Help this is hard

Let S be the set of all real numbers of the form

a1/3 + a2/3^2 + a3/3^3 + ....

where a_i \in {0, 1, 2} for all i.

(a) Is the number 1/7 in the set S?

(b) Is the number 1/4 in the set S?

Guest Mar 15, 2023

#1**0 **

(a) To determine if the number 1/7 is in the set S, we need to find a sequence of integers a_i such that:

1/7 = a1/3 + a2/3^2 + a3/3^3 + ...

Multiplying both sides by 3^2, we get:

9/7 = 3a1/3 + a2/3 + a3/3^2 + ...

Since the right-hand side of this equation is the sum of an infinite geometric series with first term a1/3 and common ratio 1/3, we can use the formula for the sum of an infinite geometric series to get:

9/7 = 3a1/3 + a2/3 + a3/3^2 + ... = a1/3 / (1 - 1/3) + a2/3^2 / (1 - 1/3) + a3/3^3 / (1 - 1/3) + ...

Simplifying, we get:

9/7 = 3a1 + a2/3 + a3/9 + ...

Since the coefficients a_i are integers between 0 and 2, the sum on the right-hand side can take on any value between 0 and 3 + 1/3 + 1/9 + ..., which is less than 4. **Therefore, there is no way to choose the coefficients a_i to make the sum equal to 9/7, so the number 1/7 is not in the set S.**

(b) Similarly, to determine if the number 1/4 is in the set S, we need to find a sequence of integers a_i such that:

1/4 = a1/3 + a2/3^2 + a3/3^3 + ...

Multiplying both sides by 3^2, we get:

9/4 = 3a1/3 + a2/3 + a3/3^2 + ...

Using the same argument as before, we get:

9/4 = 3a1 + a2/3 + a3/9 + ...

Since the coefficients a_i are integers between 0 and 2, the sum on the right-hand side can take on any value between 0 and 3 + 1/3 + 1/9 + ..., which is less than 4. **Therefore, there is no way to choose the coefficients a_i to make the sum equal to 9/4, so the number 1/4 is not in the set S.**

Guest Mar 15, 2023