Using https://socratic.org/questions/in-1-6-1-6666-repeating-6-is-called-repeatend-or-reptend-i-learn-from-https-en-w, how do you design a set of rational numbers { x } that have reptend with million digits?

2 Answers
Oct 19, 2016

See below.

Explanation:

Let's go a step further, and design a set that contains every rational number with a repetend with 10^6106 digits.

Warning: The following is highly generalized and contains some atypical constructions. It may be confusing for students not completely comfortable with constructing sets.

First, we want to construct the set of our repetends of length 10^6106. While we can start with the set {1, 2, ..., 10^(10^6+1)-1} which contains every natural number with at most 10^6 digits, we would encounter a problem. Some of these repetends could be represented with smaller strings, for example 0.bar(111...1) = 0.bar(1), or 0.bar(121212...12) = 0.bar(12). To avoid this, we first define a new term.

Consider an integer a in [1, 10^(10^6+1)-1]. Let a_1a_2...a_(10^6) be a 10^6 digit representation of that integer, possibly with leading 0s if a has fewer than 10^6 digits. We will call a useful if for every proper divisor m of 10^6, a is not of the form a_1a_2...a_ma_1a_2...a_m" "..." "a_1a_2...a_m

Now we can make our set of repetends.

Let A = {a in {1, 2, ..., 10^(10^6+1)-1} : a" is useful"}

Next, we'll construct our set of potential nonrepeating initial decimal digits. Keeping in mind that this could also have leading 0s, or consist entirely of 0s, we will represent our numbers as tuples of the form (k, b), where k will represent the length of the string of digits, and b will represent its value when evaluated as an integer. For example, the digits 00032 would pair with the tuple (5, 32).

Let B = (NNuu{0})xx(NNuu{0})

Finally, let's add our integer portion to the mix. Note that unlike the fractional portions, we will account for sign here, and use ZZ instead of NN.

Let C = A xx B xx ZZ. That is, C is the set of 3-tuples (a, (k, b), c) such that, a is a useful integer with at most 10^6 digits, (k, b) represents a k-digit string of digits whose integral value is b, and c is an integer.

Now that we have sets encompassing every possible a, b, c string with the desired properties, we will put them together using the form constructed in the referenced question.

S := {((10^kc+b)(10^(10^6)-1)+a)/(10^k(10^(10^6)-1)):(a, (k, b), c) in C}

Then S subset QQ is the set of rational numbers with 10^6 digit repetends.

Oct 19, 2016

Thanks to Sente, the theory is in his answer.

For a subset of the answer

{ x } = {I+M+(d_(msd)ddd...dddd_(lsd))/(9999...9999)},

I in N and M a proper fraction of the form m-digit

integer/10^m, d_(msd) is non-zero most significant digit. lsd

means the least significant digit..

Elucidation:

Let I =2, M =.209/1000=.209, d_(lsd)=7 and d_(msd)=3. In-

between d's are all 0..

Then.

x = 2.209+(7000...0003)/(9999...9999)

#=2.209 7000...0003 7000...0003 7000...0003 ... ad infinitum.

Note the division by 10^100001-1=9999...9999.

Both numerator and denominator have the same number of sd.

Sans msd d, d's could be any in { 0 1 2 3 4 5 6 7 8 9 }.