| 5 | Fractions |
Chapter 4 presented some concepts involving fractions that you might not yet fully understand. In this chapter, you’ll “catch up” by learning some important techniques for dealing with fractions.
SLICING UP A PIE
If you imagine fractions as slices of a pie, it can help you see how fractions work. Notice that the fraction 1/4 can be cut into smaller pieces without changing its value as part of the whole, as shown in Fig. 5-1. You can multiply 1/4 by various fractions that are always equal to 1, such as 2/2, 3/3, or 4/4, like this:
Figure 5-1
The fraction 1/4 can also be represented as 2/8, 3/12, 4/16, or any other expression of whole numbers where the denominator is exactly 4 times as big as the numerator.
When the dividend and the divisor (or numerator and denominator) are identical, the value of the quotient or fraction is equal to 1.
Whenever we multiply or divide a fraction by some other fraction that’s really equal to 1, we have the same number in a different form. Imagine, for example, that we see the fraction 20/80. We can divide this by 20/20 and get 1/4, like this:
We divide the numerator by 20 to get 1, and the denominator by 20 to get 4. Presto! We have just done a trick known as reducing or simplifying a fraction. Here, 1/4 is the lowest form or simplest form of this particular number, because we can’t find smaller whole-number numerators and denominators for it.
FACTORS HELP SIMPLIFY FRACTIONS
A fraction can sometimes contain huge numbers for both the numerator and denominator. When you see a fraction like that, you should wonder whether the fraction can be reduced to a simpler form so that the numerator and denominator are smaller but the quotient is still the same. A calculator that handles fractions will automatically find and present them in their simplest forms. If your calculator cannot handle fractions and you want to reduce a fraction to its simplest form, you can do it “by hand.”
If the numerator and denominator both end in zeros, you can strike off the same number of zeros from each. If both are even, you can divide each number by 2. (We did both of those things when we reduced 20/80 to 1/4 a moment ago.) These rules are obvious, and it’s easy to spot situations that call for them. But often the factors are harder to find. For example, in the fraction 455/462, both the numerator and the denominator can be divided by 7 and the results are whole numbers again. We can reduce 455/462 down to 65/66 that way. It isn’t immediately apparent to most people that both 455 and 462 are divisible by 7. How can we find factors in situations like this?
SPOTTING THE FACTORS
Here are a few tricks that can help you spot factors. If the last digit of a number divides by 2 without a remainder, then the whole number does. If the number consisting of the last 2 digits divides by 4 without a remainder, then the big number does. If the number consisting of the last 3 digits divides by 8 without a remainder, then the big number does.
This line of reasoning leads to a set of checks for powers of 2. A power is defined as a certain number of times that a number is multiplied by itself. For example, 2 to the second power is 4; this is written 22 = 4. The number 2 to the third power is 8; this is written 23 = 8. Continuing, we have 24 = 16, 25 = 32, 26 = 64, 27 = 128, 28 = 256, and so on.
A similar set of facts works for powers of 5. Starting with the first power and working up, we have 51 = 5, 52 = 25, 53 = 125, 54 = 625, 55 = 3,125, and so on.
Rules also exist for 3s and 9s. Add all the digits of a number together. If this sum divides by 3 without a remainder, then the whole number does. If the sum of digits divides by 9 without a remainder, then 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 any whole-number multiple of 11, then the original number divides by 11 without a remainder.
There is no easy rule that works if you want to test a number to see if it will divide by 7 without a remainder. But there’s another way to find factors when none of the above rules work and you seem to be “stuck.”
PRIME NUMBERS
A prime number is a whole number larger than 1 that can only be factored into a product of 1 and itself. (For some reason, most mathematicians do not consider 1 a prime number, even though it can be “factored” into a product of 1 and itself.) The first few prime numbers, often called simply primes, are 2, 3, 5, 7, and 11. Table 5-1 lists the first 24 prime numbers.
Prime numbers have a property that makes them ideal for finding factors. This property is so important in mathematics that it deserves a “bullet”!
• Any nonprime whole number can be factored into a product of primes. The numbers in this product are called the prime factors.
Table 5-1
The first 24 prime numbers. By convention, the number 1 is not considered prime.
FACTORING INTO PRIMES
When you want to find the prime factors of a large number, first use a calculator to find the square root of the number, and then divide the original number by all the primes less than or equal to that square root. If you ever get a whole-number quotient as you grind your way through this process, then you know that the divisor and the quotient are both factors of the original number. Sometimes the quotient is prime, and sometimes it isn’t. If it isn’t, then it can be factored down further into primes. Don’t stop dividing the original by the primes until you get all the way down to 2.
Usually, the square root of a number is not a whole number. Don’t worry about that when you’re trying to find its prime factors. Just round the square root up to the next whole number, and then look up all the primes less than or equal to that. If Table 5-1 doesn’t go far enough, you can find plenty of lists of primes on the Internet, some of which go to extremes far beyond anything you will ever need!
PRIME FACTORS OF 139
Suppose you want to find the prime factors of 139. When you find the square root of 139 using a calculator, you get a number between 11 and 12. Round it up to 12. Now find all the primes less than or equal to 12. From Table 5-1, you can see that these are 2, 3, 5, 7, and 11. Divide 139 by each of these and look for whole-number quotients:
Here, “xxx” means a string of numbers that has a recurring pattern too long to write down easily. (It doesn’t matter what the pattern is anyway, if the number is not whole.) You know from this that the only factors of 139 are 1 and itself. By definition, that means 139 is a prime number.
PRIME FACTORS OF 493
Now imagine that you are told to find the prime factors of 493. Using a calculator to find the square root and then rounding up, you get 23. From Table 5-1, you can see that the primes less than or equal to 23 are 2, 3, 5, 7, 11, 13, 17, 19, and 23. Now divide 493 by each of these and look for whole-number quotients:
You know from this that the prime factors of 493 are 17 and 29, because 29 appears in Table 5-1. You can’t factor 493 down any further than that.
PRIME FACTORS OF 546
Now suppose you want to factor 546 into primes. Using a calculator to find the square root and then rounding up, you get 24. Again, from Table 5-1, the primes less than or equal to 24 are 2, 3, 5, 7, 11, 13, 17, 19, and 23. So you divide by them all:
Now you know that 546 has prime factors of 2, 3, 7, and 13. But is that all? There’s a way to test and see. Using your calculator, you hit 2, then × (the “times” key), then 3, then ×, then 7, then ×, then 13, and finally = (the “equals” key). The result, if you hit all the right buttons and none of the wrong ones, is 546. So you know that the prime factors of 546 are 2, 3, 7, and 13.
The business of prime numbers and prime factorization can get incredibly complicated, and this chapter could go on about it for a long time. But that would be a diversion. You should now have the general idea of how prime factorization works, and that’s all you need for most practical problems.
ADDING AND SUBTRACTING FRACTIONS
When you want to add or subtract fractions “by hand,” they must all have the same denominator. For instance, to add 1/2, 2/3, and 5/12, both 1/2 and 1/3 can be changed to 12ths. You can figure out that 1/2 = 6/12 and 2/3 = 8/12. Once you have found a common denominator in this way, you can add the numerators straightaway because the denominators are all 12ths: 6 + 8 + 5 = 19. Then you get 19/12, which is more than 1. Subtract 12/12 (which is equal to 1) from this and you get 1-7/12 as the final answer in its proper form.
Now suppose you want to subtract 3-3/5 from 7-5/12. You can always find a common denominator in problems like this if you multiply the two denominators together, although the result might not be the smallest common denominator. In this case, multiplying the denominators gives you 5 × 12 = 60. Now proceed like this:
and
Note that when you multiply two fractions, the product of the whole fraction is equal to the product of the numerators divided by the product of the denominators. The subtraction problem now looks like this:
Change the whole-number parts of these numbers into fractions to get
and
The numbers in the original problem can be found by addition:
and
Now subtract the larger of these two numbers from the smaller:
In the “olden days” this sort of expression was called an improper fraction because the numerator is larger than the denominator, and it “doesn’t look right.” There’s nothing technically wrong with it. But to keep to “proper” form, you can divide out 60 from 229 three times and have a remainder of 49. That means the above improper fraction is equal to 3-49/60. You can’t reduce 49/60 to a lower form (try it and see!), so this is the final answer in its simplest form. Stated from the first to the last, the “subtraction fact” looks like this:
Again, don’t be confused by the little dashes that separate the whole-number parts of each value from the fractional parts. They aren’t minus signs! The long dash, with a space on either side of it, is the minus sign here.
FINDING THE COMMON DENOMINATOR
How do you find a reasonable common denominator when you have a long sum of fractions? Sometimes it’s easy, but sometimes it’s quite a hard trick. You can multiply all the denominators together and get a number that will work, but it will probably be a lot larger than what you need.
Older textbooks had a routine for the job, but it was difficult to follow and it confused a lot of folks. Here’s a way you can understand. Figure 5-2 illustrates an example. Suppose you have this addition problem involving fractions:
Figure 5-2
Here’s a way to find a reasonable common denominator in a long sum of fractions.
To begin, find the prime factors of each denominator in order:
Our “ideal common denominator” must contain every prime factor that is in any denominator. These are 2, 3, and 5. The common denominator is then found by multiplying all these primes together: 2 × 2 × 3 × 5 = 60. Notice that the prime number 2 is used twice here. That sort of thing is not unusual when you are finding the prime factorizations of numbers.
Now convert all the fractions to 60ths:
These fractions add up to 142/60. You can reduce this to 71/30. That’s an improper fraction, so you divide out 30 twice and get a remainder of 11. Therefore, the final answer is 2-11/30. Stated fully, here’s the fact:
SIGNIFICANT FIGURES
Devices that give you numerical information can be either analog or digital. An example of an analog device is an old-fashioned spring-loaded scale. A grandfather clock, with its hour and minute hands, is another example of an analog display. Calculators have digital displays, as do numeric-readout clocks, numeric-readout thermometers, and automotive odometers. Some people think that digital displays are more accurate than analog ones. But that isn’t always true.
Suppose a l50-pound man steps on a scale to weigh himself (Fig. 5-3A). Does he weigh exactly 150 pounds, and 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 which figures in that number 150 are “significant.” It also depends on what the man is wearing and how well-made the scale happens to be.
Figure 5-3
At A, a man weighs himself and sees that he weighs 150 pounds, more or less. How much more or less? At B, an envelope weighs 16 ounces, more or less. How much more or less? Accuracy is relative!
If the 0 were 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 were “significant,” then 150 would mean that he weighed 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. You might even specify the weight to greater accuracy, such as 150.00 pounds or even 150.000 pounds.
The same sort of situation occurs when you weigh an envelope (Fig. 5-3B) on an analog postal scale. It looks like this envelope weighs just about 16 ounces, or maybe a little less. But what if the scale is not a good one, and the envelope actually weighs a little more than 16 ounces? How much “leeway” should you allow? (And, you might wonder, what does this envelope contain that makes it so heavy for its size?)
APPROXIMATE DIVISION AND MULTIPLICATION
Suppose you divide an “approximate” 150 by an “exact” 7. If your calculator can display 10 digits, it gives you 21.42857142 (if it truncates, or simply cuts off, the digits after it runs out of space) or 21.42857143 (if it rounds off to the nearest significant digit).You are tempted to believe all those figures. But can you?
Apply what was just said about significant figures. If 150 means more than 145 and less than 155, that means only two figures are significant (the 1 and the 5). In that case, dividing by exactly 7 could yield anything between 145/7 and 155/7. Check these out on your calculator. Now, if the 0 of 150 is significant, the result can be anything between 149.5 / 7 and 150.5 / 7. Check out this range of values on your calculator. It’s a lot narrower. Now what if you had 150.000 as the specified number? That means a range of 149.995 to 150.005. When you divide these numbers by exactly 7, you get a range of 149.995 / 7 to 150.005 / 7. Check this out on your 10-digit calculator. Or better yet, use a computer calculator that can show a lot more digits.
In these division problems, an “exact 7” means a theoretically perfect number 7, which you can call 7.0, or 7.00, or 7.000, or 7.0000—with as many zeros after the decimal point as you want. In this context, 7 is a mathematical constant. But the quantity 150 (more or less), as used here, refers to an observed or measured thing. It can never be exact, because no machine or device is perfect, and if the device is analog, there can be human error in the reading, too. But if you want to find exactly 1/7 of some figure, you can divide by 7 and consider that value to have no margin of error at all.
Suppose that you want to divide an “approximate” 23,500 by an “exact” constant of 291, and you believe that the two end zeros in the dividend aren’t significant. You could assume it is between 23,450 and 23,550, perform both divisions, and then decide what is significant. But longhand, that’s a lot of work! In the olden days, the practice was to draw a vertical line where figures began to be doubtful. You could divide between the “limiting values” as the possible errors, because only so many figures are significant, and then you could guess at the most reasonable value. An example of this procedure is illustrated in Fig. 5-4.
Figure 5-4
Division of 23,500 by 291. At A, conventional long division; the digits get more and more doubtful as we churn them out. At B, an approximate method of long division. At C, a way to find the limits of accuracy.
If you are willing to play around with a calculator, it can show you how to do things that aren’t practical by using longhand methods. Take values that represent the greatest possible variation on either side of the stated value, and then deduce how accurate the answers will be by dividing them out with the calculator. Both the dividend and the divisor can have a certain number of significant figures unless one of them is known to be mathematically exact, as 7 and 291 are in the examples above.
The approximation methods you use for long division will also work for long multiplication. Figure 5-5 shows this process for the product of 2.91 and 5.32. Here, both values are expressed to three significant figures. So that’s as far as we can reasonably go with the product.
Figure 5-5
An example of approximation in long multiplication. As in division, the figures become more and more doubtful as we churn them out.
APPROXIMATE ADDITION AND SUBTRACTION
Suppose you add “exactly 55” to “roughly a million.” What does “roughly a million” mean here? It could be greater or less than an exact million (1,000,000) by 500,000, or 50,000, or 5,000, or 500. If you say “pretty close to a million,” it could be greater or less than 1,000,000 by 50, or by 5, or maybe even by 0.5. It depends on how many of the zeros after the 1 you take seriously! Mathematicians and engineers call such give-or-take by a technical name, plus or minus, symbolized as a plus sign with a minus sign underneath (±). Adding 55 doesn’t make a practical difference when we think of an approximate million that could vary by 50,000 either way, that is, 1,000,000 ± 50,000. But it makes a big difference if we take all six of the zeros seriously, in which case we mean a rather precise million—a number between 999,999.5 and 1,000,000.5!
In this chapter, you’ve had a taste of how significant figures work in practical mathematics. We’ll look more closely at this subject, and how significant figures are dealt with by scientists and engineers, in Chap. 25. For now, here is the fundamental rule for determining significant figures in any practical problem that involves quantities subject to error. It deserves another “bullet”!
• An answer cannot have more significant figures than the number with the least number of significant figures used as “input” for the problem.
QUESTIONS AND PROBLEMS
This is an open-book quiz. You may refer to the text in this chapter (and earlier ones, too, if you want) when figuring out the answers. Take your time! The correct answers are in the back of the book.
1. Arrange the following fractions into groups that have the same value.
| 1/2 | 1/3 | 2/5 | 2/3 | 3/4 | 3/6 | 4/6 |
| 4/8 | 3/9 | 4/10 | 4/12 | 8/12 | 9/12 | 5/15 |
| 6/15 | 6/18 | 9/18 | 8/20 | 10/20 | 15/20 | 7/21 |
2. Reduce each of the following fractions to its simplest form.
| 7/14 | 26/91 | 21/91 | 52/78 | 39/65 | 22/30 |
| 39/51 | 52/64 | 34/51 | 27/81 | 18/45 | 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 prime factors of the following numbers.
(a) 1,829
(b) 1,517
(c) 7,387
(d) 7,031
(e) 2,059
(f) 2,491
5. Add the following groups of fractions and reduce each answer to its simplest form.
(a) 1/5 + 1/6 + 4/15 + 3/10 + 2/3
(b) 1/8 + 1/3 + 5/18 + 7/12 + 4/9
(c) 1/4 + 1/5 + 1/6 + 1/10 + 1/12
(d) 4/7 + 3/4 + 7/12 + 8/21 + 5/6
6. Find the simplest fractional equivalents of the following decimals.
(a) 0.875
(b) 0.6
(c) 0.5625
(d) 0.741
(e) 0.128
7. Find the decimal equivalents of the following fractions.
(a) 2/3
(b) 3/4
(c) 4/5
(d) 5/6
(e) 6/7
(f) 7/8
(g) 8/9
8. Find the decimal equivalents of the following fractions.
(a) 1/3
(b) 1/4
(c) 1/5
(d) 1/6
(e) 1/7
(f) 1/8
(g) 1/9
9. Find the fractional equivalents of the following recurring decimals.
(a) 0.416416416 …
(b) 0.212121 …
(c) 0.189189189 …
(d) 0.571428571428571428 …
(e) 0.909909909 …
(f) 0.090090090 …
10. To define “significant figures,” 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 932.5 by 172.5 and 931.5 by 173.5, show how many of your figures are justified. Noting that 932 and 173 have three significant figures, what conclusion can you draw?
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 the prime numbers less than 60. (b) If you use 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 results to standard form with the smallest possible denominators.
(a) 3/4 − 1/16
(b) 11/13 − 1/7
(c) 16/20 −3/8
(d) 255/100 − 1/10
(e) 23/17 − 1/34
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