5 ms·

Where I part ways with mathematicians is the intuition we should ascribe to Cantor's argument. They say that "almost all" real numbers are irrational. What Cant

Where I part ways with mathematicians is the intuition we should ascribe to Cantor's argument. They say that "almost all" real numbers are irrational. What Cantor's argument says is that "at least one" is irrational. I know the arguments about measure and probability and such, but, there is also the fact that for every two real numbers, there is an infinity of rational numbers in between! So that makes me (personally, not mathematically) not like the "almost all" intuition, instead of an intuition of there always being at least one more, in a particularly slippery way.

ihodes 16y ago
Ah, but Cantor's proof doesn't just prove that there's "at least one", it proves that there are infinite irrationals. It's just the way this proof is often presented that leads people to believe that it only proves that at least one is irrational; but once you accept that there is one irrational number, you can "index" it and add it to your denumerable list (pretend it's rational, say) and then run the diagonlization argument again, yielding yet another irrational number. Do this ad infinitum. Or, rather, stop: this will not halt ;) Other proofs prove that the rationals are not equinumerous to the reals, but that there is a one-to-one function from the rationals to the reals… but that's another story. (EDIT: as I'm getting a lot of hate and downvotes for this, I suppose I could delete the comment. It was intended to try to clear matters up for the parent of this thread. It wasn't intended as a deep explanation of anything at all. Forgive my triviality.)
- stcredzero 16y ago
  If you said that right, my mind's blown. Wouldn't that prove the reals are countable?
  - ihodes 16y ago
    Which part? One-to-one means that there's an unique element in the range for every element in the domain. That doesn't mean there are the same number of elements. (EDIT: typo strike second range, replace with domain!) (EDIT2: To reply to a reply below: it's absolutely trivial to provide the function: to prove it's true wouldn't be, at least set theoretically, which is how I'd do it.)
    - dfranke 16y ago
      So is the proof you're referring to actually asserting anything that isn't trivial? The identity function is one-to-one from the rationals to the reals.
    - deleted 16y ago
      [deleted]
  - deleted 16y ago
    [deleted]
- larryfreeman 16y ago
  Could you provide a link to the argument that there is a one-to-one function from the rationals to the reals or at least provide more details about what you mean. Thanks.
  - ihodes 16y ago
    Absolutely: Say there's a function ƒ: Q -> R (Q: rationals, R: reals). ƒ(x) = x where x∈Q and x∈R. It's a trivial function, but clearly every rational number is also a real. I could bring it back down to set theory to prove it further, but I that should clear it up! EDIT: One-to-one is not a bijection. It is an injection. A one-to-one correspondence is a bijection. Look at Wikipedia, for instance.
    - deleted 16y ago
      [deleted]
      - jfarmer 16y ago
        "one to one" typically means injective (into). That is, every element in the image corresponds to a unique element from the domain. That's just a convention, though. To be precise one ought to use terms like injection, surjection, and bijection. When speaking informally, sometimes people say "one to one" and mean bijection. It's a catch 22 in a way, because you can only divine the precise meaning if you already understand the proof, viz., Cantor's proof isabouf bijections sp "one to one" means that.
        qwzybug 16y ago
        At my school, they said "one-to-one" and "onto" for "injective" and "surjective". But then, the linear professor started teaching the subject in 1952, so he was quirky.
- hugh3 16y ago
  It doesn't just prove that there's infinitely many irrationals -- that is trivially easy to prove. For instance we know pi is irrational, therefore 2pi, 3pi, 4pi, 5pi,... etc are all irrational as well. Cantor's proof proves that there's no one-to-one correspondence between reals and integers (whereas there is a one-to-one correspondence between integers and rationals), and therefore the infinite set of reals is "bigger" than the infinite set of rationals. That there is an infinite number of irrationals should follow as a trivial consequence of this. (Left as an exercise for the reader...)
  - ihodes 16y ago
    I understand this: I was attempting to show, intuitively, how his theory could demonstrate that there is more than one irrational number. Apologies if I wasn't clear: this was something that popped into my head to help clear this up!
  - jplewicke 16y ago
    What's also somewhat interesting is there's a non-trivial countable set of irrational numbers: those formed by taking the roots of polynomials with raitonal coefficients. http://en.wikipedia.org/wiki/Algebraic_number http://en.wikipedia.org/wiki/Algebraic_number has the low-down, but it's definitely neat that square roots, etc. are still a countable set.
- tome 16y ago
  I think ihodes is using "one-to-one" to mean injective. Sometimes "one-to-one" is used to mean bijective. This confusion is why the term "one-to-one" should never be used!
  - ihodes 16y ago
    One-to-one correspondence is bijective. One-to-one is injective. Sorry for the confusion, but this is a standard definition, even if it's one you're not used to. (EDIT: as you keep editing and replying, let me clarify: your confusion is not shared among informed others. Bijective and injective are wonderful terms, yes. One-to-one and one-to-one correspondence are equally informative and useful.)
    - tome 16y ago
      OK, well it's fine to use whatever terminology you like if there are standard definitions, but I think terms including "one-to-one" are unnecessarily confusing, whereas "injective" and "bijective" are perfectly clear.
- deleted 16y ago
  [deleted]
- WilliamLP 16y ago
  I have a degree in math, I know about Lesbegue measure and probability. My math knowledge is slightly more sophisticated than you give me credit for, if not much more! What you're describing is, to me, the "slippery way" that there is always one more. Yes, one more than a countable set implies infinitely many more, clearly, because there will always be one more even after adding another (or a countable number.) However... the point I'm attempting to articulate - which is not defined precisely and hence is not mathematical, is that almost all just doesn't work in my brain if there are an infinity of "countable" numbers between any two "uncountable" ones!
  - sajid 16y ago
    There are a countable number of countable numbers between any two uncountable ones. But there are an uncountable number of uncountable numbers between any two countable ones.
tome 16y ago
This is not correct. It says that for any countable collection of real numbers, there's another real number that doesn't occur in that collection. (Hence the collection of real numbers cannot be countable). [You don't need Cantor to show that there are irrational real numbers. sqrt(2) is irrational, for example. These kinds of numbers are called algebraic, and there are still only countably many of them.] [Edit: I corrected "transcendental" to "algebraic". Thanks to RiderOfGiraffes for the correction]
- RiderOfGiraffes 16y ago
  Er, what you have written is wrong. sqrt(2) is not transcendental, and there are uncountably many transcendentals, not countably many. You may have meant algebraic. sqrt(2) is algebraic, and there are countably many of them.
  - tome 16y ago
    Yes, I meant algebraic. Thanks for the correction. The transcendentals are the reals that are not algebraic.
RiderOfGiraffes 16y ago
Cantor's diagonalisation argument is usually presented as constructing just one number not in your contable list, but it can be seen to construct more. Take your list of numbers as decimal expansions. In the first place use any of the digits not in the first number, then in the second place use any of the digits not in the second number, etc. That immediately gives you infinitely many numbers not in your list. But we can do better. In the second place put any digit not in the second place of the first number, then in the first place put any digit not in the first place of the second number. Similarly with 3rd and 4th, and so on. But we're not done. Take any permutation of your list and do it again. In this way you can see that there isn't just one number missing, there really are lots and lots and lots and lots missing. And there's more, but this isn't the place for that. I hope this is enough that you can see why most numbers aren't in any countable list.
- WilliamLP 16y ago
  What you're describing is the "slippery" part of my "slippery one more". Yes I know the arguments. What I don't have is trust that intuitive concepts like "almost all" fit! To me the real numbers are something quite alien. The fact that they are Well-Ordered is one weird part, as is the fact that for any two there is a countable infinity of rationals in between. I just think that if I have "almost all" of the stuff, you shouldn't be able to have an infinite amount of stuff between every given two pieces of my stuff! So I prefer another intuition. And I know mathematicians disagree with me and assume I don't understand the basic facts about cardinality, bijections, measure, etc. I do! I know there are a ton of cranks who doubt Cantor's argument. Yes I've been to sci.math before! I like to think I'm not one of them so I'm careful to state I'm not talking within math itself but rather the about way that mathematicians describe an intuition about math.
  - RiderOfGiraffes 16y ago
    The reals are not Well-Ordered under the usual ordering, and they arenot Well Ordered at all. Certainly if you create a Well Ordering of the reals then it's not the usual ordering. And you are free to work with your own intuition, but it's very, very likely that your intuition will be inconsistent with other axioms that you accept to be true. Once you have worked with them enough there is nothing bizarre about every two reals having coutnably many rationals between them, and every two rationals having uncountably many reals between them. I just think that if I have "almost all" of the stuff, you shouldn't be able to have an infinite amount of stuff between every given two pieces of my stuff! This strikes me as very strange. If the rationals (or algebraics) are countable, then "almost all" of the reals are not rationals. Why should it not be that between any two transcendentals (of which there are uncountably many) there are "a few" (by which I mean only countably many) rationals? I think the point to make is that for centuries, perhaps millennia, people grappled unsuccessfully with the ideas of infinite cardinalities. Many, many mistakes were made, and finally there is now a coherent, consistent way of working with them. To say they don't match your intuition is simply to say that your intuition is at odds with the centuries of work done by people who devoted their lives to working on it. In short, you are welcome to your intuitions, but I suspect they are hindering you substantially from understanding things better. You can develop better intuition by working with these ideas "properly." But that's harder than just saying "My intuition doesn't agree with what the mathematicians say."
jfarmer 16y ago
No, that's not right. Cantor's argument says that it's impossible to create a bijection between the rationals and the irrationals. It does this by considering an arbitrary bijection and then showing there is at least one irrational not in that bijections image (a contradiction). In that sense there are "more" irrationals than rationals. There is another field of mathematics called measure theory in which the notion of "almost all" is made precise and in that case, yes, almost all reals are irrational. But that has nothing to do with Cantor's proof, which is pure set theory.

ihodes 16y ago

Ah, but Cantor's proof doesn't just prove that there's "at least one", it proves that there are infinite irrationals. It's just the way this proof is often presented that leads people to believe that it only proves that at least one is irrational; but once you accept that there is one irrational number, you can "index" it and add it to your denumerable list (pretend it's rational, say) and then run the diagonlization argument again, yielding yet another irrational number. Do this ad infinitum. Or, rather, stop: this will not halt ;) Other proofs prove that the rationals are not equinumerous to the reals, but that there is a one-to-one function from the rationals to the reals… but that's another story. (EDIT: as I'm getting a lot of hate and downvotes for this, I suppose I could delete the comment. It was intended to try to clear matters up for the parent of this thread. It wasn't intended as a deep explanation of anything at all. Forgive my triviality.)

stcredzero 16y ago
If you said that right, my mind's blown. Wouldn't that prove the reals are countable?
- ihodes 16y ago
  Which part? One-to-one means that there's an unique element in the range for every element in the domain. That doesn't mean there are the same number of elements. (EDIT: typo strike second range, replace with domain!) (EDIT2: To reply to a reply below: it's absolutely trivial to provide the function: to prove it's true wouldn't be, at least set theoretically, which is how I'd do it.)
  - dfranke 16y ago
    So is the proof you're referring to actually asserting anything that isn't trivial? The identity function is one-to-one from the rationals to the reals.
  - deleted 16y ago
    [deleted]
- deleted 16y ago
  [deleted]
larryfreeman 16y ago
Could you provide a link to the argument that there is a one-to-one function from the rationals to the reals or at least provide more details about what you mean. Thanks.
- ihodes 16y ago
  Absolutely: Say there's a function ƒ: Q -> R (Q: rationals, R: reals). ƒ(x) = x where x∈Q and x∈R. It's a trivial function, but clearly every rational number is also a real. I could bring it back down to set theory to prove it further, but I that should clear it up! EDIT: One-to-one is not a bijection. It is an injection. A one-to-one correspondence is a bijection. Look at Wikipedia, for instance.
  - deleted 16y ago
    [deleted]
    - jfarmer 16y ago
      "one to one" typically means injective (into). That is, every element in the image corresponds to a unique element from the domain. That's just a convention, though. To be precise one ought to use terms like injection, surjection, and bijection. When speaking informally, sometimes people say "one to one" and mean bijection. It's a catch 22 in a way, because you can only divine the precise meaning if you already understand the proof, viz., Cantor's proof isabouf bijections sp "one to one" means that.
      - qwzybug 16y ago
        At my school, they said "one-to-one" and "onto" for "injective" and "surjective". But then, the linear professor started teaching the subject in 1952, so he was quirky.
hugh3 16y ago
It doesn't just prove that there's infinitely many irrationals -- that is trivially easy to prove. For instance we know pi is irrational, therefore 2pi, 3pi, 4pi, 5pi,... etc are all irrational as well. Cantor's proof proves that there's no one-to-one correspondence between reals and integers (whereas there is a one-to-one correspondence between integers and rationals), and therefore the infinite set of reals is "bigger" than the infinite set of rationals. That there is an infinite number of irrationals should follow as a trivial consequence of this. (Left as an exercise for the reader...)
- ihodes 16y ago
  I understand this: I was attempting to show, intuitively, how his theory could demonstrate that there is more than one irrational number. Apologies if I wasn't clear: this was something that popped into my head to help clear this up!
- jplewicke 16y ago
  What's also somewhat interesting is there's a non-trivial countable set of irrational numbers: those formed by taking the roots of polynomials with raitonal coefficients. http://en.wikipedia.org/wiki/Algebraic_number http://en.wikipedia.org/wiki/Algebraic_number has the low-down, but it's definitely neat that square roots, etc. are still a countable set.
tome 16y ago
I think ihodes is using "one-to-one" to mean injective. Sometimes "one-to-one" is used to mean bijective. This confusion is why the term "one-to-one" should never be used!
- ihodes 16y ago
  One-to-one correspondence is bijective. One-to-one is injective. Sorry for the confusion, but this is a standard definition, even if it's one you're not used to. (EDIT: as you keep editing and replying, let me clarify: your confusion is not shared among informed others. Bijective and injective are wonderful terms, yes. One-to-one and one-to-one correspondence are equally informative and useful.)
  - tome 16y ago
    OK, well it's fine to use whatever terminology you like if there are standard definitions, but I think terms including "one-to-one" are unnecessarily confusing, whereas "injective" and "bijective" are perfectly clear.
deleted 16y ago
[deleted]
WilliamLP 16y ago
I have a degree in math, I know about Lesbegue measure and probability. My math knowledge is slightly more sophisticated than you give me credit for, if not much more! What you're describing is, to me, the "slippery way" that there is always one more. Yes, one more than a countable set implies infinitely many more, clearly, because there will always be one more even after adding another (or a countable number.) However... the point I'm attempting to articulate - which is not defined precisely and hence is not mathematical, is that almost all just doesn't work in my brain if there are an infinity of "countable" numbers between any two "uncountable" ones!
- sajid 16y ago
  There are a countable number of countable numbers between any two uncountable ones. But there are an uncountable number of uncountable numbers between any two countable ones.

tome 16y ago

This is not correct. It says that for any countable collection of real numbers, there's another real number that doesn't occur in that collection. (Hence the collection of real numbers cannot be countable). [You don't need Cantor to show that there are irrational real numbers. sqrt(2) is irrational, for example. These kinds of numbers are called algebraic, and there are still only countably many of them.] [Edit: I corrected "transcendental" to "algebraic". Thanks to RiderOfGiraffes for the correction]

RiderOfGiraffes 16y ago
Er, what you have written is wrong. sqrt(2) is not transcendental, and there are uncountably many transcendentals, not countably many. You may have meant algebraic. sqrt(2) is algebraic, and there are countably many of them.
- tome 16y ago
  Yes, I meant algebraic. Thanks for the correction. The transcendentals are the reals that are not algebraic.

RiderOfGiraffes 16y ago

Cantor's diagonalisation argument is usually presented as constructing just one number not in your contable list, but it can be seen to construct more. Take your list of numbers as decimal expansions. In the first place use any of the digits not in the first number, then in the second place use any of the digits not in the second number, etc. That immediately gives you infinitely many numbers not in your list. But we can do better. In the second place put any digit not in the second place of the first number, then in the first place put any digit not in the first place of the second number. Similarly with 3rd and 4th, and so on. But we're not done. Take any permutation of your list and do it again. In this way you can see that there isn't just one number missing, there really are lots and lots and lots and lots missing. And there's more, but this isn't the place for that. I hope this is enough that you can see why most numbers aren't in any countable list.

WilliamLP 16y ago
What you're describing is the "slippery" part of my "slippery one more". Yes I know the arguments. What I don't have is trust that intuitive concepts like "almost all" fit! To me the real numbers are something quite alien. The fact that they are Well-Ordered is one weird part, as is the fact that for any two there is a countable infinity of rationals in between. I just think that if I have "almost all" of the stuff, you shouldn't be able to have an infinite amount of stuff between every given two pieces of my stuff! So I prefer another intuition. And I know mathematicians disagree with me and assume I don't understand the basic facts about cardinality, bijections, measure, etc. I do! I know there are a ton of cranks who doubt Cantor's argument. Yes I've been to sci.math before! I like to think I'm not one of them so I'm careful to state I'm not talking within math itself but rather the about way that mathematicians describe an intuition about math.
- RiderOfGiraffes 16y ago
  The reals are not Well-Ordered under the usual ordering, and they arenot Well Ordered at all. Certainly if you create a Well Ordering of the reals then it's not the usual ordering. And you are free to work with your own intuition, but it's very, very likely that your intuition will be inconsistent with other axioms that you accept to be true. Once you have worked with them enough there is nothing bizarre about every two reals having coutnably many rationals between them, and every two rationals having uncountably many reals between them. I just think that if I have "almost all" of the stuff, you shouldn't be able to have an infinite amount of stuff between every given two pieces of my stuff! This strikes me as very strange. If the rationals (or algebraics) are countable, then "almost all" of the reals are not rationals. Why should it not be that between any two transcendentals (of which there are uncountably many) there are "a few" (by which I mean only countably many) rationals? I think the point to make is that for centuries, perhaps millennia, people grappled unsuccessfully with the ideas of infinite cardinalities. Many, many mistakes were made, and finally there is now a coherent, consistent way of working with them. To say they don't match your intuition is simply to say that your intuition is at odds with the centuries of work done by people who devoted their lives to working on it. In short, you are welcome to your intuitions, but I suspect they are hindering you substantially from understanding things better. You can develop better intuition by working with these ideas "properly." But that's harder than just saying "My intuition doesn't agree with what the mathematicians say."

jfarmer 16y ago

No, that's not right. Cantor's argument says that it's impossible to create a bijection between the rationals and the irrationals. It does this by considering an arbitrary bijection and then showing there is at least one irrational not in that bijections image (a contradiction). In that sense there are "more" irrationals than rationals. There is another field of mathematics called measure theory in which the notion of "almost all" is made precise and in that case, yes, almost all reals are irrational. But that has nothing to do with Cantor's proof, which is pure set theory.