There are obviously infinitely many irrationals between any two rationals, I think there are also infinitely many rationals between any two (different) irrationals. How would you decide whether the sets of rationals and irrationals are equal or not?
How about Reals and irrationals?? Or do I need to go and do some further reading to understand these problems?
Cheers
I'm sure AN will give us the technical dope, but I have it that the infinity of the Real numbers is greater than the infinity of the natural numbers. Since the rationals can be put into a one-to-one with the naturals, their infinity is the same as the naturals.
However, there is no function that will map the naturals to the irrationals, so that infinity is a larger infinity.
However, there is no function that will map the naturals to the irrationals, so that infinity is a larger infinity.
QUOTE (NoCleverName+Jun 19 2007, 10:35 AM)
I'm sure AN will give us the technical dope, but I have it that the infinity of the Real numbers is greater than the infinity of the natural numbers. Since the rationals can be put into a one-to-one with the naturals, their infinity is the same as the naturals.
However, there is no function that will map the naturals to the irrationals, so that infinity is a larger infinity.
Yep, that makes sense. Do you know how the rationals are mapped onto the natural numbers? Or just know they are and leaving that bit for AN?
However, there is no function that will map the naturals to the irrationals, so that infinity is a larger infinity.
Yep, that makes sense. Do you know how the rationals are mapped onto the natural numbers? Or just know they are and leaving that bit for AN?
QUOTE (bm1957+Jun 19 2007, 10:41 AM)
Yep, that makes sense. Do you know how the rationals are mapped onto the natural numbers? Or just know they are and leaving that bit for AN?
Well, quite a few of them can be easily mapped using the function y = 1/x. In other words {1,2,3, ...} quickly map to {1/1,1/2,1/3...}. I think there's a similar trick that escapes me at the moment that handles all the rest {2/2,2/3,3/3,3/4,4/4,...}.
Well, quite a few of them can be easily mapped using the function y = 1/x. In other words {1,2,3, ...} quickly map to {1/1,1/2,1/3...}. I think there's a similar trick that escapes me at the moment that handles all the rest {2/2,2/3,3/3,3/4,4/4,...}.
QUOTE (NoCleverName+Jun 19 2007, 10:49 AM)
Well, quite a few of them can be easily mapped using the function y = 1/x. In other words {1,2,3, ...} quickly map to {1/1,1/2,1/3...}. I think there's a similar trick that escapes me at the moment that handles all the rest {2/2,2/3,3/3,3/4,4/4,...}.
Ok; are 1/2 and 2/4 different rational numbers?
Either way, it appears that the rationals can be ordered: {1/1, 1/2, 2/2, 1/3, 2/3, 3/3, 1/4, ...}
Would this allow them to be simply mapped in that order? (Leaving out the repeated fractions if they do not count as distinct rationals)
Ok; are 1/2 and 2/4 different rational numbers?
Either way, it appears that the rationals can be ordered: {1/1, 1/2, 2/2, 1/3, 2/3, 3/3, 1/4, ...}
Would this allow them to be simply mapped in that order? (Leaving out the repeated fractions if they do not count as distinct rationals)
The rationals are countable.
This is provable from the fact that the rationals are defined in terms of two integers, A and B (B not 0), so a rational number is A/B. Therefore, the set of all rationals is a subset of the square of the number of integers. This is because given an order pair of integers (A,B), some pairs of A and B will give the same value for A/B. For instance (1,1) is equivalent to the same rational as (2,2).
Therefore the size of rationals is less than or equal to the size of N^2.
So let's consider N^2. Can I put this on a bijective correspondence with a subset of N? Well you'd think not but you'd be wrong. Since prime factorisation is unique for any given integer, I can relate bijectively the element (A,B) in N^2 to the number (2^A)(3^B) in N. Therefore N^2 has the same cardinality as N. Since the rationals are a subset of N^2, it turns out the rationals must be countable.
However, by Cantor's diagonal argument, the Reals are not countable, they are a larger, inequivalent, set from the integers (and hence rationals). Since "Is X rational?" is a dichotomy, since a number is either rational or irrational, there's no third choice, this implies that the set of irrational numbers is the compliement of the rationals within the Reals. Therefore, since I cannot add together Aleph Null and Aleph Null to get Aleph One (add in the set union sense), the set of irrationals must be uncountable.
Therefore the set of irrationals is strictly larger and inequivalent from the set of rationals.
QED
This is provable from the fact that the rationals are defined in terms of two integers, A and B (B not 0), so a rational number is A/B. Therefore, the set of all rationals is a subset of the square of the number of integers. This is because given an order pair of integers (A,B), some pairs of A and B will give the same value for A/B. For instance (1,1) is equivalent to the same rational as (2,2).
Therefore the size of rationals is less than or equal to the size of N^2.
So let's consider N^2. Can I put this on a bijective correspondence with a subset of N? Well you'd think not but you'd be wrong. Since prime factorisation is unique for any given integer, I can relate bijectively the element (A,B) in N^2 to the number (2^A)(3^B) in N. Therefore N^2 has the same cardinality as N. Since the rationals are a subset of N^2, it turns out the rationals must be countable.
However, by Cantor's diagonal argument, the Reals are not countable, they are a larger, inequivalent, set from the integers (and hence rationals). Since "Is X rational?" is a dichotomy, since a number is either rational or irrational, there's no third choice, this implies that the set of irrational numbers is the compliement of the rationals within the Reals. Therefore, since I cannot add together Aleph Null and Aleph Null to get Aleph One (add in the set union sense), the set of irrationals must be uncountable.
Therefore the set of irrationals is strictly larger and inequivalent from the set of rationals.
QED
QUOTE (AlphaNumeric+Jun 19 2007, 11:09 AM)
QED
I knew he'd come through. It's like blood in the water for that shark.
I knew he'd come through. It's like blood in the water for that shark.
QUOTE (NoCleverName+Jun 19 2007, 12:12 PM)
I knew he'd come through. It's like blood in the water for that shark.
This section of mathematics was one of my poorest when I was in my first year. I just couldn't get my head around it. Even now, it's one of my weaker areas, it's just that this stuff has had so much time to bubble away in the back of my mind I grasp a decent chunk of the methodology now. Still not even close to being a decent mathematician in the area though, people who do this stuff day in day out are miles ahead of me, but I'm quite pleased with just how much I can recall and prove now.
Shame it took me 4 years to get there!
This section of mathematics was one of my poorest when I was in my first year. I just couldn't get my head around it. Even now, it's one of my weaker areas, it's just that this stuff has had so much time to bubble away in the back of my mind I grasp a decent chunk of the methodology now. Still not even close to being a decent mathematician in the area though, people who do this stuff day in day out are miles ahead of me, but I'm quite pleased with just how much I can recall and prove now.
Shame it took me 4 years to get there!
Card(Z) = Card(Z*Z) : nice proof, with the prime factorization. I knew only the one with N and N*N and putting numbers on bigger and bigger triangles. Prime factorization is definitely more elegant.
Card( R) = Card(R*R) since Card([0:1]) = Card([0:1]*[0:1]) through x=tg(y)
and from a number in [0:1], make a number from the odd positions and another from the even ones.
And anyway, Card(S) = Card(S^n) for any infinite set S.
But... Hi AlphaNumeric, you wrote Aleph One as the number of irrationals...
Can you prove that it's One and not Aleph Two or Three...?
http://en.wikipedia.org/wiki/Continuum_hypothesis
You would become really famous
Card( R) = Card(R*R) since Card([0:1]) = Card([0:1]*[0:1]) through x=tg(y)
and from a number in [0:1], make a number from the odd positions and another from the even ones.
And anyway, Card(S) = Card(S^n) for any infinite set S.
But... Hi AlphaNumeric, you wrote Aleph One as the number of irrationals...
Can you prove that it's One and not Aleph Two or Three...?
http://en.wikipedia.org/wiki/Continuum_hypothesis
You would become really famous
QUOTE
not to mention all the sub-links
QUOTE (->
| QUOTE |
| Enthalpy http://en.wikipedia.org/wiki/Continuum_hypothesis |
not to mention all the sub-links
http://us.metamath.org/mpegif/aleph1irr.html
There are at least Alph-1 irrationals.
Also: http://us.metamath.org/mpegif/resdomq.html
http://us.metamath.org/mpegif/aleph1re.html
http://us.metamath.org/mpegif/qnnen.html
http://us.metamath.org/mpegif/nnenom.html
http://us.metamath.org/mpegif/aleph0.html
Please! please!
One link per week! I'm not a super-hyper absorbent sponge here...
One of my all-time favorite books is Cantor's original monograph "Contributions to the Foundation of a Theory of Transfinite Numbers." It's quite readable, available from Dover paperbacks for a low price, and is historically interesting, because it was the first introduction of many of the foundational features of 20th centure mathematics.
Try it, you'll like it!
--Stuart Anderson
Try it, you'll like it!
--Stuart Anderson
I may well do
Thanks
Thanks
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