Sets Containing an Infinite Number of Members
Date: 02/03/98 at 19:33:30 From: Kate Subject: Sets Containing an Infinite No. of Members Which of these 2 sets is larger? A. The set of all rational numbers B. The set of all irrational numbers
Date: 02/06/98 at 10:56:10 From: Doctor Celko Subject: Re: Sets Containing an Infinite No. of Members The set of irrational numbers is larger, but it is harder to find them than it is to find rational numbers. The way that you prove one set is the same size as another is to map each element of set A onto one element of set B; this is really a fancy version of a caveman matching his arrowheads to his seashells to see if he has more of one thing than another. The rational numbers can be matched to the integers using a trick called Cantor's diagonalization. Make an infinite table of the rational numbers (p/q): 1/1 1/2 1/3 ... 2/1 2/2 2/3 ... 3/1 ... ... Rows are increasing values of q and the columns are increasing values of p. You can dance down this table by starting at 1/1, then go to 1/2, to 2/1, to 3/1 and back up the diagonal. You can repeat this pattern up and down all the diagonals of the table forever, counting the (p/q) as you go. With irrational numbers, assume that you can make a countable list of them as infinitely long decimal fractions and make each row of the list unique: .0218498756985 ... .1849875698598 ... .98756985785899... ... Look down the diagonal of this list (.087...) and you will see an infinite decimal fraction. Change all the digits in this fraction to something else and you will have a number that cannot be in the list because it is different from the n-th number of the list in the n-th position. So this set must be bigger than the set of countable numbers. -Doctor Celko, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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