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kevinatilusa
Daedalian Member

Leptonn
Guest

 Posted: Sat Jul 03, 2004 6:13 pm    Post subject: 2 That was very good, Kevin. (Thank you) infinity
pikachamp
swore in chat!

 Posted: Sat Jul 03, 2004 11:02 pm    Post subject: 3 *proudly trumpets that he actually read all the way through this overly long post* What are complex numbers?
mith
Pitbull of Truth

 Posted: Sun Jul 04, 2004 12:28 am    Post subject: 4 Yum, infinity. pika, I'll answer if no one has by the time I get back. I have to go now though.
Daedalian Member

 Posted: Sun Jul 04, 2004 12:29 am    Post subject: 5 From very far away. Infinitely far, really, otherwise, they're just seeing part of it.
Lucky Wizard
Daedalian Member

 Posted: Sun Jul 04, 2004 12:46 am    Post subject: 6 Complex numbers are the sums of real numbers and imaginary numbers. (An imaginary number is a number that is the square root of a negative number; it can also be viewed as the sine of a number greater than 1.) Imaginary numbers take the form ai, where a is the square root of the absolute value of the negative number that has this imaginary number as its square root, so an imaginary number line can be imagined. The complex numbers take the form a+bi, and one can visualize a complex number plane where each complex number is at (a,b). Back to the subject, John Allen Paulos' book Beyond Numeracy gives a great, fairly elementary introduction to infinity (in the chapter titled "Infinite Sets"). It includes a proof that the set of all reals is bigger than the set of all positive integers.
Guest

 Posted: Sun Jul 04, 2004 3:27 am    Post subject: 7 And you were alluding to...what now? I'm infinitly bored. I scrolled over it earlier but I think I'll go read Antrax's sig now. Perhaps I'll find some wisdom therein.
lostdummy
Daedalian Member

 Posted: Wed Jul 07, 2004 2:45 pm    Post subject: 8 interesting post kevin :) That is interesting definition for "size" of infinite sets, but I must say that it seems too ..er... broad definition of size for my expectation, at least from two reasons: 1) First, it seems that most if not all logic problems/paradoxes related to infinity all use infinite sets that would go in your A0 classification. So it appears we would need more 'refined' definition of "size" if we want to use it to solve such problems. 2) Second, I have some intuitive problem to accept that set of all integers is same size as set of even integers, or even more that set of all rational numbers is same size as set of integers. Or, to relate to #1, it seems to me that it is possible to differenciate in size among those infinite sets that belong to A0 Examples for my second reason: a) A=set of all integers, B=set of even integers it seems intuitive for me that A>B and not B>A. It further seems intuitive to me that there is double number of elements in set A, compared to set B. Or, to use example that could arise in some puzzle, it seems to me that there is about 50% chance to pick even number out of integer numbers, regardless of size of integer numbers set In this case I would say (intuitively) that size of A is some constant (2) times size of set B. IF B = set of all integers divisible by 5, then Size(A)=5*Size(B) So, if O(set) is approximation of set size, and const O(A) = 2 * O(B) ---> A>B b) A=set of all integers, B=set of rational numbers Here, it seems intuitive to me that A AB since there are some numbers that exist in A while not existing in B (for example 8), while there is no number that exist in B which also does not exist in A. but difference to #b is in fact that number of integers containing 3 is rising, so that in reality size of B approaches size of A, so O(B) ~ O(A) Or, in our shady probablility area, it seems to me that if we randomly pick one integer number, chance for it to contain 3 approach 100% --------------------------- I used word "intuitive" a lot above, so this reasoning may not be valid, but it seems acceptable for me that if we are able to find ratio of "sizes" for two sets A and B for limited number of elements N, and if we can get meaningful result even if N approach infinity, then such result shows relative size of sets. So, if p= SizeA(n)/SizeB(n), and p is meaningful number when n-> inf, I would consider p to represent how much set A is bigger/smaller than set B. For previous examples, a) sizeA / sizeB = 2 (for any n) b) sizeA / sizeB = 0 (when n-> inf) c) sizeA / sizeB = 1 (when n-> inf) maybe some of this reasonings are wrong, but i do believe that there is difference in size of sets that would all go into A0 classification from original post.
eruonna
Icarian Member

 Posted: Sat Jul 10, 2004 5:58 am    Post subject: 9 lostdummy: The main flaw I can see with your intuitive argument that some of the A0 sets have different sizes is that it uses limits and the limits depend on the path used to approach them. If I understand your argument correctly, you are saying that if we let P(n) be the probability that a number picked from the first n natural numbers is even then P(n) --> 1/2 as n --> infinity, and that we should conclude from this that there are somehow twice as many natural numbers as even numbers. What this fails to consider is the possibility of taking the numbers in a different order. For example, we could arrange the natural numbers in the order 0, 1, 3, 2, 5, 7, 4, 9, 11, .... This sequence eventually includes any number you want, so it is a valid reordering of the natural numbers, but if we now let P(n) be the probability that a number selected from among the first n of this sequence is even, then P(n) --> 1/3 as n --> infinity, so we could just as well conclude that there are three times as many natural numbers as even numbers. Similarly we can find orderings so that P(n) goes to any value from 0 to 1, so this way of defining a size for the sets is not consistent. There is a valid way, as you suggest, to say that the set of even numbers contains only elements of the set of natural numbers but the set of natural numbers contains elements that are not even, and that way is to say that the even numbers are a proper subset of the natural numbers. One of the counterintuitive properties of an infinite set is that it can have proper subsets the same size as the set itself. (Indeed, this is one definition of an infinite set.) Anyway, I hope this explanation makes some kind of sense to people who aren't already familiar with this kind of thing. -- Eruonna
CrystyB
Misunderstood Guy

 Posted: Sat Jan 01, 2005 9:38 am    Post subject: 10 This reminds me of the How Many Threes? puzzle... Too bad the GLP thread is no more Last edited by CrystyB on Tue Jan 04, 2005 7:05 pm; edited 1 time in total
Doc Borodog
Guest

Posted: Sat Jan 01, 2005 7:26 pm    Post subject: 11

 Quote: Complex numbers are the sums of real numbers and imaginary numbers. (An imaginary number is a number that is the square root of a negative number; it can also be viewed as the sine of a number greater than 1.)

CrystyB
Misunderstood Guy

 Posted: Sat Jan 01, 2005 11:51 pm    Post subject: 12 He probably meant arcsine. Though i could be mistaken, my intuition tells me it's gotta be about the e^ix=cosx+isinx...
Lucky Wizard
Daedalian Member

 Posted: Sun Jan 02, 2005 1:31 am    Post subject: 13 What CrystyB said. Let sin x=a. cos x = sqrt(1-sin 2 x) = sqrt(1-a 2 ) = i*sqrt(a 2 -1) e ix = cos x + i sin x e ix = i*(a+sqrt(a 2 -1)) ix = ln(i*(a+sqrt(a 2 -1))) ix = ln i + ln(a+sqrt(a 2 -1)) ix = (ln (-1))/2 + ln(a+sqrt(a 2 -1)) ix = pi*i/2 + ln(a+sqrt(a 2 -1)) x = pi/2 + ln(a+sqrt(a 2 -1))/i x = pi/2 - i*ln(a+sqrt(a 2 -1))
CrystyB
Misunderstood Guy

 Posted: Tue Jan 04, 2005 8:47 pm    Post subject: 14 Addressing lostdummy's post, i'd like to point out the way we compare sets. For finite sets, people who can't count can still compare them by looking for a 1-to-1 correspondance between the elements of the two sets. For two finite piles that each contain identical objects (e.g., a pile of manatees and a pile of aardvarks), the order of selecting objects out of the piles does not matter. All that matters is that there will at each step be exactly one object selected from each pile. Turning to infinite sets, mathematicians continued to ignore the order of selecting objects from each set. If at each and every single step both sets still have elements in them, then there won't be a smaller one and a bigger one. Writing this down, it seems less substantial. I'll go for the more accurate description. For the finite sets, when every element of one set can be paired up with some element from the other set, then clearly the first set is smaller than or equal to the second (after finishing up all the elements in the first set there just might be some elements left over in the second set, making it bigger). This can be extended to infinite sets: if every element of the first infinite set can be paired up with some element from the second set, then the first set is considered smaller than or equal to the second. But what if both sets have this property? Both are smaller than or equal to the other. Then they should be considered equally infinite, since labelling any of them as bigger would disagree with the fact that it was paired up with a subset of the other... To Be Continued... I need to look deeper into this! I feel like my arguments are not solid enough.
Bicho the Inhaler
Daedalian Member

Guest

Posted: Tue Nov 08, 2005 11:36 pm    Post subject: 16

 Lucky Wizard wrote: Complex numbers are the sums of real numbers and imaginary numbers. (An imaginary number is a number that is the square root of a negative number; it can also be viewed as the sine of a number greater than 1.) Imaginary numbers take the form ai, where a is the square root of the absolute value of the negative number that has this imaginary number as its square root, so an imaginary number line can be imagined. The complex numbers take the form a+bi, and one can visualize a complex number plane where each complex number is at (a,b). Back to the subject, John Allen Paulos' book Beyond Numeracy gives a great, fairly elementary introduction to infinity (in the chapter titled "Infinite Sets"). It includes a proof that the set of all reals is bigger than the set of all positive integers.
Jack_Ian
Big Endian

 Posted: Wed Nov 09, 2005 4:48 pm    Post subject: 17 I've always thought of Cantor numbers as describing the size class of a particular infinity rather than the actual size and I therefore have a difficulty believing proofs that cancel infinities out when they have the same Cantor number. Somehow it just seems wrong to me. Is there some criteria that says when such cancellations are allowable and when not? Also, how do you go about answering a question like the following: How many positive integers have an infinite number of digits when expressed in base 10 with no leading zeros? a) None. All integers have a finite number of digits. b) One. The final integer. c) Infinite. 111..., 222..., 333..., ..., 999..., 101010..., 121212..., 131313..., ... d) Invalid question. It seems to me that any of the above answers could be correct depending upon how you view the question. (BTW Thanks for starting this thread, I love trying to wrap my head around such concepts)
extro...
Guest

 Posted: Wed Nov 09, 2005 8:36 pm    Post subject: 18 All positive integers are finite, by definition. a) None.
Chuck
Daedalian Member

 Posted: Wed Nov 09, 2005 8:38 pm    Post subject: 19 a) None. There is no last integer and an infinite string of digits doesn't represent an integer because you can't count to it.
old grey mare
Guest

 Posted: Wed Nov 09, 2005 11:21 pm    Post subject: 20 If sum(k = 1 to infinity) (9 * 10^k) can't be a number, does that mean that sum(k = 1 to infinity) (9 * 10^-k) isn't a number either?
Guest

 Posted: Thu Nov 10, 2005 1:12 am    Post subject: 21 no
Bicho the Inhaler
Daedalian Member

Posted: Sun Nov 13, 2005 7:51 am    Post subject: 22

 old grey mare wrote: If sum(k = 1 to infinity) (9 * 10^k) can't be a number, does that mean that sum(k = 1 to infinity) (9 * 10^-k) isn't a number either?
Actually, the former is interpreted as -10 in some systems (like the 10-adic system). (Intuition: the decimal expansion of this number is ...9999990. If we add 10 to this number, what do we get? 0 in the ones place; 0 in the 10s place and carry a 1; 0 in the 100s place and carry a 1; 0 in the 1000s place and carry a 1; ad infinitum, yielding ...0000000.) So the premise of your argument isn't true

Even in the 10-adics, the answer to Jack_Ian's question is (a), since no infinite-length decimal expansions are positive integers unless they begin with an infinite string of leading zeros.
old grey mare
Guest

Posted: Mon Nov 14, 2005 2:11 am    Post subject: 23

 Bicho the Inhaler wrote: the premise of your argument isn't true

Hey, it wasn't my premise. It was Chuck's. I was just noting the similarity of the expressions. If you think an infinite string of digits can represent an integer then take it up with him.
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