# 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

See below.

#### Explanation:

Let's go a step further, and design a set that contains *every* rational number with a repetend with

**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

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#0# s if#a# has fewer than#10^6# digits. We will call#a# usefulif 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

Next, we'll construct our set of potential nonrepeating initial decimal digits. Keeping in mind that this could also have leading

Let

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

Let

Now that we have sets encompassing every possible

Then

Thanks to Sente, the theory is in his answer.

For a subset of the answer

integer/

means the least significant digit..

Elucidation:

Let I =2, M =.209/1000=.209,

between d's are all 0..

Then.

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

Note the division by

Both numerator and denominator have the same number of sd.

Sans msd d, d's could be any