Prove that the dyadic rationals are dense in R?

Question

Prove that the dyadic rationals are dense in R. Use this to show that fn(x) is not integrable,
where fn(x) = 1 if x = k/(2^n) where k is a natural number
0 otherwise

Answer 1

1) The main idea is to replace n with 2^n everywhere in the classic density of Q in R proof.



We want to show that for any a, b in R with a < b,

we can produce integers m, n such that a < m/2^n < b.



Assume without loss of generality that a, b > 0.

[If a <= 0 and b > 0, consider the subinterval (b/2, b), and find a dyadic rational

there using the discussion below.

If a,< 0 and b <= 0, then by below we can find a dyadic rational c in (-b, -a);

multiply that by -1 to get one in (a, b).]

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(i) By Archimedes, pick n large enough so that 1/2^n < b - a.



(ii) So, a < m/2^n < b <==> a * 2^n < m < b * 2^n.



With n now chosen, the idea is to choose m to be the smallest positive integer

greater than a * 2^n. i.e, pick m in M so that m - 1 <= a * 2^n < m. (**)



The right half of (**) implies that a < m/2^n, giving half the desired inequality.



Next, rewriting 1/2^n < b - a in the form a < b - 1/2^n, use the left half of (**)

m <= a * 2^n + 1 < (b - 1/2^n) * 2^n + 1 = b * 2^n.



Thus, m < b * 2^n ==> m/2^n < b.



(iii) Putting this altogether, we get a < m/2^n < b, as required.

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2) For fixed n, fn is certainly Riemann integrable (as the set of discontinuities are countable); however {f_n} converges (as n tends to infinity) to a function f which is not Riemann integrable. That's where the density arguments can enter the problem.



Link:

http://books.google.com/books?id=4hhFXPdTXwoC&pg=PA243&lpg=PA243&dq=fn%28x%29+is+not+integrable,+where+fn%28x%29+%3D+1+if+x+%3D+k/%282^n%29+where+k+is+a+natural+number+0+otherwise&source=bl&ots=b9AgL9TWXJ&sig=AwGFRbfMUxIc_o2AZitmPYvXjug&hl=en&ei=-oXcTpyWN6iXiAL97qzGCQ&sa=X&oi=book_result&ct=result&resnum=8&sqi=2&ved=0CFgQ6AEwBw#v=onepage&q&f=false



More precisely, f(x) = 1 if x is dyadic, and f(x) = 0 otherwise.

This function is discontinuous everywhere. (Since the set of discontinuities is not of measure 0, f can;t be Riemann integrable on any closed interval.)



(i) f is discontinuous at any x = c, a dyadic number.

By density arguments, we have a sequence of dyadics {x(n)} as well as irrationals {y(n)} converging to c. On one hand, {f(x(n))} = {1} converges to 1, while {f(y(n))} converges to 0. Hence, the limit does not exist.



(ii) Similarly, f is discontinuous at any x = c, non-dyadic number.



I hope this helps!

Answer 2

As far as I can see, this problem suggests a novel elementary way of proving that Arccos(-1/3) is incommensurate with pi and perhaps other statements like it: As was already shown, the orthonormal transformation [1 -2 -2] |-2 1 -2| (1/3) =: M [2 2 -1] rotates the plane (a, a, b) by Arccos(-1/3) so that M(a, a, b) = (1/3)(-a-2b, -a-2b, 4a-b). Define a sequence (p_n) of points in R^3 where p_(0) = (1, 1, 1) and p_(k+1) = M(p_k). Notice that p_1 = (-1, -1, 1). We claim that for all k in Z+, we have p_k = (A_k, A_k, B_k) / 3^(k-1), where A_k = -1 (mod 3) and B_k = 1 (mod 3). The proof is straightforward using induction. Hence for all k > 1, the point p_k is never in Z^3. In particular, p_k = p_0 iff k=0. This proves that Acos(-1/3) is incommensurate with pi. ****** The density of S ={p_0, p_1, p_2, . . . } in the circle C = {(x,y,z) : x^2+y^2+z^2=3, x=y} now follows from these 2 facts: (1) M(S) = S \ {p_0}, and (2) the sequence p(n) is an injective function from {0, 1, 2, . . } into C. (1) holds because we've just shown p_0 = p_k iff k=0. (2) is true because supposing p_m = p_n for some m and n, 0 < m < n, we would have the contradiction p_0 = (M^(-m))(p_m) = (M^(-m))(p_n) = p_(n-m) with n-m > 0. A corollary of (1) is that -- save for a single point -- S and C \ S are both invariant under M. Now suppose for the sake of contradiction that there is an open nonempty arc interval H contained in C \ S. Let G be the largest arc interval contained in C \ S, such that H is contained in G. Then endpoints p and q of G will be points or limit points of S. Since G is a subset of C \ S and since C \ S is unchanged under M, it follows that M(G), M^2 (G), M^3 (G), ... are all subintervals of C \ S, whose endpoints are points or limit points of S. But the circumference of C is less than N times the arc length of G for some sufficiently large natural number N. Hence, for some m and n, 0

Answer 3

Dyadic Rationals