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.
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.)
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.
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]
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.
This is an interesting article. But I think he's being too mysterious about this notion of "cheating".
Consider a far smaller set: {1, 2, 3}. Bob wants to prove that there are at most 3 elements in this set. (Yes, I know this is trivial.) His strategy: name 3 elements in the set (1, 2, 3), and no matter what Alice picks from the set, he will name her choice. Obviously, this strategy works.
Now suppose that Bob wants to prove that there is only 1 element in the set {1, 2, 3}. His strategy is similar: name 1 element in the set (2, say), and -- he hopes -- no matter what Alice picks from the set, he will name her choice. Of course, this doesn't work. However, in order to be sure that Bob does not name her choice: Alice must know his strategy. Does that mean she is "cheating"? Hardly.
The above "cheating" issue is the same as that in the article. So we're not talking about some mysterious property of infinite sets.
The "real" problem here is that the POV the post argues against appears to include the idea that mathematical proofs must be isomorphic to some game between two (or perhaps more) people that a human brain would consider "fair". This is not a generally useful way to approach proofs.
(The word "generally" is not extraneous in this post.)
[+] [-] mjtokelly|16 years ago|reply
http://scienceblogs.com/goodmath/bad_math/cantor_crankery/
Especially frustrating is the credibility Knol can provide to crackpots like this.
[+] [-] WilliamLP|16 years ago|reply
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|16 years ago|reply
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.)
[+] [-] RiderOfGiraffes|16 years ago|reply
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.
[+] [-] tome|16 years ago|reply
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]
[+] [-] jfarmer|16 years ago|reply
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.
[+] [-] ggchappell|16 years ago|reply
Consider a far smaller set: {1, 2, 3}. Bob wants to prove that there are at most 3 elements in this set. (Yes, I know this is trivial.) His strategy: name 3 elements in the set (1, 2, 3), and no matter what Alice picks from the set, he will name her choice. Obviously, this strategy works.
Now suppose that Bob wants to prove that there is only 1 element in the set {1, 2, 3}. His strategy is similar: name 1 element in the set (2, say), and -- he hopes -- no matter what Alice picks from the set, he will name her choice. Of course, this doesn't work. However, in order to be sure that Bob does not name her choice: Alice must know his strategy. Does that mean she is "cheating"? Hardly.
The above "cheating" issue is the same as that in the article. So we're not talking about some mysterious property of infinite sets.
[+] [-] jerf|16 years ago|reply
(The word "generally" is not extraneous in this post.)