# Fractions

### Different fractions with the same value

Seeing a fraction as a piece of pie helps. Notice that the simple fraction $\frac{1}{4}$ can be cut in smaller pieces without changing its value as part of the whole.

### Factors help find the simplest form-cancellation

A fraction might have large numbers for both the numerator and denominator. See whether the fraction can be reduced to a simpler form. A calculator that handles fractions will automatically find and present them in their simplest form. If your calculator cannot handle fractions, you must know how to calculate them yourself.

#### Always Find the Factors in a Fraction

If the numerator and denominator both end in zeros, you can strike off the same number of zeros from each. If both are even, divide both by 2.

Sometimes the factors aren't so obvious. Here, for the fraction $\frac{455}{462}$, is one way. After checking the factors of each, they both contain a 7. If 7 is a factor of both the numerator and denominator, divide both by 7.

### Spotting the factors

Here are rules for spotting factors. If the last digit of a number divides by 2, then the whole number does. This rule can go further. If the last 2 digits divide by 4, the whole number does. If the last 3 digits divided by 8, the whole number does. That leads to a whole series of checks for powers of 2:4, 8, 16, 32, etc. A similar set works for powers of 5:25, 125, 625, 3125, etc.

Rules also exist for 3s and 9s. Add the digits together. If the sum of the digits divides by 3, the whole number does. If the sum of digits divides by 9, the whole number does.

The check for dividing by 11 is more complicated. Add alternate digits in two sets. If the sums are identical, differ by 11, or differ by a multiple of 11, the whole original number divides by 11.

### Rules for finding factors

That covers finding simple numbers as factors: 2, 3, 4, 5, 8, 9, 11, etc. If 2 and 3 both "go," then 6 is a factor. No easy check exists for 7.

When finding factors, try primes - numbers that won't factorize into smaller numbers.

This table shows factors of numbers and identifies primes up to the number 20. We have to try each prime when looking for factors. Isn't there an easier way?

## HOW FAR SHOULD YOU TRY FACTORS

### How far to try

A useful principle can save you from wasting time when looking for factors. Remember the multiplication table-that line of squares down the diagonal? It goes far beyond the numbers in the table. Look at it this way: If a number does not have at least two factors, it must be prime. So, if it has any factors at all, at least one of them must be smaller than the nearest square.

Look at these two examples. The nearest square to 139 is 144. If 139 has any factors, at least one of them must be less than 12. After trying up to 11, none of them work, so 139 must be prime.

The square of 23 is larger than 493. Try squaring each number before you try it as a factor. 23 squared is 529. Try all the primes up to 23. When you reach 17, you will find that 17 divides into 483, 29 times. So, the factors of 493 are 17 and 29.

### Squares and primes

Suppose you want the factors of 8,249. The table of squares show that the square of 91 is 8,281. Try primes up to 89, the last one before 91. You will find that 73 and 113 are factors. If you had reached 89 with no factor, the original number (8,249) would have been prime.

When making a table of squares, look for patterns, like you did with the much simpler multiplication table. Look at the middle column. The last two digits run the squares of 9 down to 0 and then up to 9 again for the square of 59. This occurs because the square of 50 ends in double zero and twice 5 (the first digit) is 10. A similar sequence begins at 91 and runs up to 109.

### Factoring with a calculator

Of course, a calculator can help you make such a table. It can also help you find factors. The easy way is to put the number of which you want factors in memory, then keep withdrawing it with the MR (memory recall) key to try the next prime. If it displays a number with a decimal, you don't have a factor. When you hit a factor, the readout is a whole number.

Thus, in the next example, when you get to 73, the calculator reads 113. Here's how it goes:
Enter 8249 Press M in or M+ (making sure that the memory is empty)

If your calculator handles fractions, it keeps finding the simplest form for you. But it helps if you know what it's doing. If your calculator doesn't handle fractions, it will calculate everything in decimals. This system is difficult to verify, unless you do it yourself, the old way.

Remember: in adding or subtracting fractions, they must have the same denominator. For instance, to add $\frac{1}{2}, \frac{2}{3},$ and $\frac{5}{12}$, both $\frac{1}{2}$ and $\frac{1}{3}$ can be changed to 12ths. $\frac{1}{2}$ is $\frac{6}{12}$ and $\frac{2}{3}$ is $\frac{8}{12}$. Now just add numerators, because they are all 12ths: 6 + 8 + 5 = 19. $\frac{19}{12}$ is more than 1. Subtract $\frac{12}{12}$ for 1. $\frac{17}{12}$ is the answer.

Suppose you must subtract $3\frac{3}{5}$ from $7\frac{5}{12}$. 60 is the common denominator. $3\frac{3}{5}$ is $3\frac{36}{60}$ and $3\frac{5}{12}$ is $3\frac{25}{60}$. You cannot subtract 36 from $25$. So you borrow $1$, converting it to $60$ths. Now, subtract $3\frac{36}{60}$ from $6\frac{85}{60}$. 36 from 85 is 49. 3 from 6 is 3 and the complete answer is $3\frac{49}{60}$.
To Add Fractions, Each One Must Have the SAME DENOMINATOR

## Subtract $3\frac{3}{5}$ from $7\frac{5}{12}$

1. Common denominator = 60.
2. $\frac{3}{5}=\frac{3\cdot 12}{5 \cdot 12}=\frac{36}{60}$ and $\frac{5}{12}=\frac{5 \cdot 5}{12 \cdot 5}=\frac{25}{60}$
3. Because $\frac{36}{60}$ is bigger than $\frac{25}{60}$, change $7\frac{25}{60}$ to $6 + 1\frac{25}{60} = 6\frac{60 + 25}{60} = 6\frac{85}{60}$
4. $6\frac{85}{60}-3\frac{36}{60} = 3\frac{49}{60}$ OR $7\frac{5}{12}-3\frac{3}{5} = 3\frac{49}{60}$

### Finding the common denominator

How do you find a common denominator? Sometimes it's easy. It's not always so obvious. Older textbooks had a routine for the job, but it wasn't easy to understand. Here's a way you can understand. Suppose you have:
$\frac{1}{4} + \frac{1}{3} + \frac{2}{5} + \frac{1}{6} + \frac{5}{12} + \frac{3}{10} + \frac{7}{30} + \frac{4}{15}$.
Find the factors of each denominator: $4 = 2 \times 2; 3$ and $5$ are both prime; $6$ is $2 \times 3$; $12$ is $2 \times 2 \times 3; 10$ is $2 \times 5; 30$ is $2 \times 3 \times 5$; and $15$ is $3 \times 5$.

Our common denominator must contain every factor that is in any denominator. The factors are 2, 3, and 5. The common denominator is $2\times2\times3\times5 = 60$. Now convert all the fractions to 60ths. The numerators are: 15 + 20 + 24 + 10 + 25 + 18 + 14 + 16, which adds up to 142. Reduce the final form, because both divide by 2, yielding $\frac{71}{30}$. Taking out $\frac{30}{30}$ for a whole 1, twice, the answer is $2\frac{11}{30}$.

### Calculators that "do" fractions

Some pocket calculators "do" fractions. Knowing how to use fractions and knowing calculator limitations can help you understand more about fractions and decimals. Individual calculators might use different ways of keying the problem in, but the way it works is similar. The readout might also differ.

Enter 4 and $\frac{2}{7}$. Other keys change the readout first to 30.7, called in the old parlance an improper fraction, meaning that its value is more than 1; then to $4.28571428 \cdots,$ its decimal form.

Now, multiply by 7. The calculator will read 30. Divide by 7. It does not give you 4 and $\frac{2}{7}$ again, but the decimal form, $4.28571428\cdots$ Do some of the things, adding, subtracting, multiplying, or dividing fractions that this book has shown you. The calculator will give the same answers, always presenting them in their simplest form.

### Significant figures

Just how accurate is the figure you look at? Ask yourself how reliable the number is that you look at. Devices that give you numerical information might be either analog or digital. Some think that digital is much more accurate. However, numbers can be deceptive.

Suppose you talk about a 150-pound man. Does he weigh exactly 150 pounds-not an ounce more or less? Is he between 149.5 and 150.5 pounds, or is he between $145$ and $155$ pounds? The answer depends on what figures in that number 150 are "significant."

If the 0 is just a placeholder, then if he weighed less than 145 or more than 155, the number would be stated as 140 or 160. If the 0 is "significant," then 150 means that he weighs from 149.5 to 150.5 pounds. If he weighed less than 149.5 or more than 150.5, the number would be 149 or 151.

For it to mean not an ounce more or less than 150 pounds, the weight should be written as 150 pounds, 0 ounces.

The word significant should tell you that other figures are not significant. The following text will show why this point can be important.

### Approximate long division: why use it

Long division is something you can't "tell' a calculator. Most students had trouble with it because they didn't understand it. Suppose you divide 150 by 7. Your calculator gives you something like $21.4285714\cdots$. You are tempted to believe all those figures. Now apply the previously said rule. If 150 means more than 145 and less than 155, which means only two figures are significant (the 1 and 5), then dividing by 7 could yield something between $\frac{145}{7}$ and $\frac{155}{7}$, which read out as $20.71527571\cdots$ and $22.14285714\cdots$ .The only figure that doesn't change is the 2 of 20!

The other figures, in any of these answers, can't be reliable. You could say the possible "spread" is from 20.7 to 22.1. Now, if the 0 of 150 was significant, the result can be from $149\frac{5}{7}$ to $150\frac{5}{7}$, which read out as 21.3571428 and 21.5 (surprise, 150.5 divides by 7 exactly!) Now the "spread" is reduced to between 21.357 and 21.50.

Suppose you have 153 (an extra significant figure) to divide by 7. Examine the possibilities here.

## APPROXIMATE LONG DIVISION

### Longhand procedure

Suppose that you have to divide 23,500 by 291 and you believe those end zeros in the dividend aren't significant. You could assume it was between 23,450 and 23,550, perform both divisions, and then decide what was significant. But longhand, that's a lot of work! The practice was to draw a vertical line where figures begin to be progressively more doubtful.

You could divide between the "limiting values" as the possible errors, because only so many figures are significant, then you could guess at the most probable value.

### Using a calculator to find significance

Exploring shows what you can do with a calculator that wouldn't have been practical longhand. Take values that represent the biggest variation on either side of the stated value, then deduce how accurate the answers will be. Notice that in

division, both numbers (the dividend and the divisor) can have a number of significant figures. An answer cannot have more significant figures than the number with the least number of significant figures used as "input" for the problem.

### Approximate long multiplication

The same methods for long division work for multiplication, as shown here for multiplying 5.32 by 2.91. It also applies to addition or subtraction. Suppose you add 55 to 1,000,000. What does that million mean? It could mean "give or take" 500,000. More likely, it would mean "give or take" 50,000. Even at that figure, adding 55 is unlikely to make a difference. That is an extreme example. The result of any operation can only be as accurate as the "weakest link" which in that case would be the million figure. Only in rare instances would the number 1,000,000 mean exactly that number, not 999,999 or 1,000,001.

### Questions and problems

1. Arrange the following fractions in groups that have the same value:
$\frac{1}{2}, \frac{1}{3}, \frac{2}{5}, \frac{2}{3}, \frac{3}{4}, \frac{3}{6}, \frac{4}{6},$ $\frac{4}{8}, \frac{3}{9}, \frac{4}{10}, \frac{4}{12}$,
$\frac{8}{12}, \frac{9}{12}, \frac{5}{15}, \frac{6}{15}, \frac{6}{18}, \frac{9}{18}, \frac{8}{20}, \frac{10}{20},$ $\frac{15}{20}, \frac{7}{21}$

2. Reduce each of the following fractions to its simplest form:
$\frac{7}{14}, \frac{26}{91}, \frac{21}{91}, \frac{52}{78}, \frac{39}{65}, \frac{22}{30},$ $\frac{39}{51}, \frac{52}{64}, \frac{34}{51}, \frac{27}{81}, \frac{18}{45}, \frac{57}{69}$

3. Without actually performing the divisions, indicate which of the following numbers divide exactly by 3,4, 8, 9, or 11:
(a) 10,452     (b) 2,088     (c) 5,841     (d) 41,613
(e) 64,572     (f) 37,848

4. Find the factors of the following:
(a) 1,829     (b) 1,517     (c) 7,387
(d) 7,031     (e) 2,059     (f) 2,491

5. Add together the following groups of fractions:
(a) $\frac{1}{5} + \frac{1}{6} + \frac{4}{15} + \frac{3}{10} + \frac{2}{3}$
(b) $\frac{1}{8} + \frac{1}{3} + \frac{5}{18} + \frac{7}{12} + \frac{4}{9}$
(c) $\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{10} + \frac{1}{12}$
(d) $\frac{4}{7} + \frac{3}{4} + \frac{7}{12} + \frac{8}{21} + \frac{5}{6}$
and reduce each to its simplest form.

6. Find the simplest fractional equivalent for the following decimals:
(a) 0.875     (b) 0.6     (c) 0.5625    (d) 0.741     (e) 0.128

7. Find the decimal equivalent of the following fractions:
(a) $\frac{2}{3}$     (b) $\frac{3}{4}$     (c) $\frac{4}{5}$     (d) $\frac{5}{6}$
(e) $\frac{6}{7}$     (f) $\frac{7}{8}$     (g) $\frac{8}{9}$

8. Find the decimal equivalent of the following fractions:
(a) $\frac{1}{3}$     (b) $\frac{1}{4}$     (c) $\frac{1}{5}$     (d) $\frac{1}{6}$
(e) $\frac{1}{7}$     (f) $\frac{1}{8}$     (g) $\frac{1}{9}$

9. Find the fraction equivalent to the following recurring decimals:
(a) 0.416     (b) 0.21     (c) 0.189
(e) 0.571428     (e) 0.909     (f) 0.090

10. What is meant by significant figures? To illustrate, show the limits of possible meaning for measurements given as 158 feet and 857 feet.

11. Using the approximate method, divide 932 by 173. Then by dividing (a) 932.5 by 172.5 and (b) 931.5 by 173.5, show how many of your figures are justified. Noting that 932 and 173 have three significant figures, what conclusion would you draw from these calculations?

12. Divide 93,700 by 857 using an approximate method. Then by dividing 93,750 by 856.5 and 93,650 by 857.5, show how many of your figures are justified. Can you shorten your method still further to avoid writing down meaningless figures?

13. (a) List all of the prime numbers less than 60. (b) Using this list, how can you test a given number to determine whether or not it is prime? (c) What is the largest number you can test in this way, using this list?

14. Find the differences of the following pairs of fractions. Reduce the result to standard form with the lowest possible denominator.
(a) $\frac{3}{4}-\frac{1}{16}$     (b) $\frac{11}{13}-\frac{1}{7}$     (c) $\frac{16}{20}-\frac{3}{8}$
(d) $\frac{255}{100} - \frac{1}{10}$     (e) $\frac{23}{17} - \frac{1}{34}$

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