0.3 Fractions
SLO 6.NS
The Number System. Divide fractions by fractions; compute fluently with
SLO 6.EE
Expressions and Equations. Write, read and evaluate expressions in which
SLO 7.NS
The Number System. Add, subtract, multiply and divide rational numbers,
SLO 7.EE
Expressions and Equations. Apply properties of operations to add, subtract,
SLO 8.NS
The Number System. Know that numbers which are not rational are irrational;
Learning Objectives
By the end of this section, you will be able to:
- explain what a fraction means, both as a part of a whole and as a division;
- build equivalent fractions and simplify a fraction to lowest terms;
- convert between improper fractions and mixed numbers, and place the sign of a negative fraction correctly;
- multiply and divide fractions, and explain why dividing by a fraction means multiplying by its reciprocal;
- add and subtract fractions with like and unlike denominators using the least common denominator;
- simplify complex fractions and apply the order of operations to expressions with fractions in them.
The integers from Section 0.2 count things that come in whole pieces. Most of what you measure does not. Two and a half hours of practice, three quarters of a tank of gas, a recipe scaled down to five sixths — none of those can be written with an integer, and rounding them off throws away information you may need later.
Fractions handle those in-between amounts exactly. That word matters. Section 0.4 covers decimals, which describe the same quantities in a different notation but sometimes only approximately: one third is \(0.333\dots\) forever, so any decimal you actually write down is a little bit wrong. A fraction is never a little bit wrong. That is why algebra keeps answers in fraction form and converts to decimals at the very end, if at all.
Here is the thing to watch for as you read. Most fraction mistakes are not arithmetic mistakes. They happen when a rule from one operation leaks into a different operation. Multiplication really does go straight across; addition does not, and never has. Cancelling works on factors; it does not work on terms. This section is careful to tell you not just what each rule says, but exactly where it stops.
0.3.1 Visualizing and Simplifying Fractions
Cut a pan of cornbread into 8 equal pieces and take 3 of them. You have taken \(\dfrac{3}{8}\) of the pan. The bottom number says how many equal pieces the whole was cut into, and the top number says how many of those pieces you are talking about.
A fraction is a number written \(\dfrac{a}{b}\), where \(b \neq 0\). The denominator \(b\) is the number of equal parts the whole has been divided into, and the numerator \(a\) is how many of those parts are being counted.
The word equal is doing real work in that definition. If you hack the pan into 8 pieces of wildly different sizes and grab 3, you do not have \(\dfrac{3}{8}\) of the pan. Every part-of-a-whole reading of a fraction assumes the pieces are the same size.
Two conditions in the definition are worth pulling out on their own, because both come back later. First, the denominator can never be zero: \(\dfrac{5}{0}\) asks you to cut a whole into zero equal parts, which describes nothing at all. That is the same undefined division you met in Section 0.2, and spotting when a variable denominator could turn into zero is a job you will do over and over in Chapter 3.
Second, a fraction is also a division. The expression \(\dfrac{a}{b}\) means \(a \div b\). That is not a separate meaning bolted on; it is the same idea coming at you from the other direction. Split 3 pans of cornbread evenly among 8 people and each person gets \(3 \div 8 = \dfrac{3}{8}\) of a pan, which is exactly what the picture already showed.
| Fraction | As parts of a whole | As a division | Value |
|---|---|---|---|
| \(\dfrac{3}{8}\) | 3 of 8 equal pieces | \(3 \div 8\) | \(0.375\) |
| \(\dfrac{6}{6}\) | all 6 of 6 equal pieces | \(6 \div 6\) | \(1\) |
| \(\dfrac{0}{9}\) | none of 9 equal pieces | \(0 \div 9\) | \(0\) |
| \(\dfrac{11}{4}\) | 11 quarter-pieces, more than one whole | \(11 \div 4\) | \(2.75\) |
| \(\dfrac{7}{0}\) | 0 equal pieces, which is meaningless | \(7 \div 0\) | undefined |
The second row generalizes into a rule you will use constantly. Any nonzero number divided by itself is 1, so a fraction whose numerator and denominator match is one whole:
$$\frac{a}{a} = 1 \qquad (a \neq 0)$$That small fact is the engine behind everything else in this subsection.
Definition 0.3.1 — A fraction: the denominator names the equal pieces, the numerator counts them.
Equivalent fractions
Now slice each of those 8 cornbread pieces in half. There are 16 pieces on the table, and the 3 you took have become 6. Nothing about how much cornbread you have changed — only how finely it was cut. So
$$\frac{3}{8} = \frac{6}{16}$$Fractions that name the same amount are called equivalent, and every fraction has infinitely many equivalent forms. You generate them by cutting the pieces finer or by grouping them coarser, and the rule below does both.
If \(a\), \(b\), and \(c\) are numbers with \(b \neq 0\) and \(c \neq 0\), then
$$\frac{a}{b} = \frac{a \cdot c}{b \cdot c} \qquad \text{and} \qquad \frac{a}{b} = \frac{a \div c}{b \div c}$$Cutting a pizza into 16 slices instead of 8 does not give you more pizza. It gives you more pieces, each one half as big. Equivalent fractions are the same amount described with a different piece size.
Read left to right, that property builds a fraction up to a bigger denominator. Read right to left, it reduces one. Both directions are allowed for the same reason: multiplying the numerator and denominator by the same \(c\) is really multiplying by \(\dfrac{c}{c}\), which is 1, and multiplying by 1 cannot change what a number is worth.
| Start with | Multiply top and bottom by | Equivalent fraction |
|---|---|---|
| \(\dfrac{2}{5}\) | 3 | \(\dfrac{6}{15}\) |
| \(\dfrac{2}{5}\) | 4 | \(\dfrac{8}{20}\) |
| \(\dfrac{2}{5}\) | 9 | \(\dfrac{18}{45}\) |
| \(\dfrac{7}{6}\) | 5 | \(\dfrac{35}{30}\) |
| \(\dfrac{1}{4}\) | 12 | \(\dfrac{12}{48}\) |
The requirement that \(c \neq 0\) is not fussiness. Multiplying top and bottom by 0 would turn every fraction into \(\dfrac{0}{0}\), which is undefined and tells you nothing about what you started with.
Definition 0.3.2 — Equivalent Fractions Property: finer pieces, the same amount.
Simplifying to lowest terms
A fraction is in lowest terms when its numerator and denominator share no common factor other than 1.
To simplify a fraction, divide the numerator and the denominator by their greatest common factor. If the greatest common factor does not jump out at you, divide by any common factor you do see and repeat — you land in the same place, it just takes more steps.
Definition 0.3.3 — Lowest terms: the point where no common factor is left to group by.
Simplify \(\dfrac{42}{70}\).
Solution
Step 1 — Look for common factors. Both 42 and 70 are even, so 2 divides both. Both are also divisible by 7. Putting those together, 14 divides both, and nothing larger does.
Step 2 — Divide the top and the bottom by 14.
$$\frac{42}{70} = \frac{42 \div 14}{70 \div 14} = \frac{3}{5}$$Step 3 — Check that you are done. The numbers 3 and 5 share no factor but 1, so the fraction is in lowest terms.
Answer: \(\dfrac{3}{5}\).
| Fraction | Greatest common factor | Lowest terms |
|---|---|---|
| \(\dfrac{18}{24}\) | 6 | \(\dfrac{3}{4}\) |
| \(\dfrac{42}{70}\) | 14 | \(\dfrac{3}{5}\) |
| \(\dfrac{27}{45}\) | 9 | \(\dfrac{3}{5}\) |
| \(\dfrac{16}{40}\) | 8 | \(\dfrac{2}{5}\) |
| \(\dfrac{30}{6}\) | 6 | \(\dfrac{5}{1} = 5\) |
| \(\dfrac{13}{20}\) | 1 | already lowest |
Two of those rows are worth a comment. The last one shows that simplifying does not always change the fraction: when the greatest common factor is 1, you are already finished. The row above it shows a fraction simplifying all the way down to a whole number, which happens whenever the denominator divides the numerator evenly.
Cancel factors, not terms
\(\dfrac{3 \cdot 5}{3 \cdot 7}\) really is \(\dfrac{5}{7}\), but \(\dfrac{3 + 5}{3 + 7}\) is not. In the first the 3 is being multiplied, so it is a factor. In the second it is being added, so it is a term. Only factors cancel.
This is the most common fraction error in all of algebra, so it is worth stating exactly. Simplifying divides the whole numerator and the whole denominator by a common factor. A factor is something being multiplied. If a piece of the numerator is being added instead, it is a term, and it cannot be cancelled.
Check the bad one with actual numbers: \(\dfrac{3+5}{3+7} = \dfrac{8}{10} = \dfrac{4}{5}\), which is nowhere near \(\dfrac{5}{7}\). The two 3s look identical on the page, but in one expression the 3 is a factor and in the other it is a term.
| Expression | Can the 4s cancel? | Why | Simplified |
|---|---|---|---|
| \(\dfrac{4 \cdot 9}{4 \cdot 11}\) | yes | 4 is a factor of both | \(\dfrac{9}{11}\) |
| \(\dfrac{4 + 9}{4 + 11}\) | no | 4 is a term, not a factor | \(\dfrac{13}{15}\) |
| \(\dfrac{4x}{4y}\) | yes | 4 multiplies each | \(\dfrac{x}{y}\) |
| \(\dfrac{4 + x}{4}\) | no | the numerator is a sum | \(\dfrac{4+x}{4}\) |
Here is the habit that protects you. Before cancelling anything, ask whether the numerator and the denominator are each written as a single product. If either one has a \(+\) or a \(-\) sitting at the top level, nothing inside it can be cancelled until you factor it first, which is a skill Chapter 8 builds properly.
Improper fractions and mixed numbers
A proper fraction has a numerator smaller than its denominator, so its value is less than 1. An improper fraction has a numerator greater than or equal to its denominator, so its value is 1 or more.
Definition 0.3.4 — The one-whole mark is what sorts a proper fraction from an improper one.
A mixed number is a whole number written beside a proper fraction, and it means their sum: \(2\frac{3}{4}\) means \(2 + \frac{3}{4}\).
Each whole holds 4 quarters, so 5 wholes hold 20 of them. That is all the "multiply by the denominator" step is doing — counting the quarter-pieces buried inside the whole numbers before adding the 3 loose ones.
Neither form is more correct than the other, and they are useful in different places. Mixed numbers read the way people talk, which is what you would say at a lumber yard. Improper fractions are much easier to compute with, which is why algebra almost always converts to them first.
To go from improper to mixed, divide the numerator by the denominator. The quotient is the whole-number part. The remainder is the new numerator, and the denominator stays put. To go the other way, multiply the whole number by the denominator, then add the numerator. The denominator does not change.
That second procedure is not a recipe you have to take on faith. Since \(5\frac{3}{4}\) means \(5 + \frac{3}{4}\), and \(5 = \frac{20}{4}\), the sum is \(\frac{20}{4} + \frac{3}{4} = \frac{23}{4}\).
Definition 0.3.5 — A mixed number is the same pieces gathered into wholes plus a remainder.
Write \(\dfrac{23}{4}\) as a mixed number, then convert your answer back to an improper fraction.
Solution
Step 1 — Divide the numerator by the denominator. How many times does 4 go into 23?
$$23 \div 4 = 5 \text{ with a remainder of } 3$$Step 2 — Assemble the mixed number. The quotient 5 is the whole-number part, the remainder 3 is the new numerator, and the denominator stays 4.
$$\frac{23}{4} = 5\frac{3}{4}$$Step 3 — Convert back. Multiply the whole number by the denominator, then add the numerator.
$$5 \cdot 4 + 3 = 23$$Keeping the same denominator gives \(\dfrac{23}{4}\), which is what we started with.
Answer: \(5\frac{3}{4}\), and converting back gives \(\dfrac{23}{4}\).
| Improper fraction | The division | Mixed number |
|---|---|---|
| \(\dfrac{23}{4}\) | \(23 \div 4 = 5\) remainder \(3\) | \(5\dfrac{3}{4}\) |
| \(\dfrac{17}{5}\) | \(17 \div 5 = 3\) remainder \(2\) | \(3\dfrac{2}{5}\) |
| \(\dfrac{31}{8}\) | \(31 \div 8 = 3\) remainder \(7\) | \(3\dfrac{7}{8}\) |
| \(\dfrac{9}{9}\) | \(9 \div 9 = 1\) remainder \(0\) | \(1\) |
| \(\dfrac{40}{6}\) | simplify to \(\dfrac{20}{3}\), then \(20 \div 3 = 6\) remainder \(2\) | \(6\dfrac{2}{3}\) |
One warning about mixed-number notation. Writing a whole number next to a fraction means addition, but writing a number next to a variable means multiplication, as Section 0.1 established. So \(2\frac{3}{4}\) is \(2 + \frac{3}{4}\), while \(2x\) is \(2 \cdot x\). The two notations look parallel and mean opposite things, which is one more reason algebra prefers improper fractions — \(\frac{11}{4}\) carries no such ambiguity.
Negative fractions and where the sign goes
A fraction can be negative, and the sign can be written in three different places. All three mean the same number.
$$-\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b} \qquad (b \neq 0)$$The sign rules from Section 0.2 are what make that true. A fraction is a division, and a division with exactly one negative sign among its parts comes out negative. It makes no difference whether that sign is sitting on the numerator, on the denominator, or out in front.
| Where the sign is | Example | Value |
|---|---|---|
| Out in front | \(-\dfrac{3}{7}\) | negative |
| On the numerator | \(\dfrac{-3}{7}\) | negative |
| On the denominator | \(\dfrac{3}{-7}\) | negative |
| On both | \(\dfrac{-3}{-7} = \dfrac{3}{7}\) | positive |
That last row is the like-signs-give-a-positive rule from Section 0.2, and it is the fraction version of \(-(-6) = 6\).
\(-\dfrac{2}{3}\), \(\dfrac{-2}{3}\), and \(\dfrac{2}{-3}\) are all the same number, but only the first is easy to keep track of. A minus sign buried in a denominator gets lost partway through a long problem, and once it is lost you will not find it by rereading.
Make moving the sign to the front an automatic habit. When you simplify \(\dfrac{-24}{36}\), for instance, the greatest common factor of 24 and 36 is 12, so the fraction reduces to \(\dfrac{-2}{3}\), and you write that as \(-\dfrac{2}{3}\).
Simplify \(\dfrac{54}{72}\). Then write \(\dfrac{19}{5}\) as a mixed number, and rewrite \(\dfrac{6}{-13}\) with the sign in front.
Solution
Part 1 — Simplify \(\dfrac{54}{72}\). Both numbers are divisible by 18, and nothing larger divides both.
$$\frac{54}{72} = \frac{54 \div 18}{72 \div 18} = \frac{3}{4}$$Part 2 — Write \(\dfrac{19}{5}\) as a mixed number. Divide 19 by 5.
$$19 \div 5 = 3 \text{ with a remainder of } 4$$The quotient is the whole number, the remainder is the new numerator, and the denominator stays 5.
$$\frac{19}{5} = 3\frac{4}{5}$$Part 3 — Move the sign on \(\dfrac{6}{-13}\). One negative sign among the parts makes the fraction negative, and the standard place to write it is out front.
$$\frac{6}{-13} = -\frac{6}{13}$$Answer: \(\dfrac{3}{4}\); \(3\dfrac{4}{5}\); \(-\dfrac{6}{13}\).
0.3.2 Multiplying and Dividing Fractions
If \(a\), \(b\), \(c\), and \(d\) are numbers with \(b \neq 0\) and \(d \neq 0\), then
$$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$$Multiply the numerators, multiply the denominators, then simplify. That really is the whole rule:
$$\frac{2}{5} \cdot \frac{3}{7} = \frac{2 \cdot 3}{5 \cdot 7} = \frac{6}{35}$$Here is the picture behind it. You have \(\frac{3}{7}\) of a pan of cornbread, and you want \(\frac{2}{5}\) of that. Cutting the pan into 7 strips and then cutting each strip into 5 pieces makes \(5 \cdot 7 = 35\) small pieces in the whole pan, and you are keeping \(2 \cdot 3 = 6\) of them. The denominators multiply because the two rounds of cutting compound, and the numerators multiply for the same reason. Notice also that the word "of" is signaling multiplication here, exactly as the translation table in Section 0.1 promised.
Signs work the way Section 0.2 said: like signs give a positive product, unlike signs give a negative one. A whole number joins in by writing it over 1, since \(n = \dfrac{n}{1}\).
| Product | Straight across | Simplified |
|---|---|---|
| \(\dfrac{2}{5} \cdot \dfrac{3}{7}\) | \(\dfrac{6}{35}\) | \(\dfrac{6}{35}\) |
| \(-\dfrac{3}{4} \cdot \dfrac{8}{9}\) | \(-\dfrac{24}{36}\) | \(-\dfrac{2}{3}\) |
| \(\dfrac{5}{6} \cdot 12\) | \(\dfrac{60}{6}\) | \(10\) |
| \(-\dfrac{2}{3} \cdot \left(-\dfrac{9}{10}\right)\) | \(\dfrac{18}{30}\) | \(\dfrac{3}{5}\) |
Definition 0.3.6 — Two rounds of cutting compound, so both numerators and both denominators multiply.
Simplify before you multiply
Every product in that table needed simplifying afterward, and the numbers you had to simplify were bigger than the ones you started with. There is a better order to work in: cancel the common factors first, then multiply.
That is allowed because the Equivalent Fractions Property does not care which numerator sits above which denominator. Once the product is written as one fraction, \(\dfrac{a \cdot c}{b \cdot d}\), any factor on top can be cancelled against any factor on the bottom.
Two ways through the same product. Both columns compute \(\dfrac{14}{15} \cdot \dfrac{25}{21}\):
| Multiply first | Simplify first |
|---|---|
| Step 1. \(\dfrac{14 \cdot 25}{15 \cdot 21} = \dfrac{350}{315}\) | Step 1. 14 and 21 share 7, becoming 2 and 3. |
| Step 2. Divide by 5 to get \(\dfrac{70}{63}\). | Step 2. 25 and 15 share 5, becoming 5 and 3. |
| Step 3. Divide by 7 to get \(\dfrac{10}{9}\). | Step 3. \(\dfrac{2}{3} \cdot \dfrac{5}{3} = \dfrac{10}{9}\) |
| Biggest number handled: 350. | Biggest number handled: 25. |
Same answer, but the largest product the right-hand column ever formed was \(2 \cdot 5 = 10\), against \(14 \cdot 25 = 350\) on the left. The bigger the numbers get, the more that gap matters. It matters most once the fractions contain variables: a variable factor cancels cleanly, while a variable product has to be expanded and then factored all over again.
Multiply \(\dfrac{14}{15} \cdot \dfrac{25}{21}\) by simplifying first.
Solution
Step 1 — Write it as one fraction. Every number here is a factor, so any of them can cancel against any other.
$$\frac{14 \cdot 25}{15 \cdot 21}$$Step 2 — Cancel the 7. The 14 on top and the 21 on the bottom are both multiples of 7, so they become 2 and 3.
$$\frac{2 \cdot 25}{15 \cdot 3}$$Step 3 — Cancel the 5. The 25 on top and the 15 on the bottom are both multiples of 5, so they become 5 and 3.
$$\frac{2 \cdot 5}{3 \cdot 3}$$Step 4 — Multiply what is left.
$$\frac{10}{9}$$Answer: \(\dfrac{10}{9}\).
One caution follows straight from Section 0.3.1: this cancelling is still cancelling factors. It is allowed here only because multiplication is the sole operation in sight. The moment a sum shows up in a numerator, the same-looking move becomes wrong.
Reciprocals
The reciprocal of a nonzero number is the number you multiply it by to get 1. For a fraction, swap the numerator and the denominator.
$$\frac{5}{8} \cdot \frac{8}{5} = \frac{40}{40} = 1$$| Number | Reciprocal | Their product |
|---|---|---|
| \(\dfrac{4}{7}\) | \(\dfrac{7}{4}\) | 1 |
| \(-\dfrac{5}{3}\) | \(-\dfrac{3}{5}\) | 1 |
| \(11\) | \(\dfrac{1}{11}\) | 1 |
| \(-\dfrac{1}{6}\) | \(-6\) | 1 |
| \(0\) | none exists | — |
Two facts in that table matter later on. A reciprocal keeps the sign of the original number, since flipping a fraction over does not change whether it is positive or negative, and \(-\frac{5}{3}\) times \(-\frac{3}{5}\) is \(+1\) by the like-signs rule. And zero has no reciprocal at all, because no number times 0 gives 1. That is the same fact as "you cannot divide by zero" wearing a different hat, and it is why every division rule below comes with a nonzero condition attached.
Definition 0.3.7 — A reciprocal is the number that turns a product into 1.
Dividing by a fraction
If \(b\), \(c\), and \(d\) are all nonzero, then
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$$To divide by a fraction, multiply by its reciprocal instead. Only the second fraction gets flipped — flipping both, or flipping the first one, is a common slip that throws the answer off badly enough that a rough estimate will catch it.
Definition 0.3.8 — Dividing counts how many fit inside, which is what the reciprocal does.
Divide \(\dfrac{3}{4} \div \dfrac{2}{5}\).
Solution
Step 1 — Find the reciprocal of the divisor. The divisor is \(\dfrac{2}{5}\), so its reciprocal is \(\dfrac{5}{2}\).
Step 2 — Change the division to a multiplication.
$$\frac{3}{4} \cdot \frac{5}{2}$$Step 3 — Check for common factors. The numbers 3 and 5 on top share nothing with 4 and 2 on the bottom, so there is nothing to cancel.
Step 4 — Multiply straight across.
$$\frac{15}{8}$$Answer: \(\dfrac{15}{8}\).
| Division | Rewritten as a multiplication | Result |
|---|---|---|
| \(\dfrac{3}{8} \div \dfrac{1}{4}\) | \(\dfrac{3}{8} \cdot \dfrac{4}{1}\) | \(\dfrac{3}{2}\) |
| \(-\dfrac{5}{9} \div \dfrac{10}{3}\) | \(-\dfrac{5}{9} \cdot \dfrac{3}{10}\) | \(-\dfrac{1}{6}\) |
| \(6 \div \dfrac{2}{5}\) | \(\dfrac{6}{1} \cdot \dfrac{5}{2}\) | \(15\) |
| \(\dfrac{7}{12} \div \left(-\dfrac{7}{12}\right)\) | \(\dfrac{7}{12} \cdot \left(-\dfrac{12}{7}\right)\) | \(-1\) |
Why multiplying by the reciprocal works
Memorizing "flip and multiply" will get you through the exercises. Understanding it keeps you from using it in the wrong place, so here are two explanations — one you can picture, one you can prove.
How many quarters fit inside 3? Twelve, because each whole holds four of them. So \(3 \div \frac{1}{4} = 12\), which is just \(3 \cdot 4\). Dividing by something smaller than 1 gives a bigger answer, and multiplying by its reciprocal is the only move that does the same.
The picture version is about counting. Division asks how many copies of the divisor fit inside the dividend, and when the divisor is small, a lot of them fit. The reciprocal of a small number is a large number, so multiplying by it produces the same growth that dividing by a small piece does.
The proof version is short. Write the division as a fraction, then multiply the top and the bottom by the reciprocal of the divisor. That move is legal — it is the Equivalent Fractions Property, multiplying by a well-chosen form of 1.
$$\frac{a}{b} \div \frac{c}{d} \;=\; \frac{\;\frac{a}{b}\;}{\;\frac{c}{d}\;} \;=\; \frac{\;\frac{a}{b} \cdot \frac{d}{c}\;}{\;\frac{c}{d} \cdot \frac{d}{c}\;} \;=\; \frac{\;\frac{a}{b} \cdot \frac{d}{c}\;}{1} \;=\; \frac{a}{b} \cdot \frac{d}{c}$$The reciprocal gets chosen for exactly one reason: it is the one multiplier that turns the denominator into 1 so it disappears. Nothing else about it is special. Hold onto that argument, because Section 0.3.4 uses the same move on complex fractions, and it is the reason both topics live in one section.
Multiply \(\dfrac{9}{16} \cdot \dfrac{4}{15}\), cancelling first. Then divide \(-\dfrac{3}{10} \div \dfrac{9}{20}\).
Solution
Part 1 — Multiply \(\dfrac{9}{16} \cdot \dfrac{4}{15}\). Write it as a single fraction and look for common factors.
$$\frac{9 \cdot 4}{16 \cdot 15}$$The 9 and the 15 share a factor of 3, becoming 3 and 5. The 4 and the 16 share a factor of 4, becoming 1 and 4.
$$\frac{3 \cdot 1}{4 \cdot 5} = \frac{3}{20}$$Part 2 — Divide \(-\dfrac{3}{10} \div \dfrac{9}{20}\). Multiply by the reciprocal of the second fraction.
$$-\frac{3}{10} \cdot \frac{20}{9}$$The 3 and the 9 share a factor of 3, and the 20 and the 10 share a factor of 10.
$$-\frac{1}{1} \cdot \frac{2}{3} = -\frac{2}{3}$$Unlike signs give a negative result, which is why the answer stays negative.
Answer: \(\dfrac{3}{20}\) and \(-\dfrac{2}{3}\).
0.3.3 Adding and Subtracting Fractions
If \(a\), \(b\), and \(c\) are numbers with \(c \neq 0\), then
$$\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c} \qquad \text{and} \qquad \frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}$$When the denominators already match, add or subtract the numerators and leave the denominator alone:
$$\frac{2}{9} + \frac{5}{9} = \frac{7}{9} \qquad\qquad \frac{11}{12} - \frac{5}{12} = \frac{6}{12} = \frac{1}{2}$$\(3x + 5x = 8x\) works because both terms count the same kind of thing. \(\frac{2}{9} + \frac{5}{9} = \frac{7}{9}\) works for exactly the same reason, with "ninths" playing the role of \(x\). Section 0.1 called those like terms.
The denominator does not change because it is not a quantity you are adding. It is the name of the unit. Two ninths plus five ninths is seven ninths for the same reason that 2 feet plus 5 feet is 7 feet and not 7 square feet. You count the objects; you do not count the noun.
Subtraction needs no separate treatment here either. The numerators are integers, so Section 0.2 already covers them, negative results included.
Definition 0.3.9 — Like denominators add their counts; the denominator names the unit.
Why you never add the denominators
$$\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$Adding the denominators instead would give \(\frac{2}{8}\), which is \(\frac{1}{4}\) — the same quarter you started with. That rule says adding a quarter to a quarter changes nothing at all, and two quarters of a pizza is plainly half a pizza. Any rule that returns a sum equal to one of the things being added is broken on its face.
| Sum | Correct | Denominators added, which is wrong | How you can tell |
|---|---|---|---|
| \(\dfrac{1}{4} + \dfrac{1}{4}\) | \(\dfrac{1}{2}\) | \(\dfrac{2}{8} = \dfrac{1}{4}\) | a sum cannot equal one of its addends |
| \(\dfrac{3}{5} + \dfrac{1}{5}\) | \(\dfrac{4}{5}\) | \(\dfrac{4}{10} = \dfrac{2}{5}\) | a sum cannot be less than \(\dfrac{3}{5}\) |
| \(\dfrac{1}{2} + \dfrac{1}{2}\) | \(1\) | \(\dfrac{2}{4} = \dfrac{1}{2}\) | two halves make a whole |
The confusion is understandable, because multiplication does multiply the denominators. But multiplication is answering a different question — a part of a part — and parts of parts really are smaller. Addition combines two pieces of the same-size whole, and the size of that whole does not change just because you put pieces together.
The least common denominator
Unlike denominators cannot be added until you make them alike. Any common denominator gets the job done, but the smallest one keeps the numbers manageable.
The least common denominator, or LCD, is the smallest positive number that every denominator in the problem divides into evenly.
There are two ways to find it. You can go by inspection, running through multiples of the larger denominator until you hit one that the smaller denominator also divides: for 6 and 8, try 8, then 16, then 24, and 24 works because it is \(6 \cdot 4\). Or you can use prime factorization, which keeps working when the numbers get ugly. Factor each denominator into primes, then take each prime the greatest number of times it shows up in any single factorization.
For 12 and 18, note that \(12 = 2^2 \cdot 3\) and \(18 = 2 \cdot 3^2\). Take two 2s, which is the most in either, and two 3s for the same reason, giving \(2^2 \cdot 3^2 = 36\).
| Denominators | Prime factorizations | LCD | Note |
|---|---|---|---|
| 4 and 6 | \(2^2\), \(2 \cdot 3\) | 12 | not \(4 \cdot 6 = 24\) |
| 12 and 18 | \(2^2 \cdot 3\), \(2 \cdot 3^2\) | 36 | |
| 6 and 18 | \(2 \cdot 3\), \(2 \cdot 3^2\) | 18 | one denominator divides the other |
| 5 and 9 | \(5\), \(3^2\) | 45 | no shared factor, so the LCD is the product |
| 10 and 15 | \(2 \cdot 5\), \(3 \cdot 5\) | 30 | |
| 14 and 21 | \(2 \cdot 7\), \(3 \cdot 7\) | 42 |
Two patterns in that table are worth memorizing. When the denominators share no factor, the LCD is just their product. When one denominator divides the other, the larger one is already the LCD.
Definition 0.3.10 — The LCD is the first value both ladders of multiples land on.
Adding with unlike denominators
The procedure has four steps: find the LCD, build each fraction up to it with the Equivalent Fractions Property, add the numerators, and simplify.
Add \(\dfrac{5}{12} + \dfrac{7}{18}\).
Solution
Step 1 — Find the LCD. Since \(12 = 2^2 \cdot 3\) and \(18 = 2 \cdot 3^2\), take two 2s and two 3s.
$$\text{LCD} = 2^2 \cdot 3^2 = 36$$Step 2 — Build the first fraction up to 36. Since \(12 \cdot 3 = 36\), multiply the top by 3 as well.
$$\frac{5}{12} = \frac{5 \cdot 3}{12 \cdot 3} = \frac{15}{36}$$Step 3 — Build the second fraction up to 36. Since \(18 \cdot 2 = 36\), multiply the top by 2 as well.
$$\frac{7}{18} = \frac{7 \cdot 2}{18 \cdot 2} = \frac{14}{36}$$Step 4 — Add the numerators. The denominators match now, so they stay put.
$$\frac{15}{36} + \frac{14}{36} = \frac{29}{36}$$Step 5 — Simplify. 29 is prime and does not divide 36, so this is already in lowest terms.
Answer: \(\dfrac{29}{36}\).
Whatever you multiply a denominator by, you have to multiply its numerator by as well. Multiplying only the bottom changes the number instead of renaming it, and that mistake hides well — nothing about \(\frac{5}{36}\) looks wrong on the page, it simply is not equal to \(\frac{5}{12}\).
Subtract \(\dfrac{7}{10} - \dfrac{5}{6}\).
Solution
Step 1 — Find the LCD. Since \(10 = 2 \cdot 5\) and \(6 = 2 \cdot 3\), the LCD is \(2 \cdot 3 \cdot 5 = 30\).
Step 2 — Build both fractions up to 30.
$$\frac{7}{10} = \frac{21}{30} \qquad \frac{5}{6} = \frac{25}{30}$$Step 3 — Subtract the numerators.
$$\frac{21 - 25}{30} = \frac{-4}{30}$$The numerator subtraction is ordinary integer work from Section 0.2: \(21 - 25 = -4\).
Step 4 — Simplify and move the sign out front. Both 4 and 30 are divisible by 2.
$$\frac{-4}{30} = -\frac{2}{15}$$Answer: \(-\dfrac{2}{15}\).
You are not actually required to use the least common denominator. Running that same problem over 60 gives \(\frac{42}{60} - \frac{50}{60} = \frac{-8}{60} = -\frac{2}{15}\), which is the same answer reached through bigger numbers and one extra round of simplifying. The LCD is a convenience, not a law.
Signed fractions
Nothing new is required here. Convert each fraction to the common denominator, then let the integer rules from Section 0.2 handle the numerators. Writing negative fractions with the sign up in the numerator during this step keeps the bookkeeping where you can see it.
| Problem | Over the LCD | Numerator arithmetic | Result |
|---|---|---|---|
| \(-\dfrac{3}{8} + \dfrac{5}{6}\) | \(\dfrac{-9}{24} + \dfrac{20}{24}\) | \(-9 + 20 = 11\) | \(\dfrac{11}{24}\) |
| \(\dfrac{1}{3} - \dfrac{7}{9}\) | \(\dfrac{3}{9} - \dfrac{7}{9}\) | \(3 - 7 = -4\) | \(-\dfrac{4}{9}\) |
| \(-\dfrac{2}{5} + \left(-\dfrac{1}{2}\right)\) | \(\dfrac{-4}{10} + \dfrac{-5}{10}\) | \(-4 + (-5) = -9\) | \(-\dfrac{9}{10}\) |
| \(-\dfrac{5}{6} - \left(-\dfrac{1}{4}\right)\) | \(\dfrac{-10}{12} - \dfrac{-3}{12}\) | \(-10 + 3 = -7\) | \(-\dfrac{7}{12}\) |
That last row uses the rule from Section 0.2 that subtracting a negative is the same as adding, so \(-10 - (-3)\) becomes \(-10 + 3 = -7\). Rewriting the subtraction as an addition before you touch the numbers costs one extra line and prevents most of the sign errors that happen here.
Knowing which procedure you need
The two procedures are so different that the most valuable habit is checking which one the problem is asking for before you write anything down.
| Question | Multiplying | Adding |
|---|---|---|
| Do the denominators have to match? | no | yes |
| What happens to the denominators | multiply them | keep the common one |
| What happens to the numerators | multiply them | add them |
| Can you cancel across the two fractions? | yes, before multiplying | no, never |
| Rough size of the answer | smaller than both, if both are positive and proper | bigger than both, if both are positive |
That fourth row is the one to watch. Cancelling across an addition is the fraction version of cancelling a term, and it is wrong for exactly the reason Section 0.3.1 gave.
Add \(\dfrac{7}{12} + \dfrac{5}{8}\). Then subtract \(-\dfrac{1}{6} - \dfrac{3}{4}\).
Solution
Part 1 — Add \(\dfrac{7}{12} + \dfrac{5}{8}\). Since \(12 = 2^2 \cdot 3\) and \(8 = 2^3\), take three 2s and one 3, so the LCD is 24.
$$\frac{7}{12} = \frac{14}{24} \qquad \frac{5}{8} = \frac{15}{24}$$Add the numerators and keep the denominator:
$$\frac{14 + 15}{24} = \frac{29}{24}$$Since 29 is prime, this is in lowest terms.
Part 2 — Subtract \(-\dfrac{1}{6} - \dfrac{3}{4}\). The LCD of 6 and 4 is 12.
$$\frac{-2}{12} - \frac{9}{12}$$Now the numerators are ordinary integers, and subtracting 9 from \(-2\) moves further negative:
$$\frac{-2 - 9}{12} = \frac{-11}{12} = -\frac{11}{12}$$Answer: \(\dfrac{29}{24}\) and \(-\dfrac{11}{12}\).
0.3.4 Complex Fractions and Order of Operations
Section 0.1 listed the fraction bar among the grouping symbols, right alongside parentheses and brackets. That was not a technicality. It changes the order you have to work in.
Simplify the entire numerator. Simplify the entire denominator. Then divide.
Treat the numerator and the denominator as though each one were wrapped in invisible parentheses. Written on one line with a division sign, such an expression would need those parentheses spelled out: \((5 + 3 \cdot 4) \div (2^3 - 6)\). The bar supplies the grouping for free, which is exactly why algebra prefers it to \(\div\).
Simplify \(\dfrac{5 + 3(4)}{2^3 - 6}\).
Solution
Step 1 — Simplify the numerator by itself. Multiplication comes before addition.
$$5 + 3(4) = 5 + 12 = 17$$Step 2 — Simplify the denominator by itself. The exponent comes before the subtraction.
$$2^3 - 6 = 8 - 6 = 2$$Step 3 — Now divide.
$$\frac{17}{2}$$Since 17 and 2 share no factor, this is in lowest terms.
Answer: \(\dfrac{17}{2}\).
Here is a second one with signs in it. In \(\dfrac{6 - 10}{-2 - 2}\) the numerator works out to \(-4\) and the denominator also works out to \(-4\), so the value is \(\dfrac{-4}{-4} = 1\) by the like-signs rule from Section 0.2.
Complex fractions
A complex fraction is a fraction whose numerator, denominator, or both are themselves fractions.
$$\frac{\;\dfrac{3}{4}\;}{\;\dfrac{9}{10}\;} \qquad\qquad \frac{\;\dfrac{2}{3} + \dfrac{1}{4}\;}{\;\dfrac{5}{6}\;} \qquad\qquad \frac{\;\dfrac{7}{8}\;}{\;4\;}$$These look intimidating and are not. That main bar is a division sign, so a complex fraction is just a division problem written vertically, and Section 0.3.2 already told you what to do with a division problem.
Definition 0.3.11 — The main bar of a complex fraction is the division that happens last.
Simplify \(\dfrac{\;\frac{3}{4}\;}{\;\frac{9}{10}\;}\).
Solution
Step 1 — Read the main bar as a division.
$$\frac{3}{4} \div \frac{9}{10}$$Step 2 — Multiply by the reciprocal of the divisor.
$$\frac{3}{4} \cdot \frac{10}{9}$$Step 3 — Cancel before multiplying. The 3 and the 9 share a factor of 3, and the 10 and the 4 share a factor of 2.
$$\frac{1}{2} \cdot \frac{5}{3}$$Step 4 — Multiply straight across.
$$\frac{5}{6}$$Answer: \(\dfrac{5}{6}\).
When the numerator or the denominator has an operation inside it, the grouping rule comes first: finish the top, finish the bottom, and only then divide. Do not try to combine an addition in the numerator with the division in one move — the addition happens inside the grouping, and the division happens after. Skipping that order is the usual failure here, and it produces an answer that looks plausible but is not.
Simplify \(\dfrac{\;\frac{2}{3} + \frac{1}{4}\;}{\;\frac{5}{6}\;}\).
Solution
Step 1 — Simplify the numerator first. The LCD of 3 and 4 is 12.
$$\frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$$Step 2 — Rewrite the complex fraction. The top is now a single fraction.
$$\frac{\;\frac{11}{12}\;}{\;\frac{5}{6}\;}$$Step 3 — Multiply by the reciprocal.
$$\frac{11}{12} \cdot \frac{6}{5}$$Step 4 — Cancel and multiply. The 6 and the 12 share a factor of 6.
$$\frac{11}{2} \cdot \frac{1}{5} = \frac{11}{10}$$Answer: \(\dfrac{11}{10}\), which is \(1\dfrac{1}{10}\) as a mixed number.
| Complex fraction | Rewritten | Simplified |
|---|---|---|
| \(\dfrac{\;\frac{7}{8}\;}{\;4\;}\) | \(\dfrac{7}{8} \cdot \dfrac{1}{4}\) | \(\dfrac{7}{32}\) |
| \(\dfrac{\;6\;}{\;\frac{3}{5}\;}\) | \(6 \cdot \dfrac{5}{3}\) | \(10\) |
| \(\dfrac{\;-\frac{2}{9}\;}{\;\frac{4}{3}\;}\) | \(-\dfrac{2}{9} \cdot \dfrac{3}{4}\) | \(-\dfrac{1}{6}\) |
| \(\dfrac{\;\frac{1}{2} - \frac{1}{6}\;}{\;\frac{2}{3}\;}\) | \(\dfrac{1}{3} \cdot \dfrac{3}{2}\) | \(\dfrac{1}{2}\) |
The second row makes a point about size worth repeating. Dividing 6 by a number smaller than 1 gives an answer bigger than 6. If you divide a positive number by a positive proper fraction and your answer comes out smaller than what you started with, you flipped the wrong fraction. (The size check needs both numbers positive — with negatives in play, compare magnitudes instead.)
Order of operations with fractions
The four-level order from Section 0.1 does not change at all. Fractions do not get their own rules; they just make each step take more care, the same way Section 0.2 said about negative numbers.
One fact deserves its own line before the example. An exponent on a fraction applies to the numerator and the denominator both, since \(\left(\frac{2}{5}\right)^2\) means \(\frac{2}{5} \cdot \frac{2}{5}\), which multiplies straight across to \(\frac{4}{25}\). The placement warnings from Section 0.1 still apply, so \(\left(\frac{2}{5}\right)^2\) and \(\frac{2^2}{5}\) are different numbers.
Simplify \(\;\dfrac{3}{4} + \dfrac{1}{2}\left(\dfrac{2}{5}\right)^2\).
Solution
Step 1 — Exponents first. Square the numerator and the denominator both.
$$\left(\frac{2}{5}\right)^2 = \frac{4}{25}$$Step 2 — Then multiplication. Cancel the 2 into the 4 before multiplying.
$$\frac{1}{2} \cdot \frac{4}{25} = \frac{2}{25}$$Step 3 — Addition last, so find the LCD. Since \(4 = 2^2\) and \(25 = 5^2\) share no factor, the LCD is \(4 \cdot 25 = 100\).
$$\frac{3}{4} = \frac{75}{100} \qquad \frac{2}{25} = \frac{8}{100}$$Step 4 — Add the numerators.
$$\frac{75 + 8}{100} = \frac{83}{100}$$Answer: \(\dfrac{83}{100}\).
Evaluating expressions containing fractions
To evaluate an expression, substitute the given value and simplify, wrapping the substituted value in parentheses exactly as Sections 0.1 and 0.2 insisted. Those parentheses are not decoration. Without them, \(2x\) with \(x = -\frac{1}{2}\) invites the reading \(2 - \frac{1}{2}\), and the whole problem goes wrong on the first line.
Evaluate \(\;\dfrac{2x}{x+5}\;\) when \(x = -\dfrac{1}{2}\).
Solution
Step 1 — Substitute, with parentheses around the value.
$$\frac{2\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right) + 5}$$Step 2 — Simplify the numerator. Unlike signs give a negative product.
$$2 \cdot \left(-\frac{1}{2}\right) = -1$$Step 3 — Simplify the denominator. Write 5 as \(\frac{10}{2}\) so the denominators match.
$$-\frac{1}{2} + \frac{10}{2} = \frac{9}{2}$$Step 4 — Divide. This is now a complex fraction, so multiply by the reciprocal.
$$-1 \cdot \frac{2}{9} = -\frac{2}{9}$$Answer: \(-\dfrac{2}{9}\).
The expression \(\dfrac{2x}{x+5}\) means nothing when \(x = -5\), because the denominator becomes zero. Every fractional expression carries a hidden restriction like this one, and finding it is the first move in almost every Chapter 3 problem.
A value that makes a denominator zero is not merely awkward to work with. It is not in the expression's domain at all, so the expression has no value there — the same conclusion Section 0.2 reached about \(\frac{5}{0}\).
Fractions in a real setting
A crew is repainting a hallway. Materials for the whole job cost $48, and the crew has finished \(\frac{5}{8}\) of the hallway. To find what they have used so far, take \(\frac{5}{8}\) of 48, cancelling the 8 into the 48 before multiplying:
$$\frac{5}{8} \cdot 48 = \frac{5}{8} \cdot \frac{48}{1} = 5 \cdot 6 = 30$$So they have used $30 of materials, and the remaining \(\frac{3}{8}\) of the job accounts for the other $18. Notice the check built into that: \(\frac{5}{8} + \frac{3}{8} = \frac{8}{8} = 1\), so the two pieces add to one whole job.
Now flip the question. Suppose you know that $30 of materials covered \(\frac{5}{8}\) of the hallway and you want the cost of the whole job. That is a division:
$$30 \div \frac{5}{8} = 30 \cdot \frac{8}{5} = 48$$Same two numbers, opposite operation, because the unknown moved from the part to the whole. Deciding which of those two setups a situation calls for is the real skill, and Section 0.5 gives you the properties that let you rearrange relationships like this one on purpose instead of by guessing.
Simplify \(\dfrac{\;\frac{3}{4} + \frac{1}{2}\;}{\;\frac{5}{8}\;}\). Then evaluate \(\;\dfrac{4x}{x+3}\;\) when \(x = -\dfrac{1}{2}\).
Solution
Part 1 — Simplify the complex fraction. Start with the numerator, where the LCD of 4 and 2 is 4.
$$\frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$$Now the main bar is a division, so multiply by the reciprocal.
$$\frac{5}{4} \div \frac{5}{8} = \frac{5}{4} \cdot \frac{8}{5}$$Cancel the 5s, and cancel the 4 into the 8.
$$1 \cdot 2 = 2$$Part 2 — Evaluate \(\dfrac{4x}{x+3}\) at \(x = -\dfrac{1}{2}\). Substitute with parentheses.
$$\frac{4\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right) + 3}$$The numerator is \(4 \cdot \left(-\frac{1}{2}\right) = -2\). The denominator is \(-\frac{1}{2} + \frac{6}{2} = \frac{5}{2}\).
$$\frac{-2}{\;\frac{5}{2}\;} = -2 \cdot \frac{2}{5} = -\frac{4}{5}$$Answer: \(2\) and \(-\dfrac{4}{5}\).
Problem Set 0.3
Problem 1. Simplify \(\dfrac{24}{36}\).
Solution
Step 1 — Break both numbers into prime factors: Prime-factoring makes the greatest common factor (GCF) easy to spot. $$24 = 2^3 \cdot 3, \qquad 36 = 2^2 \cdot 3^2$$ The shared factors are \(2^2\) and \(3^1\), so \(\text{GCF}(24,36) = 2^2\cdot 3 = 12\).
Step 2 — Apply the Equivalent Fractions Property: Dividing numerator and denominator by the same nonzero number (here, the GCF) produces an equivalent fraction in lowest terms — this is cancelling a common factor of the whole numerator and whole denominator, not cancelling a term. $$\frac{24}{36} = \frac{24 \div 12}{36 \div 12} = \frac{2}{3}$$
Answer: \(\dfrac{2}{3}\)
Problem 2. Simplify \(\dfrac{45}{60}\).
Solution
Step 1 — Find the greatest common factor of 45 and 60: \(45 = 3^2\cdot 5\) and \(60 = 2^2\cdot 3\cdot 5\), so the common factors are \(3\) and \(5\), giving \(\text{GCF}(45,60) = 3\cdot 5 = 15\).
Step 2 — Divide both parts by the GCF: By the Equivalent Fractions Property, dividing top and bottom by the same factor keeps the value unchanged while reaching lowest terms. $$\frac{45}{60} = \frac{45 \div 15}{60 \div 15} = \frac{3}{4}$$
Answer: \(\dfrac{3}{4}\)
Problem 3. Simplify \(\dfrac{-28}{63}\) and write the sign in front.
Solution
Step 1 — Simplify the magnitude first: Ignore the sign for a moment and reduce \(\dfrac{28}{63}\). Since \(28 = 2^2\cdot 7\) and \(63 = 3^2\cdot 7\), the only shared factor is \(7\), so \(\text{GCF}(28,63) = 7\).
Step 2 — Divide numerator and denominator by the GCF: Cancelling the common factor of \(7\) (a factor of both entire numbers, not just a piece of one) gives $$\frac{28}{63} = \frac{28 \div 7}{63 \div 7} = \frac{4}{9}$$
Step 3 — Move the sign out front: A fraction's negative sign belongs out in front of the fraction bar, not buried in the numerator or denominator. $$\frac{-28}{63} = -\frac{4}{9}$$
Answer: \(-\dfrac{4}{9}\)
Problem 4. Write three fractions equivalent to \(\dfrac{3}{7}\).
Solution
Step 1 — Recall the Equivalent Fractions Property: Multiplying (or dividing) the numerator and denominator of a fraction by the same nonzero number produces a new fraction with the same value.
Step 2 — Pick three different multipliers and apply them to \(\dfrac{3}{7}\): $$\frac{3}{7} = \frac{3 \cdot 2}{7 \cdot 2} = \frac{6}{14}, \qquad \frac{3}{7} = \frac{3 \cdot 3}{7 \cdot 3} = \frac{9}{21}, \qquad \frac{3}{7} = \frac{3 \cdot 4}{7 \cdot 4} = \frac{12}{28}$$
Answer: \(\dfrac{6}{14}\), \(\dfrac{9}{21}\), and \(\dfrac{12}{28}\) are all equivalent to \(\dfrac{3}{7}\) (many other correct choices exist).
Problem 5. Is \(\dfrac{9}{15}\) equal to \(\dfrac{3}{5}\)? Explain how you know.
Solution
Step 1 — Reduce \(\dfrac{9}{15}\) to lowest terms: Find the GCF of \(9\) and \(15\): both share a factor of \(3\), so \(\text{GCF}(9,15) = 3\). $$\frac{9}{15} = \frac{9 \div 3}{15 \div 3} = \frac{3}{5}$$
Step 2 — Compare the reduced form to the target fraction: Since dividing numerator and denominator of \(\dfrac{9}{15}\) by their GCF lands exactly on \(\dfrac{3}{5}\), the two fractions are the same number by the Equivalent Fractions Property, just written with different-sized pieces.
Answer: Yes — \(\dfrac{9}{15}\) equals \(\dfrac{3}{5}\), because dividing \(9\) and \(15\) by their greatest common factor, \(3\), reduces \(\dfrac{9}{15}\) exactly to \(\dfrac{3}{5}\).
Problem 6. Write \(\dfrac{29}{6}\) as a mixed number.
Solution
Step 1 — Divide the numerator by the denominator: \(29 \div 6\) gives a whole-number quotient and a remainder: since \(6 \cdot 4 = 24\) and \(29 - 24 = 5\), the quotient is \(4\) with remainder \(5\).
Step 2 — Write the mixed number: The quotient becomes the whole-number part, and the remainder is placed over the original denominator. $$\frac{29}{6} = 4\frac{5}{6}$$
Answer: \(4\dfrac{5}{6}\)
Problem 7. Write \(7\frac{1}{3}\) as an improper fraction.
Solution
Step 1 — Multiply the whole number by the denominator: For \(7\frac{1}{3}\), multiply the whole part \(7\) by the denominator \(3\): \(7 \cdot 3 = 21\).
Step 2 — Add the numerator, keeping the same denominator: Add the original numerator \(1\) to get the new numerator, and keep the denominator \(3\) unchanged. $$7\frac{1}{3} = \frac{7 \cdot 3 + 1}{3} = \frac{22}{3}$$
Answer: \(\dfrac{22}{3}\)
Problem 8. Explain in one sentence why \(\dfrac{5+2}{5+9}\) is not equal to \(\dfrac{2}{9}\).
Solution
Step 1 — Actually simplify \(\dfrac{5+2}{5+9}\): Add inside the numerator and denominator first: \(5+2 = 7\) and \(5+9=14\), so \(\dfrac{5+2}{5+9} = \dfrac{7}{14}\). Since \(\text{GCF}(7,14)=7\), this reduces to \(\dfrac{7}{14} = \dfrac{1}{2}\), not \(\dfrac{2}{9}\).
Step 2 — Identify why cancelling the 5s is invalid: The Equivalent Fractions Property only allows dividing out a factor that multiplies the entire numerator and the entire denominator. Here the \(5\)s are terms being added, not factors being multiplied, so they cannot be cancelled — you cancel factors, not terms.
Answer: \(\dfrac{5+2}{5+9}\) is not \(\dfrac{2}{9}\) because the \(5\)s are added terms rather than common factors of the whole numerator and denominator, so they cannot be cancelled; the fraction actually simplifies to \(\dfrac{1}{2}\).
Problem 9. Rewrite \(\dfrac{4}{-11}\) with the sign in front.
Solution
Step 1 — Locate the negative sign: In \(\dfrac{4}{-11}\), the negative sign sits on the denominator rather than out in front of the fraction.
Step 2 — Move the sign in front: A fraction with exactly one negative part (numerator or denominator) is negative overall, and convention places that sign in front of the fraction bar rather than inside it. $$\frac{4}{-11} = -\frac{4}{11}$$
Answer: \(-\dfrac{4}{11}\)
Problem 10. Multiply: \(\dfrac{3}{8} \cdot \dfrac{2}{9}\).
Solution
Step 1 — Look for common factors to cancel before multiplying: In \(\dfrac{3}{8} \cdot \dfrac{2}{9}\), compare each numerator with the other fraction's denominator. The \(3\) and the \(9\) share a factor of \(3\): the \(3\) becomes \(1\) and the \(9\) becomes \(3\). The \(2\) and the \(8\) share a factor of \(2\): the \(2\) becomes \(1\) and the \(8\) becomes \(4\). Cancelling first keeps the numbers small.
$$\frac{3}{8} \cdot \frac{2}{9} = \frac{1}{4} \cdot \frac{1}{3}$$Step 2 — Multiply straight across: With common factors already removed, multiply numerator by numerator and denominator by denominator.
$$\frac{1}{4} \cdot \frac{1}{3} = \frac{1 \cdot 1}{4 \cdot 3} = \frac{1}{12}$$Answer: \(\dfrac{1}{12}\)
Problem 11. Multiply: \(-\dfrac{5}{12} \cdot \dfrac{8}{15}\).
Solution
Step 1 — Determine the sign of the product: One factor, \(-\dfrac{5}{12}\), is negative and the other, \(\dfrac{8}{15}\), is positive. Unlike signs give a negative product, so the answer will be negative once the fraction part is found.
Step 2 — Cancel common factors first: Working with the fraction part \(\dfrac{5}{12} \cdot \dfrac{8}{15}\), compare each numerator to the other denominator. The \(5\) and the \(15\) share a factor of \(5\): the \(5\) becomes \(1\) and the \(15\) becomes \(3\). The \(8\) and the \(12\) share a factor of \(4\): the \(8\) becomes \(2\) and the \(12\) becomes \(3\).
$$\frac{5}{12} \cdot \frac{8}{15} = \frac{1}{3} \cdot \frac{2}{3}$$Step 3 — Multiply straight across and attach the sign:
$$\frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}, \qquad \text{so} \qquad -\frac{5}{12} \cdot \frac{8}{15} = -\frac{2}{9}$$Answer: \(-\dfrac{2}{9}\)
Problem 12. Multiply: \(\dfrac{7}{10} \cdot 25\).
Solution
Step 1 — Write the whole number as a fraction: Every whole number \(n\) can be written as \(\dfrac{n}{1}\), so \(25 = \dfrac{25}{1}\). This lets us treat the problem as a fraction times a fraction.
$$\frac{7}{10} \cdot 25 = \frac{7}{10} \cdot \frac{25}{1}$$Step 2 — Cancel common factors before multiplying: The \(25\) on top and the \(10\) on the bottom share a factor of \(5\): the \(25\) becomes \(5\) and the \(10\) becomes \(2\).
$$\frac{7}{10} \cdot \frac{25}{1} = \frac{7}{2} \cdot \frac{5}{1}$$Step 3 — Multiply straight across:
$$\frac{7}{2} \cdot \frac{5}{1} = \frac{35}{2}$$Step 4 — Write it as a mixed number as well. Since \(35 \div 2 = 17\) with remainder \(1\), the improper fraction is \(17\dfrac{1}{2}\).
Answer: \(\dfrac{35}{2}\), which is \(17\dfrac{1}{2}\)
Problem 13. Find the reciprocal of \(-\dfrac{8}{3}\).
Solution
Step 1 — Recall what a reciprocal is: The reciprocal of a number is what you multiply it by to get \(1\). For a fraction, that means swapping the numerator and the denominator.
Step 2 — Flip the fraction, keeping the sign: Flipping does not change whether a number is positive or negative, so the reciprocal of \(-\dfrac{8}{3}\) is \(-\dfrac{3}{8}\).
Step 3 — Check the result: Multiplying the original number by this candidate should give \(1\), and like signs give a positive product.
$$-\frac{8}{3} \cdot \left(-\frac{3}{8}\right) = \frac{8 \cdot 3}{3 \cdot 8} = \frac{24}{24} = 1$$Answer: \(-\dfrac{3}{8}\)
Problem 14. Explain why 0 has no reciprocal.
Solution
Step 1 — Recall the definition of a reciprocal: The reciprocal of a number \(n\) is the number you multiply \(n\) by to get a product of \(1\).
Step 2 — Test whether such a number exists for \(0\): Multiplying \(0\) by any number always gives \(0\), since \(0 \cdot x = 0\) for every \(x\). There is no value of \(x\) that makes \(0 \cdot x = 1\) true.
Step 3 — Connect this to division: This is the same fact as "you cannot divide by zero" wearing a different hat, which is why every division rule in this section carries a nonzero condition.
Answer: \(0\) has no reciprocal because no number multiplied by \(0\) can ever produce \(1\).
Problem 15. Divide: \(\dfrac{5}{6} \div \dfrac{10}{9}\).
Solution
Step 1 — Rewrite the division as multiplication by the reciprocal: To divide by a fraction, multiply by the reciprocal of the second fraction only. The reciprocal of \(\dfrac{10}{9}\) is \(\dfrac{9}{10}\).
$$\frac{5}{6} \div \frac{10}{9} = \frac{5}{6} \cdot \frac{9}{10}$$Step 2 — Cancel common factors before multiplying: The \(5\) on top and the \(10\) on the bottom share a factor of \(5\), becoming \(1\) and \(2\). The \(9\) on top and the \(6\) on the bottom share a factor of \(3\), becoming \(3\) and \(2\).
$$\frac{5}{6} \cdot \frac{9}{10} = \frac{1}{2} \cdot \frac{3}{2}$$Step 3 — Multiply straight across:
$$\frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$$Answer: \(\dfrac{3}{4}\)
Problem 16. Divide: \(9 \div \dfrac{3}{4}\).
Solution
Step 1 — Write the whole number over 1: Since any whole number \(n\) can be written as \(\dfrac{n}{1}\), the dividend \(9\) becomes \(\dfrac{9}{1}\).
Step 2 — Multiply by the reciprocal of the divisor: The reciprocal of \(\dfrac{3}{4}\) is \(\dfrac{4}{3}\).
$$9 \div \frac{3}{4} = \frac{9}{1} \cdot \frac{4}{3}$$Step 3 — Cancel before multiplying: The \(9\) on top and the \(3\) on the bottom share a factor of \(3\), becoming \(3\) and \(1\).
$$\frac{3}{1} \cdot \frac{4}{1} = \frac{12}{1} = 12$$Step 4 — Sanity-check the size. Dividing by a positive number smaller than \(1\) must give an answer bigger than \(9\), and \(12\) is.
Answer: \(12\)
Problem 17. Add: \(\dfrac{4}{15} + \dfrac{7}{15}\).
Solution
Step 1 — Check the denominators: Both fractions, \(\dfrac{4}{15}\) and \(\dfrac{7}{15}\), already have the same denominator, \(15\), so no building up is needed.
Step 2 — Add the numerators and keep the denominator: The denominator names the size of the pieces, so it stays put while the counts combine.
$$\frac{4}{15} + \frac{7}{15} = \frac{4+7}{15} = \frac{11}{15}$$Step 3 — Check for simplification: \(11\) is prime and does not divide \(15\), so the fraction is already in lowest terms.
Answer: \(\dfrac{11}{15}\)
Problem 18. Find the least common denominator of 8 and 20.
Solution
Step 1 — Find the prime factorization of each number:
$$8 = 2 \cdot 2 \cdot 2 = 2^3 \qquad\qquad 20 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 5$$Step 2 — Take each prime the greatest number of times it appears in any one factorization: The prime \(2\) appears at most three times (in \(8 = 2^3\)); the prime \(5\) appears at most once (in \(20 = 2^2 \cdot 5\)).
Step 3 — Multiply those prime powers together:
$$\text{LCD} = 2^3 \cdot 5 = 8 \cdot 5 = 40$$Answer: The least common denominator of \(8\) and \(20\) is \(40\).
Problem 19. Add: \(\dfrac{3}{8} + \dfrac{5}{12}\).
Solution
Step 1 — Break each denominator into primes: \(8 = 2^3\) and \(12 = 2^2 \cdot 3\). Taking each prime the greatest number of times it appears gives an LCD of \(2^3 \cdot 3 = 24\).
Step 2 — Build each fraction up to the LCD using the Equivalent Fractions Property: $$\frac{3}{8} = \frac{3 \cdot 3}{8 \cdot 3} = \frac{9}{24}, \qquad \frac{5}{12} = \frac{5 \cdot 2}{12 \cdot 2} = \frac{10}{24}$$ Multiplying top and bottom by the same number renames the fraction without changing its value.
Step 3 — Add the numerators over the shared denominator: $$\frac{9}{24} + \frac{10}{24} = \frac{19}{24}$$
Step 4 — Simplify. \(19\) is prime and does not divide \(24\), so this is in lowest terms.
Answer: \(\dfrac{19}{24}\)
Problem 20. Subtract: \(\dfrac{5}{9} - \dfrac{5}{6}\).
Solution
Step 1 — Find the LCD by prime factorization: \(9 = 3^2\) and \(6 = 2 \cdot 3\). Taking the highest power of each prime gives \(\text{LCD} = 2 \cdot 3^2 = 18\).
Step 2 — Rewrite each fraction with denominator 18: $$\frac{5}{9} = \frac{5 \cdot 2}{9 \cdot 2} = \frac{10}{18}, \qquad \frac{5}{6} = \frac{5 \cdot 3}{6 \cdot 3} = \frac{15}{18}$$
Step 3 — Subtract the numerators as ordinary signed integers: $$\frac{10}{18} - \frac{15}{18} = \frac{10 - 15}{18} = \frac{-5}{18}$$
Step 4 — Move the sign out front. \(5\) and \(18\) share no common factor, so nothing reduces.
$$\frac{-5}{18} = -\frac{5}{18}$$Answer: \(-\dfrac{5}{18}\)
Problem 21. Add: \(-\dfrac{7}{10} + \dfrac{2}{5}\).
Solution
Step 1 — Locate the LCD: Since \(10 = 2 \cdot 5\) and \(5\) is already a factor of \(10\), the larger denominator is itself the LCD, so \(\text{LCD} = 10\).
Step 2 — Build the second fraction up to a denominator of 10: $$\frac{2}{5} = \frac{2 \cdot 2}{5 \cdot 2} = \frac{4}{10}$$ The first term, \(-\dfrac{7}{10}\), already has the LCD, so it needs no change.
Step 3 — Combine the numerators as signed integers over the common denominator: $$-\frac{7}{10} + \frac{4}{10} = \frac{-7 + 4}{10} = \frac{-3}{10} = -\frac{3}{10}$$
Answer: \(-\dfrac{3}{10}\)
Problem 22. Explain in one sentence why \(\dfrac{1}{3} + \dfrac{1}{3}\) is not \(\dfrac{2}{6}\).
Solution
Step 1 — Recall what the denominator is doing: the denominator names the size of the pieces being counted, and only the numerators — the counts — combine when you add.
Step 2 — Apply that to \(\dfrac{1}{3} + \dfrac{1}{3}\): both terms already share the denominator \(3\), so the numerators add directly: \(1 + 1 = 2\), giving \(\dfrac{2}{3}\).
Step 3 — See why \(\dfrac{2}{6}\) has to be wrong: \(\dfrac{2}{6}\) comes from also adding the denominators, and it simplifies back to \(\dfrac{1}{3}\) — a sum that equals one of the things being added, which cannot happen.
Answer: Adding fractions combines only the numerators once the denominators match, so \(\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\), not \(\dfrac{2}{6}\), which is just \(\dfrac{1}{3}\) again.
Problem 23. Simplify the complex fraction \(\dfrac{\;\frac{2}{5}\;}{\;\frac{8}{15}\;}\).
Solution
Step 1 — Read the main bar as a division: the long bar separating the two small fractions is a division symbol, so the complex fraction means \(\dfrac{2}{5} \div \dfrac{8}{15}\).
Step 2 — Multiply by the reciprocal of the divisor: $$\frac{2}{5} \div \frac{8}{15} = \frac{2}{5} \cdot \frac{15}{8}$$
Step 3 — Cancel before multiplying: the \(15\) and the \(5\) share a factor of \(5\), becoming \(3\) and \(1\); the \(2\) and the \(8\) share a factor of \(2\), becoming \(1\) and \(4\).
$$\frac{1}{1} \cdot \frac{3}{4} = \frac{3}{4}$$Answer: \(\dfrac{3}{4}\)
Problem 24. Simplify \(\dfrac{7 - 3(4)}{2^2 + 1}\).
Solution
Step 1 — Treat the fraction bar as a grouping symbol: the entire numerator and the entire denominator each get simplified on their own, as if wrapped in invisible parentheses, before any dividing happens.
Step 2 — Simplify the numerator, multiplication before subtraction: $$7 - 3(4) = 7 - 12 = -5$$
Step 3 — Simplify the denominator, exponent before addition: $$2^2 + 1 = 4 + 1 = 5$$
Step 4 — Now divide. Unlike signs give a negative quotient. $$\frac{-5}{5} = -1$$
Answer: \(-1\)
Problem 25. Simplify \(\;\dfrac{2}{3} + \dfrac{1}{4}\left(\dfrac{2}{3}\right)^2\).
Solution
Step 1 — Exponents first, applied to numerator and denominator both: $$\left(\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \frac{4}{9}$$
Step 2 — Then the multiplication, cancelling the 4 against the 4 first: $$\frac{1}{4} \cdot \frac{4}{9} = \frac{1}{9}$$
Step 3 — Addition last, so find the LCD. Since \(9 = 3^2\) and \(3\) divides \(9\), the LCD is \(9\); build the first fraction up with the Equivalent Fractions Property: \(\dfrac{2}{3} = \dfrac{6}{9}\).
Step 4 — Add the numerators over the common denominator: $$\frac{6}{9} + \frac{1}{9} = \frac{7}{9}$$
Answer: \(\dfrac{7}{9}\)
Problem 26. Evaluate \(\;\dfrac{3x}{x+2}\;\) when \(x = -\dfrac{1}{3}\).
Solution
Step 1 — Substitute \(x = -\dfrac{1}{3}\), wrapped in parentheses, everywhere \(x\) appears: $$\frac{3x}{x+2} = \frac{3\left(-\frac{1}{3}\right)}{\left(-\frac{1}{3}\right)+2}$$ The parentheses keep the negative sign attached to the value as it is carried through.
Step 2 — Simplify the numerator. Unlike signs give a negative product. $$3\left(-\frac{1}{3}\right) = -1$$
Step 3 — Simplify the denominator by writing 2 over the same denominator: $$-\frac{1}{3} + \frac{6}{3} = \frac{5}{3}$$
Step 4 — The bar is a division, so multiply by the reciprocal: $$-1 \cdot \frac{3}{5} = -\frac{3}{5}$$
Answer: \(-\dfrac{3}{5}\)
Problem 27. Evaluate \(\;\dfrac{1}{2}n + \dfrac{3}{4}\;\) when \(n = -\dfrac{1}{2}\).
Solution
Step 1 — Substitute \(n = -\dfrac{1}{2}\), wrapped in parentheses: $$\frac{1}{2}n + \frac{3}{4} = \frac{1}{2}\left(-\frac{1}{2}\right) + \frac{3}{4}$$
Step 2 — Multiplication before addition. Unlike signs give a negative product. $$\frac{1}{2}\left(-\frac{1}{2}\right) = -\frac{1}{4}$$
Step 3 — Add. The two fractions already share the denominator \(4\), so combine the numerators as signed integers. $$-\frac{1}{4} + \frac{3}{4} = \frac{-1+3}{4} = \frac{2}{4}$$
Step 4 — Simplify. Both \(2\) and \(4\) are divisible by \(2\). $$\frac{2}{4} = \frac{1}{2}$$
Answer: \(\dfrac{1}{2}\)
Key Terms
fraction — a number written \(\dfrac{a}{b}\) with \(b \neq 0\), naming \(a\) parts of a whole that was divided into \(b\) equal parts.
numerator — the top number of a fraction, counting how many parts are being taken.
denominator — the bottom number of a fraction, naming how many equal parts the whole was divided into.
equivalent fractions — two fractions that name the same amount, such as \(\dfrac{3}{8}\) and \(\dfrac{6}{16}\).
lowest terms — the form of a fraction whose numerator and denominator share no common factor but 1.
proper fraction — a fraction whose numerator is smaller than its denominator, so its value is less than 1.
improper fraction — a fraction whose numerator is greater than or equal to its denominator, so its value is 1 or more.
mixed number — a whole number written beside a proper fraction, meaning their sum.
reciprocal — the number you multiply a given nonzero number by to get 1, found by swapping numerator and denominator.
least common denominator — the smallest positive number that every denominator in a problem divides into evenly.
complex fraction — a fraction whose numerator, denominator, or both are themselves fractions.