Rational/Irrational Numbers
alok bakshi
- An irrational number exist between any two rational numbers.
Given that both these assertions are true, I cannot see how there are
lot more irrational numbers than rational numbers. By intuition,
shouldn't both be same(having same cardinality) because of these two
above mentioned facts...
Thanks,
Alok Bakshi
William Elliot
> - A rational number exist between any two irrational numbers.
>
In fact a lot of rational numbers.
> - An irrational number exist between any two rational numbers.
>
In fact, a lot more irrational numbers.
> Given that both these assertions are true, I cannot see how there are
> lot more irrational numbers than rational numbers. By intuition,
> shouldn't both be same(having same cardinality) because of these two
> above mentioned facts...
>
No. You're thinking somethings like
Just one rational number exists between each pair of
irrational numbers.
Just one irrational number exists between each pair of
rational numbers.
alok bakshi
numbers, no matter how close they are. But then it looks paradoxical
that irrational numbers are more than rational numbers. Rational
numbers are everywhere, between any two irrational numbers, then
doesn't it imply that there are more rational than irrationals. I
understand the mathematical proof, but I cannot convince myself if I
think along these lines...
Thanks,
Alok Bakshi
Rupert
Perhaps there's a certain tendency to think there's something odd about
it, but the simple fact is that there are more irrational numbers than
rational numbers. There are many mathematical theorems that seem odd at
first glance.
Have you ever read a proof that there are more irrational numbers than
rational numbers? Maybe I could try to help you understand it.
alok bakshi
(1) Rational numbers are countable: by enumerating all rationals like
this:
1 2 3 4 5 6.......
2 3 4 5 6 7...
3 4 5 6 7 8..
4 5 6 7 8 9...
.......................
.......................
(2) Then there is proof that real numbers are uncountable by supposing
that a list of real numbers between 0 and 1 (in binary format) exists
and then using Cantor's diagonalization argument to show that it is
impossible.
(3) Construction of real number by Dedekind's cut. From that cut we can
see that
- between two rationals infinite irrational numbers exist. -----(a)
- Between two irrational infinite rational numbers exist. ------(b)
But just by looking at the conclusion (a), and (b) I cannot understand
that there are much more irrational numbers than rational numbers.
Because if I see any two irrational numbers, which may be arbitrarily
close, there are still many(infinite) rational numbers between them.
Then how irrationals are more in number. It looks contradictory to
me...
Thanks for your help...
Alok Bakshi
Rupert
Well, it's not contradictory. I'm not sure why you think it is. Try to
make your argument for a contradiction more explicit.
William Elliot
You agree to what? As you don't quote sufficient contexts, I choose to
end this thread. You'll get to agree to that also.
Reply only if adequate contexts is included _within_ the reply.
http://oakroadsystems.com/genl/unice.htm#quote
http://www.xs4all.nl/~hanb/documents/quotingguide.html
Otherwise essential contexts is removed from view,
the flow of thought disrupted and chaos reigns.
alok bakshi
Let Q be the set of rational numbers and R be set of irrational
numbers.
For any r1, r2 \in R (r1 < r2) we have at least one q \in Q such that
r1 < q < r2
Now I know that elements of Q can be enumerated, and this enumeration
contains all the numbers which exist between any two irrational
numbers, NO MATTER HOW CLOSE THEY ARE. Now doesn't it imply that R is
also countable.
I mean between any two numbers in R, there is at least(in fact many!)
one element of set Q. Now I have problem that I cannot really list all
the elements of R(since it is uncountable). But even then doesn't it
look like R must've same cardinality as that of Q.
Jesse F. Hughes
> You agree to what? As you don't quote sufficient contexts, I choose
> to end this thread. You'll get to agree to that also.
You choose to end this thread? How do you do that?
--
Jesse F. Hughes
"The past two days I haven't had any beer or any wine."
-- Quincy P. Hughes, Age 4 and so damn European
kilian heckrodt
resolved by reading it slightly differently.
"A rational number" has to be read as "At least one rational number"
"An irrational number" has to be read as "At least one irrational number"
If do that the vague 1-1 mapiing notion vanishes immediately, as there's
no way to tell whether one "at least" is as big as the ozher. In fact
for other reason you can show that one is "bigger" than the other.
If you are looking for a detailed argument/reason, why there are more
irrationals, check the web/literature for a proof that the reals are
uncountable.
Note: Generally speaking an unspecified existence formulation (" there
exist a..." ) in mathematical statements is always to read as at least
("there is at least one, that ...").
kilian heckrodt
existence always means the existence of at leas one (unless otherwise
specified)
fishfry
"alok bakshi" <alokb...@gmail.com> wrote:
That's exactly right. The cardinality arguments are counterintuitive.
And the proofs are easy to follow. That's the answer to another thread
-- why so many people question or oppose the basic cardinality proofs.
They're easy to understand, and they're amazing.
There is a rational between every two irrational; and an irrational
between every two rationals. Yet there are a lot more irrationals, by
both cardinality and measure. It's pretty strange.
G. Frege
wrote:
>
> [...] I cannot see how there are lot more irrational numbers than
> rational numbers. By intuition, shouldn't both be same (having same
> cardinality) because of these two above mentioned facts?
>
"The irrational numbers may just be defined as the set of gaps in Q.
Since the set of the irrational numbers has the cardinal aleph
[2^aleph_0], in our case the set of the gaps has a greater cardinal
than Q itself; this phenomenon is not surprising since the cuts are
essentially subsets and not members of the set. By mere spatial
intuition or classical geometrical attitudes without set-theoretical
methods, the phenomenon can neither be stated nor comprehended."
(A. A. Fraenkel, A. Levy, Abstract Set Theory)
Sincerely,
F.
--
E-mail: info<at>simple-line<dot>de
G. Frege
<BLOCKSPA...@your-mailbox.com> wrote:
>
> There is a rational between every two irrational; and an irrational
> between every two rationals. Yet there are a lot more irrationals, by
> both cardinality and measure. It's pretty strange.
>
Indeed!
"By mere spatial intuition or classical geometrical attitudes without
set-theoretical methods, the phenomenon can neither be stated nor
comprehended." (A. A. Fraenkel, A. Levy, Abstract Set Theory)
William Elliot
> William Elliot <ma...@hevanet.remove.com> writes:
>
> > You agree to what? As you don't quote sufficient contexts, I choose
> > to end this thread. You'll get to agree to that also.
>
> You choose to end this thread? How do you do that?
>
There, I've ended my thread to that ranting anti-Cantorian chantor.
G. Frege
wrote:
>
> [...] I mean between any two numbers in R, there is at least (in fact many!)
> one element of set Q. Now I have problem that I cannot really list all
> the elements of R (since it is uncountable). But even then doesn't it
> look like R must've same cardinality as that of Q.
>
Completely agree with you (i.e. my naive intuition would say the
same).
But remember: "By mere spatial intuition or classical geometrical
attitudes without set-theoretical methods, the phenomenon can neither
be stated nor comprehended." (A. A. Fraenkel, A. Levy, Abstract Set
Theory)
With other words, (naive) intuition is misleading in this case.
(Actually that's a common feature of naive intuition in math.)
Now there are two considerations that might help (a little bit):
1. Between any two numbers in R, there are (only) _countable_
many elements of Q. But between any two numbers in Q there are
_uncountable_ many elements of R. (It's not just a simple 1:1
relationship!)
2. The rational numbers are (lie) dense on the "number line". Now
clearly every rational number is a real number. Hence there are at
least as much real numbers as rational numbers. Still there _are_
(infinitely many) gaps on the number line (if we only consider the
rational numbers). For example:
sqrt(p) where p is a prime number.
(The proof of this fact is rather straight forward).
Imho, the intuitive consideration you mentioned above wouldn't allow
for fact (2). (Imho, it does not count for the gaps.) Hence it's clear
that this intuitive consideration (i.e. our "intuition") is (somehow)
misleading.
Rupert
alok bakshi wrote:
> Ok, Here goes my argument:
>
> Let Q be the set of rational numbers and R be set of irrational
> numbers.
>
> For any r1, r2 \in R (r1 < r2) we have at least one q \in Q such that
> r1 < q < r2
>
> Now I know that elements of Q can be enumerated, and this enumeration
> contains all the numbers which exist between any two irrational
> numbers, NO MATTER HOW CLOSE THEY ARE.
No, just the rational ones.
> Now doesn't it imply that R is
> also countable.
>
No.
> I mean between any two numbers in R, there is at least(in fact many!)
> one element of set Q. Now I have problem that I cannot really list all
> the elements of R(since it is uncountable). But even then doesn't it
> look like R must've same cardinality as that of Q.
Not to me. I still don't understand why you think this follows.
Proginoskes
alok bakshi wrote:
> - A rational number exist between any two irrational numbers.
>
> - An irrational number exist between any two rational numbers.
>
> Given that both these assertions are true, I cannot see how there are
It's a non-sequitor.
Consider the set S = {1, 2, 3, 4, 5}. Then the following statements are
true:
- An even number in S exists between any two odd numbers in S.
- An odd number in S exists between any two even numbers in S.
What makes you think the number of even numbers in S should equal the
number of odd numbers in S?
--- Christopher Heckman
Ross A. Finlayson
There aren't infinitely many more of them.
Countably many nested intervals, you think, is not enough to leave no
unmapped reals. So, index by an ordinal equivalent to c the
cardinality of the continuum those nested intervals. Each contains a
distinct rational, or countably many was enough.
It's not anti-Cantorian, it's post-Cantorian.
Warm regards,
Ross F.
G. Frege
wrote:
>>
>> - A rational number exist between any two irrational numbers.
>>
>> - An irrational number exist between any two rational numbers.
>>
>> Given that both these assertions are true, I cannot see how there
>> are lot more irrational numbers than rational numbers. [By intuition,
>> shouldn't both be same (having same cardinality)?]
>>
> It's a non-sequitor.
>
> Consider the set S = {1, 2, 3, 4, 5}. Then the following statements are
> true:
>
> - An even number in S exists between any two odd numbers in S.
>
> - An odd number in S exists between any two even numbers in S.
>
> What makes you think the number of even numbers in S should equal the
> number of odd numbers in S?
>
Consider the set S = {1, 2, 3, 4, 5, ...}
Then the following statements are true:
- An even number in S exists between any two odd numbers in S.
- An odd number in S exists between any two even numbers in S.
And the number of even numbers in S equals the number of odd numbers
in S. The OP was talking about the cardinality of _infinite_ sets.
Ross A. Finlayson
Q: What are points and lines without geometry?
A: They're points and lines with geometry.
In set theories there are only sets, in regular set theories everything
is not a set.
Cardinality is that general property which by means of our general
capability of thinking can be attached to a set if the order of its
elements and their properties are not taken into account. (G. Cantor,
Collected works p. 282, W. Mueckenheim, trans.)
The order and properties of the numbers are key to their meaning.
There are only sets in set theory. Set theory is a useful tool for
many things. Transfinite cardinals don't have known applications.
Standard measure can be replaced with nonstandard measure. Sets of
different measure may have quite the same cardinality, where
cardinality is basically meaningless. Consider a distribution over
events each with probability zero: one of them happens, viz the
uniform. Integration doesn't work with a finite differential.
"... we can usually breathe a sigh of relief when a logarithm appears
inside O-notation, because O ignores multiplicative constants. There
is no difference between O(lg n), O(ln n), and O(log n), as n -> oo;
similarly, there is no difference between O(lg lg n), O(ln ln n), O(log
log n)."
-- Graham, Knuth, and Patashnik, _Concrete Mathematics_.
There is a difference between O(lg n) and O(lg lg n). Also: half of
the integers are even. As a set of numbers, the integers have all the
properties of themselves.
The consideration of rationals and irrationals, or algebraics (real
roots of integer coefficient polynomials) and transcendentals, and
other sets that are NCD2 in the reals from "On Well-Ordering(s) of Sets
Dense in the Reals, Infinity" leads to unintuitive or counterintuitive
notions of those numbers of the real numbers, of the real number lines,
where any point on the real number interpreted as a scalar has all the
same meaning as the scalar real number.
There are counterexamples in standard real analysis that there is a
least positive real.
The notions of alternations of rationals and irrationals on the real
number line, or algebraics and transcendentals, or rationals,
irrationals algebraics, and transcendentals, or rationals, irrationals
algebraics, and S, T, and U transcendentals, although I don't know the
definition of Mahler's S, T, and U transcendentals, the notions of
alternations of those items from their properties of each being dense
in the reals, everywhere discontinuous, disjoint, and having as union
the reals is both intuitive, in that pretty much everyone has thought
that, and counterintuitive, where there is no least positive real.
Funny things happen on the way to infinity, or on arrival, re the
transfer principle and etc.
Ross F.
Proginoskes
My point was that the two statements about "betweenness" don't have any
relationship with how large certain subset of S are.
--- Christopher Heckman
fernando revilla
Even intuition shows us that there are more irrationals
than rationals.
For example, choose the real number in decimal expression
x=N'x_1x_2...x_n...
It is very intuitive to see that it is more probable for
x_1,...,x_n,... not to be repeated in periods.
Fernando.
William Elliot
> Proginoskes wrote:
> > alok bakshi wrote:
> > > - A rational number exist between any two irrational numbers.
> > > - An irrational number exist between any two rational numbers.
> > >
> > > Given that both these assertions are true, I cannot see how there
> > > are lot more irrational numbers than rational numbers. [...]
> It's not anti-Cantorian, it's post-Cantorian.
>
Naw, anti or post-Cantorian is denial of Cantor's diagonal proof.
This is neo-Cantorian showing countable is uncoutable.
William Elliot
> alok bakshi wote:
>
> > - A rational number exist between any two irrational
> > numbers.
> >
> > - An irrational number exist between any two rational
> > numbers.
>
> Even intuition shows us that there are more irrationals
> than rationals.
>
You don't need intuition, just a trip to Congress
to know there are more irrationals than rationals.
David C. Ullrich
wrote:
>Ok, Here goes my argument:
>
>Let Q be the set of rational numbers and R be set of irrational
>numbers.
>
>For any r1, r2 \in R (r1 < r2) we have at least one q \in Q such that
>r1 < q < r2
>
>Now I know that elements of Q can be enumerated, and this enumeration
>contains all the numbers which exist between any two irrational
>numbers, NO MATTER HOW CLOSE THEY ARE. Now doesn't it imply that R is
>also countable.
No.
It simply doesn't imply that. It's hard to say _why_ it doesn't
imply that, because we don't see why you think it _does_ imply
that! If you gave more explanation as to why you think this
does imply that someone could say where your error is, but
as long as you're just saying "doesn't this imply that" it
seems like the only answer people can give is just "no".
>I mean between any two numbers in R, there is at least(in fact many!)
>one element of set Q. Now I have problem that I cannot really list all
>the elements of R(since it is uncountable). But even then doesn't it
>look like R must've same cardinality as that of Q.
No. To show this you need to match up every element of R with
exactly one element of Q.
************************
David C. Ullrich
christian.bau
> - A rational number exist between any two irrational numbers.
>
> - An irrational number exist between any two rational numbers.
>
> shouldn't both be same(having same cardinality) because of these two
> above mentioned facts...
For every integer number, there is an irrational number that is larger.
And for every irrational number, there is an integer number that is
larger.
Given that both these assertions are true, I cannot see how there are a
lot more irrational numbers than integer numbers... Does this sound
convincing to you?
Here is where you went wrong: The range "between two rational numbers"
or "between two irrational numbers" is huge. There is not just one
irrational or rational number in that range, there are not just many,
there are infinitely many. And here is the difference: Between any two
rational numbers (or between any two irrational numbers) there is an
_uncountable infinite number of irrational numbers_ but only a
_countable finite number of rational numbers_.
(Ok, you don't have to believe this. You'd have to prove it. But it
shows where your intuition went wrong. The proof goes almost exactly
like the proof that the irrational numbers are not countable).
Jesse F. Hughes
Oh, you mean you've ended *your* contributions. At least your
responses to him. Man, that is pure, raw power.
And yet the thread goes on, even more pointlessly than ever.
--
Jesse F. Hughes
"To be honest, I don't have enough interest in math to spend the time
it would take to clean up the mess that I believe has been created in
the past 100 or so years." -- Curt Welch lets the world down.
Russell
> Ok, Here goes my argument:
>
> Let Q be the set of rational numbers and R be set of irrational
> numbers.
>
> For any r1, r2 \in R (r1 < r2) we have at least one q \in Q such that
> r1 < q < r2
In fact, infinitely many of them. (Countably infinite.) This
can be proved.
>
> Now I know that elements of Q can be enumerated, and this enumeration
> contains all the numbers which exist between any two irrational
^^^^^^^^^^^^^^^
> numbers, NO MATTER HOW CLOSE THEY ARE.
Not all the "numbers". All the *rational* numbers.
Did you think there are no irrational numbers between r1
and r2?
There are in fact uncountably many between r1 and r2.
This, too, can be proved.
Putting your final clause in caps does *not* change that
fact. It doesn't matter how big of a microscope you get
to look at the number line between r1 and r2, there are
still uncountably many irrational numbers between r1 and
r2, no matter how close they are.
Now doesn't it imply that R is
> also countable.
It would, if the number of irrationals between r1 and r2
were zero. Or even finite, or countably infinite. I think
your mental picture is assuming something like that, but
it's a fallacy as it amounts to assuming what you are
trying to prove.
>
> I mean between any two numbers in R, there is at least(in fact many!)
> one element of set Q. Now I have problem that I cannot really list all
> the elements of R(since it is uncountable). But even then doesn't it
> look like R must've same cardinality as that of Q.
Well, it surprised Cantor when he found that they do not.
It's a surprising result.
christian.bau
Just an observation: We have a proof that real numbers are uncountable
but rational numbers aren't, and you have an argument where intuitively
there shouldn't be more real than rational number. Now I'll give you a
similar intuitive argument: Between any two integers, there is an
infinite number of rational numbers. But between any two rational
numbers, there are very often no integers at all, and in any case at
most a finite number of integers. So intuition seems to indicate that
there are more rational numbers than integers.
That makes it two intuitive, but incorrect arguments. Interestingly,
both arguments together indicate that the reals are uncountable, but
for completely wrong reasons!
Nick
"William Elliot" <ma...@hevanet.remove.com> wrote in message
news:Pine.BSI.4.58.06...@vista.hevanet.com...
You'll get to learn that once you have posted on a newsgroup it becomes the
property of the newsgroup rather than the possession of the individual who
originally posted.
Nick
Nick
"Proginoskes" <CCHe...@gmail.com> wrote in message
news:1166859184.3...@f1g2000cwa.googlegroups.com...
>
> alok bakshi wrote:
>> - A rational number exist between any two irrational numbers.
>>
>> - An irrational number exist between any two rational numbers.
>>
>> Given that both these assertions are true, I cannot see how there are
>> lot more irrational numbers than rational numbers. [...]
>
> It's a non-sequitor.
Or a non-sequitur! And if that is a non-sequitur so be it.
Nick
Zdislav V. Kovarik
On Fri, 22 Dec 2006, alok bakshi wrote:
> I agree, there are lot of rational numbers between two irrational
> numbers, no matter how close they are. But then it looks paradoxical
> that irrational numbers are more than rational numbers. Rational
> numbers are everywhere, between any two irrational numbers, then
> doesn't it imply that there are more rational than irrationals. I
> understand the mathematical proof, but I cannot convince myself if I
> think along these lines...
>
> Thanks,
>
> Alok Bakshi
Here is something that appeals both to intuition and to rigor:
Find out about (regular) continued fractions.
Every rational number has a finite regular continued fraction (actually,
at most two of them), and every f.r.c.f. represents a rational number.
Illustration:
77/17 = 4 + 1 / ( 1 + 1 / ( 1 + 1 / 8))
In contrast, every irrational number has an infinite f.r.c.f., and every
infinite f.r.c.f. represents an irrational number.
Intuitively, you can see that you have many more chances to produce
irrational numbers than rational numbers.
And rigorously, this can be supported by a "Cantor diagonal process."
Cheers, ZVK(Slavek).
William Elliot
> William Elliot <ma...@hevanet.remove.com> writes:
> > On Fri, 22 Dec 2006, Jesse F. Hughes wrote:
> >> William Elliot <ma...@hevanet.remove.com> writes:
> >>
> >> > You agree to what? As you don't quote sufficient contexts, I choose
> >> > to end this thread. You'll get to agree to that also.
> >>
> >> You choose to end this thread? How do you do that?
> >>
> >
> Oh, you mean you've ended *your* contributions. At least your
> responses to him. Man, that is pure, raw power.
>
Were I polite about it, I would have would curtailed rudeness. However by
his second or third post, I would have evaporated because anti-, post- and
neo-Cantorian entertainments, like stale beer, are off of my diet.
> And yet the thread goes on, even more pointlessly than ever.
>
Usually I prefer to goof off elsewhere than at sci.math.
sci.space.policy has real hard core dups that are more fun.
Saurav
alok bakshi wrote:
> - A rational number exist between any two irrational numbers.
>
> - An irrational number exist between any two rational numbers.
>
> Given that both these assertions are true, I cannot see how there are
> shouldn't both be same(having same cardinality) because of these two
> above mentioned facts...
>
>
> Alok Bakshi
I'd first tell you that, to infer something about the complete ordered
field, you have got to use the whole theory of complete ordered field,
not just a few consequences of it.
For example, the theory of complete ordered fields includes the axioms
for Dense ordered fields; and if you take only them and reject others,
and claim that the uncountability of IR fails to be in tune with our
intuition, it would not look like a mathematician's statement!
David Marcus
> On 22 Dec 2006 19:07:38 -0800, "alok bakshi" <alokb...@gmail.com>
> wrote:
>
> > [...] I mean between any two numbers in R, there is at least (in fact many!)
> > one element of set Q. Now I have problem that I cannot really list all
> > the elements of R (since it is uncountable). But even then doesn't it
> > look like R must've same cardinality as that of Q.
> >
> Completely agree with you (i.e. my naive intuition would say the
> same).
>
> But remember: "By mere spatial intuition or classical geometrical
> attitudes without set-theoretical methods, the phenomenon can neither
> be stated nor comprehended." (A. A. Fraenkel, A. Levy, Abstract Set
> Theory)
>
> With other words, (naive) intuition is misleading in this case.
> (Actually that's a common feature of naive intuition in math.)
>
> Now there are two considerations that might help (a little bit):
>
> 1. Between any two numbers in R, there are (only) _countable_
> many elements of Q. But between any two numbers in Q there are
> _uncountable_ many elements of R. (It's not just a simple 1:1
> relationship!)
I think that is a good way to look at it. It stops the false symmetry
right from the start.
--
David Marcus
Reef Fish
alok bakshi wrote:
> - A rational number exist between any two irrational numbers.
>
> - An irrational number exist between any two rational numbers.
>
> Given that both these assertions are true, I cannot see how there are
> lot more irrational numbers than rational numbers.
Alok, your question sounded almost like a troll, but you proved to be
serious and there were many interesting and correct explanations of
your false "intuition" fallacy.
But the simplest explation of your fallacious reasoning is to change
your assertions slightly without changing the correctness of the
statements:
> - Only a COUNTABLE number of rational number exist between
any two irrational numbers.
>
> - An UNCOUNTABLE number of irrational number exist between
any two rational numbers.
That would put the cardinality of the sets to work.
An alternative reasoning would take the "measure theory" result
for granted (or work it out yourself) that the set of rationals on
the real line is a set of measure ZERO. All the non-zero sets
contains and consists of irrational numbers. :-)
Does that sound rational enough?
-- Reef Fish Bob.
P.S. Fell in this group accidentally.
G. Frege
<David...@alumdotmit.edu> wrote:
>>
>> [...] there [is a] consideration that might help (a little bit):
>>
>> Between any two numbers in R, there are (only) _countable_
>> many elements of Q. But between any two numbers in Q there are
>> _uncountable_ many elements of R. (It's not just a simple 1:1
>> relationship!)
>>
> I think that is a good way to look at it. It stops the false symmetry
> right from the start.
>
It seems that Cristian Bau came up (independently?) with the same
idea:
"Here is where you went wrong: The range 'between two rational
numbers' or 'between two irrational numbers' is huge. There is not
just one irrational or rational number in that range, there are not
just many, there are infinitely many. And here is the difference:
Between any two _rational numbers (or between any two irrational
numbers) there is an _uncountable infinite number of irrational
numbers_ but only a _countable finite number of rational numbers_.
(Ok, you don't have to believe this. You'd have to prove it. But it
shows where your intuition went wrong. The proof goes almost exactly
like the proof that the irrational numbers are not countable)."
:-)
Proginoskes
This is sci.math, not perfect.spelling.
After all, you understood what I meant, right?
--- Christopher Heckman
William Elliot
> alok bakshi wrote:
> > - A rational number exist between any two irrational numbers.
> > - An irrational number exist between any two rational numbers.
>
> > - Only a COUNTABLE number of rational number exist between
> any two irrational numbers.
> >
> > - An UNCOUNTABLE number of irrational number exist between
> any two rational numbers.
>
Divide the reals into the surreals and the surrationals so that between
any two surreals are uncountably many surrationals and between any two
surrationals are uncountably many surreals.
> Does that sound rational enough?
>
OP expects real rational thought?
> P.S. Fell in this group accidentally.
>
How do you know you had an accident?
Is the probability of your presence equal to or less than the probability
of an accident? No. You are here by an application of the Axiom of
Choice. Clearly, consequently you're a prochoice mathematician.
BTW, how much do the surreals and the surrationals of the unit
interval measure?
Ross A. Finlayson
That's interesting, those continued fractions. If it takes infinitely
many coin tosses or dice rolls or picks from the hat to produce an
irrational number, you can see why there are more chances to produce a
rational number.
Consider each irrational p_i on [0,1]. In [0, p_i] there's a rational
q_i that is not in [0,p_h] for each p_h < p_i. So, for each irrational
there's a rational, less than it and greater than each irrational less
than it. Then, there's a rational q_i such that q_i is not in [0,p_h]
for any irrational p_h < p_i.
Similarly for each p_j > p_i there exists a rational q_j in [0, p_j]
not in [0, p_i].
Is that disagreeable? Basically the second sentence in the paragraph
above is the point of contention. There is a rational q_i such that
q_i is not in [0,p_h] for each irrational p_h < p_i. For each there
is, for any there is, that there is one for all and for every is the
point of contention.
While the discussion has leaned towards that the focus should be that
there are infinitely many rationals between each pair of irrationals
and infinitely many irrationals between each pair of rationals, there
is a rational between each pair of irrationals.
Zdislav, there is no universe in ZF.
Ross
Ross A. Finlayson
>
> Consider each irrational p_i on [0,1]. In [0, p_i] there's a rational
> q_i that is not in [0,p_h] for each p_h < p_i. So, for each irrational
> there's a rational, less than it and greater than each irrational less
> than it. Then, there's a rational q_i such that q_i is not in [0,p_h]
> for any irrational p_h < p_i.
>
> Similarly for each p_j > p_i there exists a rational q_j in [0, p_j]
> not in [0, p_i].
>
> Is that disagreeable? Basically the second sentence in the paragraph
> above is the point of contention. There is a rational q_i such that
> q_i is not in [0,p_h] for each irrational p_h < p_i. For each there
> is, for any there is, that there is one for all and for every is the
> point of contention.
>
>
>
> While the discussion has leaned towards that the focus should be that
> there are infinitely many rationals between each pair of irrationals
> and infinitely many irrationals between each pair of rationals, there
> is a rational between each pair of irrationals.
>
So, are there rationals between any two irrationals?
Yeah, there are.
Ross
Nick
Well, surely you would like to learn - if you are going to use rather posh
expressions I suggest that you might learn how to spell them properly -
otherwise you look like a pretentious ...
Nick
rich burge
alok bakshi wrote:
> - A rational number exist between any two irrational numbers.
>
> - An irrational number exist between any two rational numbers.
>
> Given that both these assertions are true, I cannot see how there are
> shouldn't both be same(having same cardinality) because of these two
> above mentioned facts...
>
this intuition. Let R={1,2,3,4,5}, E={2,4} and O={1,3,5}. Now,
between any two (distinct) even numbers in R there is an odd number in
R. And between any two distinct odd numbers in R there is an even
number in R. But R clearly has more odd numbers than even numbers.
Rich
G. Frege
>
> alok bakshi wrote:
>
>> - A rational number exist between any two irrational numbers.
>> - An irrational number exist between any two rational numbers.
>>
>> Given that both these assertions are true, I cannot see how there are
>> _lot more_ irrational numbers than rational numbers. By intuition,
>> shouldn't both be same (having same cardinality) because of these two
>> above mentioned facts...
>>
> At first, it may seem that they should. But there is an easy test of
> this intuition. Let R = {1,2,3,4,5}, E = {2,4} and O = {1,3,5}. Now,
> between any two (distinct) even numbers in R there is an odd number in
> R. And between any two distinct odd numbers in R there is an even
> number in R. But R clearly has more odd numbers than even numbers.
>
Note that the OP was (is) talking about _infinite_ sets. Now consider:
R = {1,2,3,4,5,...}, E = {2,4,...} and O = {1,3,5,...}. Your "counter
example" doesn't work in this case. Actually, |E| = |0| in this case.
Ross A. Finlayson
In considering the positive rationals less than some irrational p_i,
for each p_j < p_i the set of rationals Q_j: q E Q+, q < j is always a
proper subset of the set of rationals Q_i, those rationals less than
p_i. So, Q_i setminus Q_j is always non-empty, for each p_j < p_i in
P+, the set of positive irrationals. The set difference is always
non-empty, there is for each p_j < p_i at least one distinct rational
q_i such that p_j < q_i < p_i. Simlarly, yet in a contradicting
manner, for each q_i there is an irrational p_h such that p_j < q_i <
p_h < p_i, so q_i is not a distinct rational over each p_h < p_i, and
there exists q_h, and around it goes.
I want to use that kind of notion to see why Q_i setminus the union of
Q_j for p_j < p_i is or is not empty.
If you would claim it empty, exchange rationals with irrationals. Then
there wouldn't be a distinct irrational for each rational, by that
argument. Where there is exchange them back and it's not.
For the difference of Q_i and union of the Q_j's for p_j < p_i to be
empty, as the union is a subset of Q_i they would be equal. Where the
union contains all the elements, how did that last one get in there,
and where it is, how is that p_j less than p_i?
The union can't contain an element not in at least one of those sets
which comprise it by their elements, unless you're looking at something
like a Russell set.
So, what is that called?
Ross
Ross A. Finlayson
Is there a rational between any two irrationals?
Yeah, there is.
Ross
Ross A. Finlayson
So, for no positive irrational p_h < p_i is it so that the set of
positive rationals less than p_h, Q_h, is equal to the set of rationals
less than p_i, Q_i. So, Q_h < Q_i, the (positive) rationals less than
p_h is a proper subset of Q_i, so Q_i setminus Q_h is non-empty, and
for each irrational p_i there's an element of the difference of Q_i and
the union of the Q_h's for p_h < p_i such that q_i is related to p_i
and not to any other p_j =/= p_i.
That Q_i, setminus the union of the Q_h's, is non-empty, is true. Do
you agree? If not there's a p_h >= p_i, yet each p_h < p_i, or a q_h
in a Q_h greater than p_h, yet
each q_h is less than p_h.
So, here's the question: is the intersection of (Q_i setminus Q_h) for
each p_h < p_i empty? Again, is the difference Q_i setminus the (union
of the Q_h)'s for each p_h < p_i empty?
There's a rational between any two irrationals.
Ross
Ross A. Finlayson
To some extent I regret to inform you: there is only the null axiom
theory. After Existence: all other scientific theories are moot,
null, and void.
That's not my fault. There is no theory of everything that is not the
null axiom theory.
So: accept.
There is only one theory the null axiom theory, it's the theory of
everything.
Besides that, my statement about the rationals is indisputable.
Ross
Ross A. Finlayson
So, the union of the Q_h's for p_h < p_i is equal to Q_i, else Q_i \
Union Q_h is non-empty and there's a distinct rational element of that
difference for each irrational p_i, and thus an injection from the
irrationals into the rationals and thus a surjection from the rationals
onto the irrationals. (I think infinite sets are equivalent so that's
not a contradiction, for Union Q_h < Q_i.)
Where the union of the Q_h's contains an (at least one) element not in
any of the Q_h's, for there to not be a distinct rational for each
irrational, that's a contradiction., the set is defined by its
elements, and p_h < p_i, not p_h = p_i.
Yet, for any q_i, there exists a p_j such that q_i < p_j < p_i, then
there wouldn't be a rational between any two irrationals,
contradiction.
The difference Q_i \ Q_h is non-empty for each p_h < p_i, and that
difference Delta(Q_i,Q_h) is a superset of Delta(Q_i,Q_j) for p_h < p_j
< p_i. So the intersection is only empty when p_h = p_i, but p_h <
p_i, the intersection is non-empty else contradiction.
Transpose p and q, (positive) P and Q, irrationals and rationals, it's
the same system.
It seems there are contradictory true statements, so one of them is
false.
Again, it seems like the Russell paradox. Where it is true that the
rationals and irrationals are each dense in the reals, it can not be
that of the set of irrationals [0,p_i) that one of them equals p_i,
because they are defined to be not. It's reminiscent of the ordinals
where the set of ordinals up to but not including n has order type n,
here the union of the Q_h's for p_h < p_i equals Q_i, else infinite
sets are equivalent, yet then p_h = p_i, a contradiction.
So, where's the contradiction?
If the union of the Q_h's for p_h < pi is equal to Q_i, else there
exists a distinct rational for each irrational, then the union of the
P_h's for q_h < q_i is equal to P_i, and there doesn't exist a distinct
irrational for each rational. Then exchange rationals and irrationals
and it's neither.
Ross
Ross A. Finlayson
i) rationals and irrationals are reals
ii) the reals are a set
iii) the reals are trichotomous
iv) the rationals and irrationals are each dense in the reals
a) between any two irrationals there is a rational
b) between any two rationals there is an irrational
Considering only the positive reals, consider a rational q_i (let q_i
be a positive rational). Is there a distinct irrational p_i for each
q_i? The answer would necessarily be yes for any who think the
cardinality of P is greater than, or equal to, the cardinality of Q.
So, let P_x be the set of irrationals less than q_x. Consider the
union of the P_h's for q_h < q_i. If the difference of P_i and that
union is empty, then there's a rational for which there is not a
distinct irrational. So, is the union of the P_h's for q_h < q_i not
equal to P_i? If so, then the union of the Q_h's for p_h < p_i is not
equal to Q_i, and there's a distinct rational for each irrational. If
not, there's not a distinct irrational for each rational, for every
q_i.
"Rational" and "irrational" could be replaced here with "algebraic" and
"transcendental", or even in a three-state case "rational", "algebraic
irrational", and "transcendental", or any other partition of the reals
into subsets dense in the reals.
Is there not a simple answer to this conundrum?
Infinite sets are equivalent.
Ross
Ross A. Finlayson
Consider again the set of irrationals P_i, those less than some
rational q_i, all positive. Is the P_i minus the union of the P_h's
for q_h < q_i empty?
If that difference is empty, for each rational q_i, does that not seem
that there is not a distinct irrational for each rational, and thus no
surjection from the irrationals over the rationals?
The irrationals and rationals are each dense in the reals.
Q_i setminus Q_h is not empty for each p_h < p_i, and Q_i setminus Q_h
is a subset of Q_i setminus Q_g for p_g < p_h, i.e., none are empty and
none are pairwise disjoint. Is the intersection of Q_i setminus Q_h
for each p_h < p_i empty? It seems that one way to argue that it is
empty is to show that for each q_h < p_i, it is not in that difference.
Then it would be so for the reversed case as well about the
irrationals.
As well, for each p_h there exists a q_i in Q_i and not in each Q_g for
p_g <= p_h, except again, for any q_i, it's not in the difference.
For each p_h < p_i, Q_i - Q_h is not the empty set. If there were a
set Q_h such that Q_i-Q_h would be empty, then p_h would equal p_i,
which would contradict that p_h < p_i.
So, it seems the difference is empty, yet cannot be empty, at the same
time. It (the density of rationals and irrationals in the reals)
defies both set membership and set non-membership, of some element.
Surjections of the irrationals over the rationals are obvious, so if
the difference is empty shows there is not a distinct irrationa for
each rational, then the difference is non-empty.
It's reminiscent of Brentano boundaries and notions of partitioning the
reals that are not standard.
So, does this describe a deep flaw in the standard construction of the
real number system?
Ross
Ross A. Finlayson
So, is it empty, or non-empty, the difference?
If it's empty, then, there seems to not be a distinct irrational for
each rational, which there obviously is. If it's non-empty, then there
does seem to be a distinct rational for each irrational. It's for both
or neither.
If the difference is empty then for each irrational there is not a
distinct rational. Agree or disagree, why?
So, which is it?
Ross A. Finlayson
If it's empty there's a p_j < p_i, such that p_j = p_i. If it's not
empty then there exists a rational q_j that is basically adjacent to
irrational p_j.
Either way a contradiction is derived.
Also, ZF is inconsistent, for it being the Russell set and denying
that.
Ross
Ross A. Finlayson
Well that is rather disconcerting. Is the difference empty or not, or
both and/or neither?
Obviously if the real numbers have these seemingly contradictory
properties, then there's something about them that defies the standard.
There's more to them, these real numbers, than the standard can
handle.
Is the set of irrational numbers less than an irrational number the
same set as the set of irrational numbers less than or equal to that
irrational number? No, it's not.
Then, with density of rationals in the reals, there is a distinct
rational for each irrational, there's a counterexample to
uncountability of the reals, similarly as to how the existence of the
universe is a counterexample to the powerset result.
Some people consider fractals in space-time, as far as I know nobody
uses transfinite cardinals in physics, and there's a nonstandard
countable measure theory, a uniform distribution over the naturals
where there is one over the reals, as I have described, and N/U EF is
its CDF.
Is the set of irrational numbers less than an irrational number that
irrational number? Perhaps it can be seen that way, similarly as to
how an ordinal may be seen as the set of ordinals less than it.
(Powerset is order type is successor in ubiquitous naturals.)
The universe of ZF is the Russell set so it's irregular, and thus ZF is
inconsistent.
The universe is infinite: infinite sets are equivalent.
Ross
Ross A. Finlayson
So, is the set difference on the open boundary empty? If it is, then,
it is. If it's not, then I wrote a proof that the rationals surject
onto the irrationals, in any interval of which the characteristic has
it that that is so.
What's your opinion?
My opinion is that these statements are cold mathematical fact.
It's basically about the difference, at the boundary. The infinite
sets, they aren't always Archimedean, where leading to infinity never
returns. Where they're non-Archimedean they might go back to zero.
You see, the boundary is defined in the real numbers, so if it's an
irrational, you take the set of all rationals less than that, and the
set of all irrationals less than that. You'd expect because the
irrational real numbers, in a post-Cantorian framework where that is
reasonable, and rational real numbers, which are disjoint and share no
elements, that all the rationals less the number less all the
irrationals less the number would not be the empty set. If it is the
empty set, then the dense elements as well alternate in the micro, the
infinimicro. Otherwise they _don't_ alternate, and then irrationals
are adjacent and thus not dense.
Ross
Ross A. Finlayson
How about that?
Ross