Proof: Rational q and Irrational pDate: 03/03/2003 at 21:36:26 From: Brandon Subject: Topology Prove that if a and b are rationals, then there is a rational q and an irrational p so that both p and q are between a and b. Also: Prove that if a and b are irrationals, then there is a rational q and an irrational p so that q and p are both between a and b. I was thinking it might have to do with taking an average or possibly taking the rational and putting it over something irrational [i.e. sqrt(2)]. I'm really not sure. Thank you! Date: 03/04/2003 at 02:24:34 From: Doctor Jacques Subject: Re: Topology Hi Brandon, First problem ------------- Let a and b be rational numbers, with a < b. To find a rational q between a and b, your idea of taking an average is correct: it should be easy to prove it. We want also to find an irrational p between a and b. You can certainly find an irrational number between 0 and 1 (for example, 1/sqrt(2) is such a number) - call that number z. For general a and b, consider the function: f(x) = (b-a)x + a The point is that f maps 0 to a and 1 to b, and any number between 0 and 1 to a number between a and b. In addition, as the coefficients are rationals, f maps rationals to rationals and irrationals to irrationals. Specifically, if we consider the number 1/sqrt(2) (an irrational between 0 and 1), its image will be: a + (b-a)*(1/sqrt(2)) and that number is irrational and between a and b. I used the letter z because there is nothing special with 1/sqrt(2) - any irrational between 0 and 1 would do. Second problem -------------- We assume now that a and b are irrational numbers, with a < b. The idea is as follows: Consider the set of numbers: A = {n/r | n in Z} where r is an integer greater than 2/(b-a). Note that all elements of A are rational. On the number line, these numbers correspond to points spaced (1/r) apart. Note that (1/r) < (b-a)/2. The length of the segment (a,b) is (b-a), and the spacing of the numbers of A is less than half that interval. It is intuitively obvious that there will be one point of A in the interval (a,b). To prove it rigorously, define n as the least integer greater than (ra). We have: n-1 <= ra < n or (n-1)/r <= a < n/r Can you prove that (n/r) < b, using the fact that 1/r < (b-a)/2 ? This will give you the rational q you are looking for. To find an irrational between a and b, consider (a+q)/2. That number is certainly between a and b; can it be rational ? Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 03/05/2003 at 11:37:49 From: Brandon Subject: Topology Thank you for what you have given me so far. But I still have a question on how to prove n/r < b using 1/r < (b-a)/2. Thank you for your help. Brandon Date: 03/06/2003 at 02:10:06 From: Doctor Jacques Subject: Re: Topology Hi Brandon, Concerning the second part, we have: (n-1)/r <= a 1/r < (b-a)/2 If we add these inequalities, we get: n/r < a + (b-a)/2 = (a+b)/2 < (b+b)/2 = b since a < b. As we also know that n/r > a and n and r are integers, n/r is a rational between a and b. Does this make sense ? - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 03/06/2003 at 12:57:09 From: Brandon Subject: Thank you (Topology) Yes, that makes perfect sense. Thank you Dr. Jacques. Brandon |
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