| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
The set of rational numbers is countable. You know; go through the
table diagonally.
The set of irrationals is not. Just about the time you think you've
got all of the irrationals listed another can be constructed which
isn't on your list.
I can show that the set of rational numbers between ANY two rationals
is infinite. Call the two numbers P and Q. Since the following set
(call it R) is infinite and the values are all rational and between P
and Q;
R(0) = (P+Q)/2
R(n) = (P+R(n-1))/2 QED.
The set R is a subset of the set of rational numbers and so countable.
What can be shown about the following sets?
irrationals between ANY two irrationals
rationals between ANY two irrationals
irrationals between ANY two rationals
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 742.1 | KIRK::KOLKER | Conan the Librarian | Tue Jul 28 1987 17:44 | 13 | |
re .0
Let's take a crack at irrationals between two irrationals. Let
a1 and a2 be irrationals. Consider (a1 + a2)/2. This is either
rational or irrational. If it is a rational (call it m1) then
m1 + a1 is irrational for sure there fore (m1 + a1)/2 is irrational.
The worst case you can get is an alternation of rational and irrational
mid points between a1 and a2 . Take the subsequence of irrational
midpoints and you get a countably infinite set of irrationals between
a1 and a2. There are more, but we have shown there is
an infinite number of them.
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| 742.2 | What if you have a non-provable case? | AKQJ10::YARBROUGH | Why is computing so labor intensive? | Wed Jul 29 1987 08:56 | 23 |
Another way of dealing with these sets is to consider their decimal-digit representations (or any other radix, for that matter). The worst case is for two irrationals: are all such pairs separated by a rational? One might argue that, since there are more irrationals than rationals, that pairing them off must leave some irrationals without partners and thus 'adjacent' pairs of irrationals with no rational number between them. However: Given two irrational numbers 0 < a < b < 1, consider their radix-R expansions .a[0]a[1]a[2]... and .b[0]b[1]b[2]...; there is some minimum value of k for which a[k] <> b[k]. Then the number .b[0]b[1]...b[k]0000000... lies between them and is rational, namely b[0]b[1]...b[k]/R**k. However, this does raise another philosophical issue. Since all irrational numbers have non-terminating n-ary expansions, one can ask the folowing questions: If there *does not exist* any finite k for which a[k] <> b[k], is it necessarily the case that a=b? If *one cannot determine* that a[k] <> b[k] for some finite k, is it necessarily the case that a=b? Lynn Yarbrough | |||||
| 742.3 | KIRK::KOLKER | Conan the Librarian | Wed Jul 29 1987 10:41 | 5 | |
re .2
L.E.J.Brauer and Heyting would have said no to both. (Intuitionist
view)
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| 742.4 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Fri Jul 31 1987 10:28 | 12 | |
Let x and y be two non-equal real numbers. Let n = 10000
[[1/abs(x-y)+1]], where [[z]] is the greatest integer not greater than
z.
([[nx]]+1) / n and ([[nx]]+sqrt(2) / n are a rational number and an
irrational number, both between x and y. This procedure can obviously
be repeated with the new numbers, so between any two numbers, whether
rational or irrational, there are an infinite number of both rational
and irrational numbers.
-- edp
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