# Fractions and Decimals Homework

Problem H1
a. Find the fractional equivalent for 0.142857.
b. Find the fractional equivalent for 0.142857142857142857….

Problem H2
Shigeto and Consuela were computing the decimal expansion of 1/19. Since Shigeto used scratch paper, he had only a little room to write his answer. He continued writing the digits on the next line, and in the end, his answer looked like this: Shigeto noticed a pattern in these numbers. Describe his pattern. Look at the columns of digits. What do you notice?

Problem H3
Consuela did her computation on a narrow notepad. Her answer looked like this: After looking at Shigeto’s pattern, Consuela tried to find a pattern in her answer. What observations can you make about Consuela’s pattern? Look at the columns of digits. What do you notice?

Problem H4
David started to compute the decimal expansion of 1/47. He got tired after computing this much of the expansion:  What are the possible periods for the decimal expansion of 1/47? Can you predict the actual period based on how many digits there are in the decimal expansion so far?
How would you arrange those digits so that the columns add up to 9?

Problem H5
Does the length of the period of your expansion make sense? Explain why or why not.

Problem H6
Consuela looked at David’s work and knew immediately that her method would not be helpful. Explain why not.

Problem H7
Mr. Teague asked the class to compute a decimal expansion with period 42. Unfortunately, his dog spilled paint on Shigeto’s and Consuela’s answers. Use the visible information in Shigeto’s and Consuela’s answers and the patterns you have seen to find the complete decimal expansion: Problem H8
Find the fraction with this particular decimal expansion.

Problem H9
Is it possible to represent the number 1 as a repeating decimal? Think about the decimal expansion for 1/3 = 0.333333…. What would 2/3 be? What fraction would be closest or equal to 1?

Problem H10
Is it possible to predict the period of 1/14 if you know the period of 1/7 (i.e., six)?

Problem H11
Is it possible to provide a convincing argument to prove that the decimal expansion of 1/n has a period that is less than n?

### Solutions

Problem H1
a. 0.142857 = 142,857/1,000,000.
b. F = 0.142857142857. . . . There are six digits repeated, so multiply by 1,000,000 (106) to get 1,000,000F = 142,857.142857142857. . . . Subtracting F from each side gives you 999,999F = 142,857, so F = 142,857/999,999, which can be reduced to F = 1/7.

Problem H2
Adding the two groups of nine numbers gives you 99,999,999.

Note that if the period length is even, you can always break the number into two rows where columns of digits will add up to 9.

Problem H3
Adding the three groups of six numbers gives you 999,999.

If the period length is a multiple of three, you can always break the number into three rows where each column will either add up to 9 or will be a two-digit number ending in 9. Note that this works only if we break a number into two or three rows. No other break-up of the number will work all the time.

Problem H4
Since the sum should be all 9s, the pattern would continue:

 0.021 276 595 744 680 851 063 82 978 723 404 255 319 148 936 17

Note that the repetition starts after 46 decimal places, so the period is 46.

Problem H5
Yes, it makes sense. As with other prime numbers, the period is a factor of one less than the number itself. In this case, with the prime number 47, the period would need to be either 46, 23, 2 or 1. We already have 23 digits and they haven’t repeated, so the period must be 46.

Problem H6
Consuela’s method is not helpful because 3 is not a factor of the period, 46. Her method relies on being able to arrange the non-repeating digits into three rows of identical lengths.

Problem H7
We can primarily refer to Consuela’s numbers, starting from the right side. If the sum is going to be all 9s, the right-most covered digits must be 775 51 (don’t forget to carry). We can then use Shigeto’s numbers 408 163 (uncovered) to find the rest of Consuela’s covered bottom row: 346 938 (the 34 is visible on Shigeto’s page). Finally, we can complete Consuela’s covered top row: 020. The completed number is .

As a check, you might verify that these numbers work by using Shigeto’s rule; i.e., that they add (in pairs) to 9s.

Problem H8
Going on the assumption that the number is in the form 1/n, n must be less than 50 (since 1/50 = 0.02) but greater than 47 (1/47 = 0.021276. . ., as seen in Problem H3). So n is either 48 or 49; 1/48 = 0.208333. . ., which is too large, and 1/49 is just right.

If we could not make this assumption, we could use the multiply-and-subtract method from Part B, but we’d need a really accurate calculator!

Problem H9
Yes, 1 can be represented as 0.99999…. The technique of Part B can be applied to this decimal; if F = 0.999…, then 10F = 9.999…, and, subtracting F from each side, 9F must equal 9. This means that F = 1!

Another way to convince yourself that this is true is that 0.333… represents 1/3. Then 2/3 is represented as 0.666…, and 3/3 is represented as 0.999. …Since 3/3 = 1, 1 and 0.999… are the same number.

All terminating decimals have an alternate representation in this form. For example, 0.25 can also be represented as 0.24999999….

Problem H10
Yes. Because 2 is a factor of 10, it will have no effect on the period; it will only delay the repetition by one decimal place. So 1/14 = 0.0714285714285. . ., which, like 1/7, has a period of six. Similarly, 1/28 will also have a period of six, delayed by two decimal places.

Problem H11
One convincing argument is that when dividing by n, there are only n possible remainders: the numbers (0, 1, 2, 3, 4, …, n – 1). If we divide and get a remainder of 0, then the division is complete, and the resulting decimal is terminating. If we divide and get a remainder that we have seen earlier in our division, this means that the decimal is about to repeat. So if we want the longest possible period for our decimal, we want to avoid 0 but run through every possible number before repeating. To do this, we would hit all the non-zero remainders (1, 2, 3, …, n – 1) while missing 0 — a total of n – 1 possible remainders. Since each remainder corresponds to continuing the decimal by one place, there is a maximum of n – 1 decimal places before repetition begins.