Density of Rational NumbersDate: 01/31/98 at 16:38:11 From: Keri McKenzie Subject: Why the set of rational numbers is dense Dear Dr. Math, I am wondering what the following statement means: "The set of rational numbers is dense." I know what rational numbers are thanks to my algebra textbook and your question sites. I have determined through brainstorming that rational numbers are dense because there are so many of them. In other words, they are densely crowded when defined as a set IE. Q={-4,0,2/3,4,5.56869}. Am I on the right track? Thanks for your help. I look forward to your input and I love your site! Sincerely, Keri Date: 01/31/98 at 19:26:52 From: Doctor Pete Subject: Re: Why the set of rational numbers is dense Hi, The statement should read: "The set of rational numbers is dense in the real numbers." (We will shortly see why we must say "...in the real numbers.") Now, let Z = the set of all integers Q = the set of all rational numbers R = the set of all real numbers R\Q = the set of all irrational numbers (namely, those elements in R which are not in Q). Then the meaning of "dense," in this case, is the following: Given any two distinct rational numbers p < q, there exists an irrational number x such that p < x < q; that is, given two distinct rational numbers we can always find an irrational number in between. And similarly, given any two distinct irrational numbers x < y, we can find a rational number q such that x < q < y. So, we see why we must append the statement "...in the real numbers" because the property of "dense" of a set is dependent on the bigger set you contain the original set in. That is, in this case, Q is a proper subset of R, and to consider if Q is dense in R, we must examine elements which are in their difference R\Q. A question: Is Z (the set of integers) dense in Q? The proof that Q is dense in R involves some basic analysis, and without knowing your background, I don't know how useful it would be for me to include it here. However, it is a somewhat "intuitively obvious" fact (but not so obvious once you see the proof). Density is a much stronger statement than just saying "there are a lot of them," because it is in effect saying, "there are a lot, *and* they're very, very closely packed together." Hence the choice of the word "dense" (as opposed to, say, "numerous"). -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/02/98 at 14:12:46 From: Joan Vilaseca Subject: Cardinality of rationals and irrationals In a section of the entry Density of Rational Numbers, you wrote the following: Given any two distinct rational numbers p < q, there exists an irrational number x such that p < x < q; that is, given two distinct rational numbers we can always find an irrational number in between. And similarly, given any two distinct irrational numbers x < y, we can find a rational number q such that x < q < y. That is, there is no pair of irrationals without an irrational in between and also no pair of irrationals without a rational in between. But then, this is a one-to-one relation (just map the first of every pair with the number in between.) And that means that cardinality of rationals and irrationals seems to be the same. Where's the fault? In geometry, that contradiction arises when we suppose a line to be equivalent to a set of points (0+0+0+....+0<>0). Joan Vilaseca Date: 09/02/98 at 16:24:35 From: Doctor Wilkinson Subject: Re: Cardinality of rationals and irrationals You have not defined a one-one relation. In the first place, every rational appears in infinitely many pairs, so you can't say for a given rational which irrational you're mapping to, and similarly for the irrational -> rational mapping. Secondly, you have not given any rule for choosing an irrational between two rationals or for choosing a rational between two irrationals. Finally, to get a one-one relation from these two mappings, even assuming they were defined, you would have to show that they were inverses of each other. - Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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