## Learning Math: Number and Operations

# 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:

Shigeto had no trouble finishing the expansion using his pattern. How about you? Can you finish the expansion and explain your answer?

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?

##### Problems H2-H11 adapted from Findell, Carol, ed. *Teaching with Student Math Notes, Volume 3.* p. 121. © 2000 by the National Council of Teachers of Mathematics. Used with permission. All rights reserved.

### 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 (10

^{6}) 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.