3 ms·

Reminds me of the proof that all natural numbers are interesting. If there is some set of uninteresting natural numbers, there must be a minimal element of that

Reminds me of the proof that all natural numbers are interesting. If there is some set of uninteresting natural numbers, there must be a minimal element of that set. It being the smallest uninteresting number is interesting which is a contradiction.

rzzzt 1y ago
Why aren't all numbers in the set uninteresting? Did someone make a mistake when defining it? Perhaps the minimal element should be removed from the set; there will be plenty of members that still remain.
- Cerium 1y ago
  Serious response? In that case the set still has a smallest member which can then be removed, if we keep going eventually there will be no uninteresting numbers remaining.
  - leereeves 1y ago
    The problem with that is the explanation of why each number is interesting becomes: the smallest member of the original set of uninteresting numbers the second smallest member of the original set of uninteresting numbers the third ... ... That version of "interesting" quickly becomes "not interesting". The concept simply defies mathematical logic.
    - kbelder 1y ago
      It reminds me about the logic puzzle of the criminal sentenced to death, where the judge says "you will be executed on or before Sunday, and you won't know what day it will be until we come for you." The criminal knows it can't be Sunday, because he would wake up on Sunday and know he was going to be executed that day. But if Sunday isn't possible, on Saturday he would know he was being executed that day; so Saturday wasn't possible either. The same reasoning can be repeatedly applied to every day between now and Sunday. It's obviously flawed reasoning (Surprise! they execute you on Thursday), but the flaw is difficult to articulate.
    - jameshart 1y ago
      This isn't how math works. When you get to the point in a proof of the irrationality of root two where you've demonstrated that if it is expressible as a fraction p/q, then both p and q have to be even, you don't then need to proceed to prove that if they're both even, then they both have to be divisible by four, and then if they're both divisible by four, that means they're both divisible by eight... I mean, you can, but you don't have to. You can just say 'if it's a rational number then it has a reduced form where p and q have gcf of 1, so if p and q would both have to be even, that is a contradiction'. Same with the 'set of uninteresting numbers'. If 'being uninteresting' is a property numbers can have, then the 'set of uninteresting numbers' exists, and it has a least member. Being the least member of the set of uninteresting numbers is interesting. You don't have to infinitely regress from here and get tied up in knots saying that surely there is some 'first truly uninteresting number' to prove that the set is actually empty - you can just see that you must have gone wrong somewhere. Either: 1) Being the least member of the set of uninteresting numbers isn't as interesting as we assume. or 2) 'Being uninteresting' is not a property numbers can have I think actually of the two, 1) is more likely the case. But that doesn't defy mathematical logic. It is a consequence of mathematical logic.
      - deleted 1y ago
        [deleted]
      - leereeves 1y ago
        There's a third option. The definition of uninteresting we're using may be flawed. Here's a quick stab at a more rigorous approach: We could start by defining a set of "all numbers that are uninteresting other than by membership or position in this set". That describes the set the proof naively called "interesting numbers" without the contradiction. Then we could create a second set with all members of the first set except those that are interesting because of where they are in that set (smallest, whatever). This is a new version of "interesting numbers" that approaches the version in the original proof but is, in human terms, less interesting. As you said, "Being the least member of the set of uninteresting numbers isn't as interesting as we assume." We could repeat that, making a sequence of sets that approach the definition of interesting in the original proof, but the definition of each set is progressively less interesting in human terms. Then if we really want to be rigorous, we could talk about "first degree interesting" (what most people mean), "nth degree interesting", or "asymptotically interesting", but the last one is an empty set.
Tade0 1y ago
My algebra 101 professor made this exact argument.
bee_rider 1y ago
Of course, it sort of a joke, and so having an element of surprise helps it. But really, the properties that make a number “interesting” should probably be defined from the outset. By including “the smallest member of any set is interesting,” at the start, the joke is kind of blown because the result becomes obvious, right? Edit: oh, are there uninteresting reals?

rzzzt 1y ago

Why aren't all numbers in the set uninteresting? Did someone make a mistake when defining it? Perhaps the minimal element should be removed from the set; there will be plenty of members that still remain.

Cerium 1y ago
Serious response? In that case the set still has a smallest member which can then be removed, if we keep going eventually there will be no uninteresting numbers remaining.
- leereeves 1y ago
  The problem with that is the explanation of why each number is interesting becomes: the smallest member of the original set of uninteresting numbers the second smallest member of the original set of uninteresting numbers the third ... ... That version of "interesting" quickly becomes "not interesting". The concept simply defies mathematical logic.
  - kbelder 1y ago
    It reminds me about the logic puzzle of the criminal sentenced to death, where the judge says "you will be executed on or before Sunday, and you won't know what day it will be until we come for you." The criminal knows it can't be Sunday, because he would wake up on Sunday and know he was going to be executed that day. But if Sunday isn't possible, on Saturday he would know he was being executed that day; so Saturday wasn't possible either. The same reasoning can be repeatedly applied to every day between now and Sunday. It's obviously flawed reasoning (Surprise! they execute you on Thursday), but the flaw is difficult to articulate.
  - jameshart 1y ago
    This isn't how math works. When you get to the point in a proof of the irrationality of root two where you've demonstrated that if it is expressible as a fraction p/q, then both p and q have to be even, you don't then need to proceed to prove that if they're both even, then they both have to be divisible by four, and then if they're both divisible by four, that means they're both divisible by eight... I mean, you can, but you don't have to. You can just say 'if it's a rational number then it has a reduced form where p and q have gcf of 1, so if p and q would both have to be even, that is a contradiction'. Same with the 'set of uninteresting numbers'. If 'being uninteresting' is a property numbers can have, then the 'set of uninteresting numbers' exists, and it has a least member. Being the least member of the set of uninteresting numbers is interesting. You don't have to infinitely regress from here and get tied up in knots saying that surely there is some 'first truly uninteresting number' to prove that the set is actually empty - you can just see that you must have gone wrong somewhere. Either: 1) Being the least member of the set of uninteresting numbers isn't as interesting as we assume. or 2) 'Being uninteresting' is not a property numbers can have I think actually of the two, 1) is more likely the case. But that doesn't defy mathematical logic. It is a consequence of mathematical logic.
    - deleted 1y ago
      [deleted]
    - leereeves 1y ago
      There's a third option. The definition of uninteresting we're using may be flawed. Here's a quick stab at a more rigorous approach: We could start by defining a set of "all numbers that are uninteresting other than by membership or position in this set". That describes the set the proof naively called "interesting numbers" without the contradiction. Then we could create a second set with all members of the first set except those that are interesting because of where they are in that set (smallest, whatever). This is a new version of "interesting numbers" that approaches the version in the original proof but is, in human terms, less interesting. As you said, "Being the least member of the set of uninteresting numbers isn't as interesting as we assume." We could repeat that, making a sequence of sets that approach the definition of interesting in the original proof, but the definition of each set is progressively less interesting in human terms. Then if we really want to be rigorous, we could talk about "first degree interesting" (what most people mean), "nth degree interesting", or "asymptotically interesting", but the last one is an empty set.

Tade0 1y ago

My algebra 101 professor made this exact argument.

bee_rider 1y ago

Of course, it sort of a joke, and so having an element of surprise helps it. But really, the properties that make a number “interesting” should probably be defined from the outset. By including “the smallest member of any set is interesting,” at the start, the joke is kind of blown because the result becomes obvious, right? Edit: oh, are there uninteresting reals?