Re: On Counting Irrational Numbers
- From: Dave Seaman <dseaman@xxxxxxxxxxxx>
- Date: Fri, 27 Jul 2007 12:12:00 +0000 (UTC)
On Thu, 26 Jul 2007 23:36:01 -0700, mthggl@xxxxxxxx wrote:
I have studied that irrational numbers are uncountable. I also have
an idea which shows that there cannot be two successive irrational
numbers, such that irrational numbers alternate with rational numbers
in occurance.
It's not an "alternation", because there is never a "next" number, either
rational or irrational. Between any two irrational numbers, there is a
countable infinity of rationals. But between any two rationals, there
are uncountably many irrationals.
If this is the case, it means that there is, at most,
an equivalence between the number of irrational and rational numbers.
An equivalence would be a bijection f: Q -> R\Q. What mapping f do you
have in mind? (Hint: there isn't one.)
Why then are irrational numbers uncountable? This idea has been
troubling me for some time and I would ask if someone could clarify
the error. The full argument is located at:
http://geocities.com/mthggl/On_Counting_Irrational_Numbers.pdf
I am an undergraduate student and would truly appreciate any and all
help. Thank you for your time.
Sincerely,
Ryan R. Zito
The first error is in your very first sentence:
The irrational number set, R\Q, has been defined as uncountable
in contrast to the countable rational number set, Q.
You seem to think that someone just woke up one morning and decided,
"Today, I think I'll define the irrationals to be uncountable." False.
It's actually a very easy proof. Do you agree to the following?
(1) Q is countable.
(2) R is uncountable.
(3) The union of two countable sets is countable.
From these facts it easily follows that the irrationals, R\Q, must beuncountable.
You argue that uncountability has something to do with "denseness".
Incorrect. For example, Q is dense in R but is countable. On the other
hand, P(N), the power set of the naturals, is uncountable but is not in
any sense dense.
You claim that the irrationals cannot be uncountable unless there are
"consecutive irrationals" somewhere in the line. You give no reason for
thinking this, and it is clearly wrong. After all, the reals are
uncountable, but there are no "consecutive reals" anywhere in the line.
--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
.
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- On Counting Irrational Numbers
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