Proof of Density: D = dyadic rationals set is dense on [0,1] | Homework Help

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The discussion focuses on proving that the dyadic rationals set, defined as D={m/2^n : n∈N, m=0,1,2,...,2^n}, is dense in the interval [0,1]. Participants emphasize the necessity of using the Archimedean property to demonstrate that for any interval (a,b) within [0,1], there exists a natural number n such that b-a < 1/2^n, ensuring that D intersects with (a,b). The use of the Pigeonhole Principle is also suggested as a method to substantiate the intersection of D with any given interval.

PREREQUISITES
  • Understanding of dyadic rationals and their properties
  • Familiarity with the Archimedean property of real numbers
  • Knowledge of the Pigeonhole Principle in mathematical proofs
  • Basic concepts of real analysis and intervals
NEXT STEPS
  • Study the Archimedean property in detail
  • Explore the Pigeonhole Principle and its applications in proofs
  • Investigate the properties of dyadic rationals and their density in real numbers
  • Review real analysis concepts related to intervals and density
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Mathematics students, educators, and anyone interested in real analysis, particularly those studying properties of rational numbers and their applications in proofs.

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Homework Statement


Prove that D={[tex]\frac{m}{2^{n}}[/tex] : n[tex]\in[/tex] N , m=0,1,2,...,2[tex]^{n}[/tex]} (dyatic rationals set) is dense on [0,1] , i.e. if (a,b) [tex]\subset[/tex] [0,1] then (a,b) [tex]\bigcap[/tex] D [tex]\neq[/tex] emptyset

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The Attempt at a Solution


Is it wrong if I just state that because a,b[tex]\in[/tex][tex]\Re[/tex] we know that there exists m,n[tex]\in[/tex]N, l=[tex]\frac{m}{2^{n+1}}[/tex], u=[tex]\frac{m}{2^{n-1}}[/tex] and r=[tex]\frac{m}{2^{n}}[/tex] such that l [tex]\leq[/tex] a [tex]\leq[/tex] r [tex]\leq[/tex] b [tex]\leq[/tex] u?
 
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Why do you need the extremal inequalities? You should only need [itex]a \leq r \leq b[/itex]. Also it seems like you are trying to say "Because a and b are real, its is obvious there are naturals m,n that allow us to conclude the result", which really is not substantial enough.

Instead, I would use the Archimedian property to argue where there exists [itex]n \in \mathbb{N}[/itex] such that [tex]b-a < \frac{1}{2^n}[/tex], then you a Pigenhole Principle type argument to see why D should intersect with (a,b) .
 

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