this one, because the union is allowed to be infinite, and you can play all kinds of games with that. It was only later that some of us came across the huge literature that alters these axioms (some by weakening the additivity axiom above), to see what more general systems come out.
- btilly 14y ago
  The third should be just additivity over countable sets of disjoint Ai's. If you leave out countable, then I can make each Ai be a single real number. Under any continuous probability distribution, they have individual probability 0. Yet all together we want them to have probability 1.
  - mturmon 14y ago
    Indeed. I used the ... notation rather than an abstract index to indicate a countable index set. If you expand out all the details, it's tedious. For instance, each Ai also has to be measurable.
  - Someone 14y ago
    Countable does not make a difference. If you pick a random integer, each has probability zero, but you will pick one of them.
    - btilly 14y ago
      Not a counterexample. There exists no continuous probability distribution over the integers, and therefore there is no way to "pick a random integer, each with probability zero".
    - mturmon 14y ago
      No, it needs to be countable unions only. And to your specific example: There is no probability measure on the integers that assigns zero probability to each integer.
      - Someone 14y ago
        Yep, that was a lapse of mind. I was thinking of divergent double sums vs divergent double integrals.

jpwagner 14y ago

Remember that this is over the positive integers. I think if you were interested enough to give it deeper thought, you would not be so disappointed.

brazzy 14y ago

Seems to me quite interesting that you need methods involving negative polynomial coefficients to prove statements that use only positive numbers. Actually, it seems to be a concrete example for Gödel's incompleteness theorems: a very simple system can be used to make statements whose veracity can be proven only by using concepts outside the system.

carstimon 14y ago
Something similar: If you want to write down a formula for solving a cubic: ax^3+bx^2+cx+d = 0, for give a,b,c,d, you can't avoid using complex numbers. This is true even if you only care about cubics which have real solutions. To clarify that last sentence, contrast it with the quadratic formula, which has no mention of the imaginary unit i. It's true that a quadratic with real coefficients might have only complex solutions, but you might just interpret that as having no solutions: once the formula tells you "take the square root of a negative", you know it has no real solutions. Maybe this could be worked into something similar to the high school algebra problem, but with a list of axioms about real numbers.
visarga 14y ago
That's what I thought too: no fixed set of axioms can prove all the possible true affirmations. There is always going to be one that is outside of their scope. That makes mathematics more interesting, almost like a living organism, full of surprises.

tlb 14y ago

It's interesting that you need negative numbers in order to prove certain statements containing only positive numbers. It's analogous to needing complex numbers to prove certain things about reals.