Grade 5 · 4.NF.A.2

Comparing by finding a common denominator

One dog ate five sixths of its bowl and another ate seven ninths. The pieces are different sizes, so how can you tell which dog ate more?

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Common denominators and the least common denominatorGrade 4 Comparing against a benchmarkGrade 4

Where you are 3 in Comparing fractions

Fractions, ratio and percent3

Fractions15

Comparing fractions3

Comparing fractions with the same denominator or numerator Comparing against a benchmark

Comparing by finding a common denominator

When you reach for it

Chiyo, Obinna and Valeria run a pet sitting club. They feed dogs and cats, fill food bins, and walk dogs along the bike trail by the park. Their notes are full of fractions: a fraction of a scoop, of a bag, of a mile. Again and again they need to know which of two amounts is more.

Some comparisons are quick. If the denominators match, the pieces are the same size, and the fraction with more pieces is larger (Comparing fractions with the same denominator or numerator). If one fraction is less than \tfrac{1}{2} and the other is more, a benchmark settles it (Comparing against a benchmark).

But here is a harder one. Biscuit the dog ate \tfrac{5}{6} of her bowl. Rocket ate \tfrac{7}{9} of his, and the two bowls are the same size. Both fractions are a little less than 1, so benchmarks do not help. The pieces are different sizes, so you cannot just count them. This lesson is about making the pieces match, so that counting works again. It has three parts, one for each day:

Part 1: rename both fractions to a common denominator and compare the counts.
Part 2: two other routes: a common numerator, or changing both fractions to decimals.
Part 3: putting a list of three or more fractions in order, and the cross product shortcut, which turns out to be the common denominator method in disguise.

The procedure

Part 1: make the pieces the same size

A fraction counts pieces. \tfrac{5}{6} means 5 pieces, each one sixth of the whole. \tfrac{7}{9} means 7 pieces, each one ninth. A sixth is bigger than a ninth, so 5 big pieces and 7 small pieces cannot be compared just by looking at 5 and 7.

The fix is to cut the pieces again until both fractions use the same size. Cut every sixth into 3 equal parts and you get eighteenths. Cut every ninth into 2 equal parts and you also get eighteenths.

Figure 1Cut both bars into eighteenths and the pieces match. \tfrac{5}{6} is 15 of them and \tfrac{7}{9} is 14 of them.

Cutting each piece into 3 makes three times as many pieces, so 5 sixths become 15 eighteenths. Cutting each piece into 2 makes twice as many, so 7 ninths become 14 eighteenths. In symbols, multiply the top and bottom by the same number, which is the same as multiplying by 1 and does not change the amount (Equivalent fractions):

\frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18} \qquad \frac{7}{9} = \frac{7 \times 2}{9 \times 2} = \frac{14}{18}

Now the pieces match, so count them. 15 eighteenths is more than 14 eighteenths, so

\frac{5}{6} > \frac{7}{9}

Biscuit ate more, by exactly \tfrac{1}{18} of a bowl. The renamed fractions tell you which is bigger and also how far apart they are.

Which common denominator? A common denominator is any number that both denominators divide into evenly. Renaming two fractions this way has its own lesson, Common denominators and the least common denominator; here is the short version. Two choices always work.

The product of the denominators. 6 \times 9 = 54, and both 6 and 9 divide 54. So \tfrac{5}{6} = \tfrac{45}{54} and \tfrac{7}{9} = \tfrac{42}{54}. Again 45 > 42, the same answer. The product never fails, and it needs no thinking, but the numbers can get large.
The least common denominator, the smallest number both denominators divide into. List multiples of the larger denominator until one is also a multiple of the smaller: 9, 18. 18 is a multiple of 6, so 18 is the least common denominator. Smaller numbers are easier to work with, and they make the gap (\tfrac{1}{18}) easier to read.

Any common denominator gives the same answer, because each one only cuts the pieces finer. If the denominators share no factor (like 4 and 7), the least common denominator is the product, so the two choices are the same.

Part 2: two other routes

A common numerator. Matching the bottoms is not the only way. You can match the tops instead. When two fractions have the same number of pieces, the one with the bigger pieces is larger, and bigger pieces means a smaller denominator. For \tfrac{5}{6} and \tfrac{7}{9}, make both numerators 35:

\frac{5}{6} = \frac{5 \times 7}{6 \times 7} = \frac{35}{42} \qquad \frac{7}{9} = \frac{7 \times 5}{9 \times 5} = \frac{35}{45}

Each has 35 pieces. A forty-second of a bowl is bigger than a forty-fifth, so 35 forty-seconds is more. Again \tfrac{5}{6} > \tfrac{7}{9}. This route is quickest when one numerator already divides the other, so only one fraction needs renaming.

Change both to decimals. A fraction is a division: \tfrac{5}{6} means 5 \div 6. Dividing gives a decimal, and decimals are compared place by place, like whole numbers (Converting between fractions and decimals shows the division). Here

\frac{5}{6} = 0.8333\ldots \qquad \frac{7}{9} = 0.7777\ldots

The tenths digits already differ: 8 tenths against 7 tenths. So \tfrac{5}{6} is larger. The decimal route shines when one amount is already a decimal, such as a distance of 0.6 mile, or when the denominators are 2, 4, 5, 8, 10, 20 or 25, whose decimals stop quickly.

All three routes must agree, because they are three ways of writing the same two amounts.

Part 3: a list, and the cross product shortcut

Ordering a list. Valeria has three food bins with \tfrac{2}{3}, \tfrac{5}{8} and \tfrac{7}{12} of a bin of food left. To put them in order, give all three one common denominator. Multiples of 12 are 12, 24, and so on; 12 is not a multiple of 8, but 24 is a multiple of 3, 8 and 12. So

\frac{2}{3} = \frac{16}{24} \qquad \frac{5}{8} = \frac{15}{24} \qquad \frac{7}{12} = \frac{14}{24}

Once every fraction counts twenty-fourths, ordering them is ordering 16, 15 and 14. From least to greatest:

\frac{7}{12} < \frac{5}{8} < \frac{2}{3}

Figure 2In twenty-fourths the three fractions are 14, 15 and 16 pieces, so they sit in that order on the line.

On the number line the three dots sit one twenty-fourth apart, which is why a quick look or a benchmark could not tell them apart. For a long list you can also compare two at a time, but one denominator for all of them is usually fewer steps.

A mixed number, such as 2\tfrac{3}{5}, is compared by its whole number part first. Only when the whole numbers match do you compare the fraction parts (Mixed numbers and improper fractions: rewriting and comparing shows the other way, turning both into improper fractions).

The cross product shortcut. Go back to \tfrac{5}{6} and \tfrac{7}{9} with the product denominator 54. To rename \tfrac{5}{6} we multiplied its top by 9, the other fraction’s denominator, and got 45. To rename \tfrac{7}{9} we multiplied its top by 6 and got 42. Both bottoms became 54, so only the tops mattered.

\frac{5}{6} \text{ and } \frac{7}{9}: \qquad 5 \times 9 = 45 \qquad 7 \times 6 = 42

The two products 5 \times 9 and 7 \times 6 are called cross products, because each one crosses from the top of one fraction to the bottom of the other. They are the numerators over the shared denominator 6 \times 9, with that denominator left unwritten. So in general, for fractions \tfrac{a}{b} and \tfrac{c}{d} with b and d positive,

\frac{a}{b} < \frac{c}{d} \text{ exactly when } a \times d < c \times b

and the same holds with > or = in both places. The shortcut is fast, but it is not a new idea, only Part 1 with the bottoms skipped. It also tells you less: it says which is bigger, but not by how much, since 45 - 42 = 3 means \tfrac{3}{54}, which you have to work out.

Worked examples

Examples 1 and 2 belong to Part 1. Examples 3 and 4 belong to Part 2. Examples 5 to 7 belong to Part 3.

Example 1

Denominators with no common factor

Chiyo gives a dog \tfrac{3}{4} of a scoop of food in the morning and \tfrac{5}{7} of a scoop in the evening. Which meal is bigger, and by how much?

Solution

Quarters and sevenths are different sizes, so first make them match. 4 and 7 share no factor other than 1, so the least common denominator is the product, 4 \times 7 = 28. To turn quarters into twenty-eighths, cut each quarter into 7 parts: multiply top and bottom by 7. To turn sevenths into twenty-eighths, cut each seventh into 4 parts: multiply top and bottom by 4.

\frac{3}{4} = \frac{21}{28} \qquad \frac{5}{7} = \frac{20}{28}

The pieces now match, and 21 > 20. So \tfrac{3}{4} > \tfrac{5}{7}: the morning meal is bigger, by \tfrac{1}{28} of a scoop. That is a tiny difference, which is why a quick look could not decide it.

Example 2

The least common denominator saves work

Obinna walked a dog named Pepper \tfrac{7}{10} of a mile on Monday and \tfrac{11}{15} of a mile on Tuesday. Which walk was longer?

Solution

The product 10 \times 15 = 150 would work, but it makes big numbers. Look for something smaller. Multiples of the larger denominator, 15, are 15, 30, and so on. 15 is not a multiple of 10, but 30 is. So 30 is the least common denominator.

\frac{7}{10} = \frac{7 \times 3}{10 \times 3} = \frac{21}{30} \qquad \frac{11}{15} = \frac{11 \times 2}{15 \times 2} = \frac{22}{30}

22 thirtieths is more than 21 thirtieths, so \tfrac{11}{15} > \tfrac{7}{10}. Tuesday’s walk was longer, by \tfrac{1}{30} of a mile.

Check with the product: \tfrac{7}{10} = \tfrac{105}{150} and \tfrac{11}{15} = \tfrac{110}{150}. Again Tuesday wins, by \tfrac{5}{150}, which is the same \tfrac{1}{30} written with smaller pieces.

Example 3

Matching the numerators

Valeria’s two cats eat from bowls of the same size. Mochi ate \tfrac{6}{7} of her bowl and Pip ate \tfrac{9}{10} of his. Which cat ate more?

Solution

The numerators 6 and 9 are both factors of 18, so a common numerator of 18 takes only small multiplications: multiply \tfrac{6}{7} by \tfrac{3}{3} and \tfrac{9}{10} by \tfrac{2}{2}.

\frac{6}{7} = \frac{18}{21} \qquad \frac{9}{10} = \frac{18}{20}

Each cat ate 18 pieces. A twentieth of a bowl is bigger than a twenty-first, because the bowl is cut into fewer parts. So 18 twentieths is more than 18 twenty-firsts, and \tfrac{9}{10} > \tfrac{6}{7}. Pip ate more.

There is a second way to see it. Mochi left \tfrac{1}{7} of her bowl and Pip left \tfrac{1}{10} of his. One tenth is less than one seventh, so Pip left less, which means he ate more. Both views agree.

Example 4

A fraction against a decimal

Obinna’s route to the dog park is \tfrac{7}{8} of a mile. Chiyo’s route is 0.85 mile. Whose route is longer?

Solution

One amount is already a decimal, so turn the other into a decimal too. \tfrac{7}{8} means 7 \div 8, and

7 \div 8 = 0.875

Now line up the places, writing 0.85 as 0.850 so both have three decimal places. The ones digits match (0) and the tenths digits match (8). In the hundredths, 7 > 5. So 0.875 > 0.850, and Obinna’s route is longer, by 0.025 mile.

Check with fractions: 0.85 = \tfrac{85}{100} = \tfrac{17}{20}, and with denominator 40, \tfrac{7}{8} = \tfrac{35}{40} while \tfrac{17}{20} = \tfrac{34}{40}. The same answer.

Example 5

Ordering four food bins

Valeria checks four food bins. They hold \tfrac{4}{5}, \tfrac{7}{9}, \tfrac{11}{15} and \tfrac{13}{18} of a bin. Put the amounts in order from least to greatest.

Solution

All four are between \tfrac{1}{2} and 1, so benchmarks will not sort them. Find one denominator for all four. Multiples of the largest denominator, 18, are 18, 36, 54, 72, 90. The first that 5, 9 and 15 all divide is 90. Rename each fraction, multiplying top and bottom by what turns its denominator into 90:

\begin{aligned} \frac{4}{5} &= \frac{4 \times 18}{5 \times 18} = \frac{72}{90} & \frac{7}{9} &= \frac{7 \times 10}{9 \times 10} = \frac{70}{90} \\ \frac{11}{15} &= \frac{11 \times 6}{15 \times 6} = \frac{66}{90} & \frac{13}{18} &= \frac{13 \times 5}{18 \times 5} = \frac{65}{90} \end{aligned}

Now order the numerators: 65 < 66 < 70 < 72. So from least to greatest,

\frac{13}{18} < \frac{11}{15} < \frac{7}{9} < \frac{4}{5}

The two smallest differ by just \tfrac{1}{90} of a bin, close enough that only matching pieces could separate them.

Example 6

Cross products on the bike trail

Chiyo has ridden \tfrac{7}{11} of the bike trail. Obinna has ridden \tfrac{5}{8} of the same trail. Who has ridden farther?

Solution

Use cross products. Multiply each numerator by the other fraction’s denominator:

7 \times 8 = 56 \qquad 5 \times 11 = 55

56 > 55, so \tfrac{7}{11} > \tfrac{5}{8}, and Chiyo has ridden farther.

Here is why the shortcut is allowed. With the common denominator 11 \times 8 = 88,

\frac{7}{11} = \frac{7 \times 8}{11 \times 8} = \frac{56}{88} \qquad \frac{5}{8} = \frac{5 \times 11}{8 \times 11} = \frac{55}{88}

The cross products are exactly these numerators. Chiyo is ahead by \tfrac{1}{88} of the trail.

Example 7

Which marks are enough?

Chiyo’s measuring cup is marked in eighths of a cup, from \tfrac{1}{8} up to \tfrac{8}{8}. A bag of puppy food says each meal must be more than \tfrac{2}{3} of a cup. Which marks give more than \tfrac{2}{3} cup?

Solution

Call the mark \tfrac{n}{8}. Compare it with \tfrac{2}{3} using the common denominator 8 \times 3 = 24:

\frac{n}{8} = \frac{3n}{24} \qquad \frac{2}{3} = \frac{16}{24}

So \tfrac{n}{8} is more than \tfrac{2}{3} exactly when 3n is more than 16. Try values: 3 \times 5 = 15, too small, and 3 \times 6 = 18, enough. Larger n make 3n larger still. So the marks \tfrac{6}{8}, \tfrac{7}{8} and \tfrac{8}{8} all work, and \tfrac{6}{8} is the smallest of them. It is more than \tfrac{2}{3} by \tfrac{2}{24}, which is \tfrac{1}{12} of a cup.

Boundaries

The same whole. Comparing \tfrac{5}{6} and \tfrac{7}{9} tells you which amount is more only when both are fractions of the same whole. Five sixths of a small bowl can be less food than seven ninths of a big one. Every comparison in this lesson assumes equal wholes.
Equal fractions. If the renamed numerators match, the fractions are equal, such as \tfrac{6}{8} = \tfrac{18}{24} and \tfrac{9}{12} = \tfrac{18}{24}. Different-looking fractions can name the same amount.
Positive denominators for the cross product rule. The rule \tfrac{a}{b} < \tfrac{c}{d} exactly when a \times d < c \times b needs b and d to be positive. Every denominator in this lesson is. With negative numbers the comparison is done differently (Comparing and ordering rational numbers).
Two at a time. Cross products compare two fractions. For a list you either compare pair by pair or give the whole list one common denominator.
Rounded decimals can tie. \tfrac{2}{3} = 0.6666\ldots rounds to 0.67, but it is less than 0.67. If two rounded decimals come out equal, the rounding has hidden the answer: use more places, or go back to fractions.

Where this goes next

Making the pieces match is the first step of adding and subtracting fractions with unlike denominators (Adding and subtracting fractions with unlike denominators): you cannot add sixths to ninths until both are eighteenths, just as you could not compare them. Comparing also feeds estimation (Estimating with fractions), where knowing that \tfrac{7}{12} is just over \tfrac{1}{2} helps judge a sum.

Later, the same ideas place fractions, decimals and negative numbers on one number line (Comparing and ordering rational numbers). The cross products come back too. Two fractions are equal exactly when their cross products are equal, and that fact is how a proportion is solved (Solving a proportion).

Exercises

Part 1: rename and compare

Write <, > or = between each pair, and show the renamed fractions.

\tfrac{2}{3} and \tfrac{3}{5}
\tfrac{3}{8} and \tfrac{5}{12}
\tfrac{4}{7} and \tfrac{5}{9}
\tfrac{7}{10} and \tfrac{3}{4}
\tfrac{5}{12} and \tfrac{4}{9}
\tfrac{8}{15} and \tfrac{11}{20}

Part 1: at the pet sitting club

Chiyo’s cat ate \tfrac{2}{3} of a can of food. Obinna’s cat ate \tfrac{5}{7} of a can of the same size. Which cat ate more, and by how much?
Valeria compares \tfrac{7}{12} of a bag of dog food with \tfrac{5}{18} of the same bag. Rename both fractions twice, once with the product of the denominators and once with the least common denominator, and show that both give the same answer.

Part 2: common numerators

Compare each pair by giving both fractions the same numerator.

\tfrac{3}{7} and \tfrac{6}{13}
\tfrac{4}{9} and \tfrac{2}{5}
\tfrac{10}{11} and \tfrac{5}{6}

Part 2: decimals

Compare \tfrac{3}{8} and \tfrac{2}{5} by changing both to decimals.
On the bike trail there is a water fountain 0.7 mile from the start and a bench \tfrac{17}{25} of a mile from the start. Which is farther from the start?
Is \tfrac{5}{9} more or less than 0.56?
Obinna’s dog walk was \tfrac{11}{20} of a mile and Valeria’s was \tfrac{13}{25} of a mile. Whose walk was longer? Use decimals.

Part 3: ordering lists

Order from least to greatest: \tfrac{1}{2}, \tfrac{3}{5}, \tfrac{4}{7}.
Order from least to greatest: \tfrac{3}{4}, \tfrac{7}{10}, \tfrac{5}{7}.
Four dogs ate \tfrac{9}{10}, \tfrac{7}{8}, \tfrac{11}{12} and \tfrac{13}{15} of their bowls, all the same size. Order the amounts from greatest to least.
Valeria walked 2\tfrac{3}{5} miles on Monday and 2\tfrac{4}{7} miles on Tuesday. Which walk was longer?

Part 3: cross products

Use cross products to write <, > or =.

\tfrac{5}{7} and \tfrac{8}{11}
\tfrac{13}{17} and \tfrac{10}{13}
\tfrac{6}{9} and \tfrac{8}{12}
Which whole numbers n make \tfrac{n}{12} less than \tfrac{4}{7}?

Reasoning

Explain why comparing 5 \times 11 with 8 \times 7 tells you which of \tfrac{5}{7} and \tfrac{8}{11} is larger. Your explanation should say what the two products count.
Chiyo always uses the least common denominator. Valeria always uses the product of the denominators. Explain why they always reach the same answer when they compare two fractions.

Answers

\tfrac{2}{3} > \tfrac{3}{5}. With denominator 15: \tfrac{2}{3} = \tfrac{10}{15} and \tfrac{3}{5} = \tfrac{9}{15}, and 10 > 9.
\tfrac{3}{8} < \tfrac{5}{12}. The least common denominator is 24: \tfrac{3}{8} = \tfrac{9}{24} and \tfrac{5}{12} = \tfrac{10}{24}.
\tfrac{4}{7} > \tfrac{5}{9}. With denominator 63: \tfrac{4}{7} = \tfrac{36}{63} and \tfrac{5}{9} = \tfrac{35}{63}.
\tfrac{7}{10} < \tfrac{3}{4}. With denominator 20: \tfrac{7}{10} = \tfrac{14}{20} and \tfrac{3}{4} = \tfrac{15}{20}.
\tfrac{5}{12} < \tfrac{4}{9}. With denominator 36: \tfrac{5}{12} = \tfrac{15}{36} and \tfrac{4}{9} = \tfrac{16}{36}.
\tfrac{8}{15} < \tfrac{11}{20}. With denominator 60: \tfrac{8}{15} = \tfrac{32}{60} and \tfrac{11}{20} = \tfrac{33}{60}.
Obinna’s cat, by \tfrac{1}{21} of a can. With denominator 21: \tfrac{2}{3} = \tfrac{14}{21} and \tfrac{5}{7} = \tfrac{15}{21}.
\tfrac{7}{12} > \tfrac{5}{18} both ways. Product 216: \tfrac{7}{12} = \tfrac{126}{216} and \tfrac{5}{18} = \tfrac{60}{216}. Least common denominator 36: \tfrac{7}{12} = \tfrac{21}{36} and \tfrac{5}{18} = \tfrac{10}{36}.
\tfrac{3}{7} < \tfrac{6}{13}. \tfrac{3}{7} = \tfrac{6}{14}, and fourteenths are smaller than thirteenths, so 6 fourteenths is less than 6 thirteenths.
\tfrac{4}{9} > \tfrac{2}{5}. \tfrac{2}{5} = \tfrac{4}{10}, and ninths are bigger than tenths.
\tfrac{10}{11} > \tfrac{5}{6}. \tfrac{5}{6} = \tfrac{10}{12}, and elevenths are bigger than twelfths.
\tfrac{3}{8} < \tfrac{2}{5}. 3 \div 8 = 0.375 and 2 \div 5 = 0.4, and 0.375 < 0.400.
The water fountain. \tfrac{17}{25} = 0.68, and 0.68 < 0.70.
Less. 5 \div 9 = 0.5555\ldots, and in the hundredths place 5 < 6, so 0.5555\ldots < 0.56.
Obinna’s. \tfrac{11}{20} = 0.55 and \tfrac{13}{25} = 0.52, and 0.55 > 0.52.
\tfrac{1}{2} < \tfrac{4}{7} < \tfrac{3}{5}. With denominator 70 the numerators are 35, 42 and 40 for \tfrac{1}{2}, \tfrac{3}{5} and \tfrac{4}{7}.
\tfrac{7}{10} < \tfrac{5}{7} < \tfrac{3}{4}. With denominator 140: \tfrac{3}{4} = \tfrac{105}{140}, \tfrac{7}{10} = \tfrac{98}{140}, \tfrac{5}{7} = \tfrac{100}{140}.
\tfrac{11}{12} > \tfrac{9}{10} > \tfrac{7}{8} > \tfrac{13}{15}. With denominator 120 the numerators are 110, 108, 105 and 104.
Monday’s. The whole numbers match, so compare \tfrac{3}{5} and \tfrac{4}{7}: with denominator 35 they are \tfrac{21}{35} and \tfrac{20}{35}.
\tfrac{5}{7} < \tfrac{8}{11}. 5 \times 11 = 55 and 8 \times 7 = 56, and 55 < 56.
\tfrac{13}{17} < \tfrac{10}{13}. 13 \times 13 = 169 and 10 \times 17 = 170.
\tfrac{6}{9} = \tfrac{8}{12}. 6 \times 12 = 72 and 8 \times 9 = 72. (Both equal \tfrac{2}{3}.)
n = 0, 1, 2, 3, 4, 5, 6. With denominator 84, \tfrac{n}{12} = \tfrac{7n}{84} and \tfrac{4}{7} = \tfrac{48}{84}, so we need 7n < 48. 7 \times 6 = 42 works and 7 \times 7 = 49 does not.
Rename both fractions over 7 \times 11 = 77. To do it, \tfrac{5}{7} is multiplied top and bottom by 11, giving \tfrac{5 \times 11}{77} = \tfrac{55}{77}, and \tfrac{8}{11} is multiplied top and bottom by 7, giving \tfrac{8 \times 7}{77} = \tfrac{56}{77}. So the two products count seventy-sevenths in each fraction. The pieces are the same size, so the larger count is the larger fraction.
Both renamings cut the whole into equal pieces of one size, so both turn the comparison into a count of matching pieces. The product is always a multiple of the least common denominator, so Valeria’s pieces are Chiyo’s pieces each cut into the same number of smaller parts. That multiplies both of Chiyo’s counts by the same number, which keeps the larger count larger. For example, with \tfrac{5}{6} and \tfrac{7}{9}, Chiyo gets 15 and 14 eighteenths and Valeria gets 45 and 42 fifty-fourths, each count three times Chiyo’s.