Dividing by multiplying by the reciprocal
A jug holds six liters of sports drink and each bottle takes three quarters of a liter. How do you find how many bottles it fills, quickly, whatever the fractions are?
Where you are 6 in Dividing fractions
When you reach for it
At the drink station of a cycling race, riders grab small bottles of sports drink as they pass, and the volunteers fill those bottles from big jugs. Valentina has a jug holding 6 liters, and each bottle holds \displaystyle \frac{3}{4} of a liter. How many bottles can she fill?
That is a division, \displaystyle 6 \div \frac{3}{4}, in the “how many fit” sense of dividing fractions: how many fit. You could draw the jug as a bar, cut it into quarter liters and count groups of three: 24 quarters make 8 groups, so 8 bottles. Drawing works for this one. It is slow for \displaystyle \frac{8}{15} \div \frac{4}{9}, and the riders will not wait.
This lesson is the fast route. Dividing by a fraction gives the same answer as multiplying by that fraction turned upside down, so every fraction division becomes a fraction multiplication, which you already know how to do. Reach for it whenever a division has a fraction in it: a fraction divided by a fraction, a whole number divided by a fraction, or a fraction divided by a whole number. It also answers the sharing question of dividing fractions: sharing into equal parts, since both meanings are the same division.
The lesson runs over two days. Day 1 is the move itself: the reciprocal, the rule, and whole numbers. Day 2 makes it quick and safe: simplifying before you multiply, judging the size of the answer before you start, and checking it by multiplying back.
The procedure
Day 1: the reciprocal, and the rule
The reciprocal. Two numbers are reciprocals of each other when their product is 1. The reciprocal of a fraction \displaystyle \frac{c}{d}, with c and d both not 0, is \displaystyle \frac{d}{c}, the same fraction turned upside down, because
A whole number n is the fraction \displaystyle \frac{n}{1}, so its reciprocal is \displaystyle \frac{1}{n}. Read the other way, the reciprocal of a unit fraction \displaystyle \frac{1}{n} is the whole number n. The number 1 is its own reciprocal, since 1 \times 1 = 1. And 0 has no reciprocal at all, because 0 times any number is 0, never 1.
| Number | \displaystyle \frac{2}{5} | \displaystyle \frac{9}{4} | 3 | \displaystyle \frac{1}{8} | 1 |
|---|---|---|---|---|---|
| Reciprocal | \displaystyle \frac{5}{2} | \displaystyle \frac{4}{9} | \displaystyle \frac{1}{3} | 8 | 1 |
The rule. For whole numbers a, b, c, d, with none of b, c, d equal to 0,
In words: dividing by a number is the same as multiplying by its reciprocal. The reason, in one paragraph: a bottle of \displaystyle \frac{c}{d} liter uses c of the d equal pieces that make up one liter, so one liter fills \displaystyle \frac{d}{c} bottles. A jug of \displaystyle \frac{a}{b} liter fills that many bottles for every liter in it, which is \displaystyle \frac{a}{b} \times \frac{d}{c} bottles. The full argument, made three different ways, is in why flipping the second fraction works. This lesson is about using it.
Doing it. Take \displaystyle \frac{2}{3} \div \frac{4}{7}. Leave the first fraction exactly as it is: it is the amount being measured, and nothing about it changes. Change the division to a multiplication and, in the same step, replace the second fraction by its reciprocal. The rule says these two changes, made together, keep the value. Only the divisor flips.
Now it is an ordinary product: numerator times numerator over denominator times denominator, as in multiplying a fraction by a fraction.
The last step puts the answer in simplest form by dividing the top and the bottom by their common factor 2, which gives an equivalent fraction. An answer bigger than 1, such as \displaystyle \frac{7}{6}, is fine left as an improper fraction, and this lesson leaves its answers that way. To say what it means in a story, split off the wholes: \displaystyle \frac{7}{6} = \frac{6}{6} + \frac{1}{6} = 1\frac{1}{6}, one whole bottle and one sixth of another.
Whole numbers. Write a whole number as a fraction over 1, and the same rule applies. Valentina’s jug gives
bottles, which matches the count of quarter liters. When the whole number is the divisor, you multiply by its reciprocal \displaystyle \frac{1}{n}, so dividing by n takes \displaystyle \frac{1}{n} of the amount. Sharing \displaystyle \frac{9}{10} of a liter equally among 3 bottles gives each
of a liter.
Day 2: simplify first, estimate, check
Simplify before you multiply. Once the division has become a multiplication, the answer is one fraction: the product of the top numbers over the product of the bottom numbers. A factor shared by any top number and any bottom number can be divided out of that fraction, wherever the two numbers sit, because dividing the numerator and the denominator by the same number gives an equivalent fraction. Dividing out first keeps the numbers small, and the answer arrives already in simplest form. In \displaystyle \frac{5}{12} \div \frac{10}{9}, flip and then look for shared factors: 5 goes into 5 and 10, and 3 goes into 9 and 12.
In this lesson the order is always the same: flip first, then simplify. After the flip the problem is a product, and the rule “any top number with any bottom number” holds. Before the flip it does not: the top of the first fraction and the bottom of the divisor both end up on top, so a factor they share cannot be divided out (Exercise 23 shows what goes wrong). Some shared factors can safely be divided out before the flip, but sorting out which ones is a second rule to carry. Flipping first means only one rule is ever needed.
Estimate first. Two quick judgments catch most slips before any arithmetic is done.
The first compares the answer with the number being divided. Dividing by b is multiplying by \displaystyle \frac{1}{b}, and multiplying as scaling says that multiplying by a number greater than 1 makes a number larger, while multiplying by a number less than 1 makes it smaller. The reciprocal of a number less than 1 is greater than 1 (a small bottle fits into a liter many times), and the reciprocal of a number greater than 1 is less than 1. So, for any positive number a:
- dividing by a number less than 1 gives an answer larger than a;
- dividing by 1 gives a itself;
- dividing by a number greater than 1 gives an answer smaller than a.
The second compares the two numbers with each other. If the jug holds more than one bottle’s worth, the answer is more than 1; if it holds less than one bottle, the answer is less than 1.
For a closer estimate, round the fractions to nearby friendly numbers and divide those. Figure 1 shows \displaystyle \frac{5}{6} \div \frac{1}{4}. Since \displaystyle \frac{5}{6} is a little under 1, and a quarter-liter bottle fits into one liter 4 times, expect a little under 4. The bar shows three full bottles and a third of a fourth. The rule agrees: \displaystyle \frac{5}{6} \times \frac{4}{1} = \frac{20}{6} = \frac{10}{3}, and \displaystyle \frac{10}{3} = 3 + \frac{1}{3} = 3\frac{1}{3}.
Check by multiplying. Multiplication and division undo each other (multiplication and division undo each other): a \div b = q means exactly q \times b = a. So multiply your answer by the divisor, and you must get back the number you divided. For the figure, \displaystyle \frac{10}{3} \times \frac{1}{4} = \frac{10}{12} = \frac{5}{6}, as it should be. A wrong answer cannot pass this check, because only one number times \displaystyle \frac{1}{4} gives \displaystyle \frac{5}{6}.
Worked examples
Day 1: reciprocals
(a) Write the reciprocal of \displaystyle \frac{4}{9}, of 7 and of \displaystyle \frac{1}{6}.
(b) Show that each number times its reciprocal is 1.
Solution(a) Turn each fraction upside down. The reciprocal of \displaystyle \frac{4}{9} is \displaystyle \frac{9}{4}. Since \displaystyle 7 = \frac{7}{1}, its reciprocal is \displaystyle \frac{1}{7}. The reciprocal of \displaystyle \frac{1}{6} is \displaystyle \frac{6}{1} = 6.
(b) Multiply each pair, numerator times numerator and denominator times denominator:
Each product is 1, which is exactly what being reciprocals means. The top of one number is always the bottom of the other, so the product is always some number over itself.
Day 1: two fractions
Find (a) \displaystyle \frac{5}{6} \div \frac{2}{9} and (b) \displaystyle \frac{3}{10} \div \frac{4}{5}. Give each answer in simplest form.
Solution(a) Keep \displaystyle \frac{5}{6}, change the division to multiplication, and flip the divisor \displaystyle \frac{2}{9} to \displaystyle \frac{9}{2}. Then multiply across:
The common factor of 45 and 12 is 3, and dividing both by it gives \displaystyle \frac{15}{4}. The answer is more than 1 because \displaystyle \frac{5}{6} is more than \displaystyle \frac{2}{9}: the jug holds more than one bottle.
(b) Flip \displaystyle \frac{4}{5} to \displaystyle \frac{5}{4}:
Here the common factor is 5. This answer is less than 1, because \displaystyle \frac{3}{10} is less than \displaystyle \frac{4}{5}: there is not enough drink for one whole bottle, only \displaystyle \frac{3}{8} of one.
Day 1: a whole number on either side
Hiro has a jug holding 4 liters and bottles that hold \displaystyle \frac{2}{5} of a liter.
(a) How many bottles can he fill?
(b) He also has \displaystyle \frac{2}{5} of a liter of drink concentrate left, which he shares equally among 4 jugs. How much does each jug get?
Solution(a) The question is how many \displaystyle \frac{2}{5}-liter bottles fit into 4 liters, so divide. Write 4 as \displaystyle \frac{4}{1} and flip the divisor:
Hiro fills 10 bottles.
(b) Now \displaystyle \frac{2}{5} of a liter is shared into 4 equal parts, so divide by 4. The reciprocal of 4 is \displaystyle \frac{1}{4}:
Each jug gets \displaystyle \frac{1}{10} of a liter. Notice that the two questions use the same two numbers in opposite orders, and the answers 10 and \displaystyle \frac{1}{10} are reciprocals. Swapping the numbers in a division always does this (Exercise 21 asks why), so the order matters: the amount being divided comes first.
Day 2: simplifying before multiplying
Find (a) \displaystyle \frac{8}{15} \div \frac{4}{9} and (b) \displaystyle 12 \div \frac{8}{3}.
Solution(a) Flip first: \displaystyle \frac{8}{15} \times \frac{9}{4}. Now look across the whole product for a top number and a bottom number with a common factor. 4 goes into 8 twice and into 4 once, and 3 goes into 9 three times and into 15 five times:
Multiplying out first gives the same answer the long way, \displaystyle \frac{72}{60}, which then needs its common factor 12 found and divided out. Simplifying first meant smaller numbers and no hunt for a large common factor at the end.
(b) Write 12 as \displaystyle \frac{12}{1} and flip \displaystyle \frac{8}{3} to \displaystyle \frac{3}{8}. The 12 and the 8 share the factor 4:
The divisor \displaystyle \frac{8}{3} is more than 1, so the answer should be less than 12, and \displaystyle \frac{9}{2} is. Check: \displaystyle \frac{9}{2} \times \frac{8}{3} = \frac{72}{6} = 12.
Day 2: estimate, divide, check
Laila’s jug has \displaystyle \frac{9}{10} of a liter left, and the bottles on her table hold \displaystyle \frac{3}{8} of a liter. Estimate how many bottles she can fill, then find the exact number and check it.
SolutionEstimate. \displaystyle \frac{9}{10} is just under 1 liter, and \displaystyle \frac{3}{8} is just under \displaystyle \frac{1}{2}, which fits into one liter twice. So expect about 2 bottles. Two safer facts back this up: the divisor is less than 1, so the answer is more than \displaystyle \frac{9}{10}, and the jug holds more than one bottle, so the answer is more than 1.
Divide. Flip \displaystyle \frac{3}{8} to \displaystyle \frac{8}{3}, then simplify across: 3 goes into 9 and 3, and 2 goes into 8 and 10:
Check. Multiply the answer by the divisor: \displaystyle \frac{12}{5} \times \frac{3}{8} = \frac{36}{40} = \frac{9}{10}, the amount in the jug.
Since \displaystyle \frac{12}{5} = \frac{10}{5} + \frac{2}{5} = 2 + \frac{2}{5} = 2\frac{2}{5}, Laila fills 2 bottles and a third bottle \displaystyle \frac{2}{5} full. That agrees with the estimate of about 2.
Day 2: a missing factor
Find the number that, multiplied by \displaystyle \frac{3}{4}, gives \displaystyle \frac{9}{14}.
SolutionCall the number n, so \displaystyle n \times \frac{3}{4} = \frac{9}{14}. By the meaning of division, the number that \displaystyle \frac{3}{4} must be multiplied by to give \displaystyle \frac{9}{14} is \displaystyle \frac{9}{14} \div \frac{3}{4}. Flip and simplify: 3 goes into 9 and 3, and 2 goes into 4 and 14.
Check: \displaystyle \frac{6}{7} \times \frac{3}{4} = \frac{18}{28} = \frac{9}{14}. The same move solves an equation such as \displaystyle \frac{3}{4}n = \frac{9}{14}: multiplying both sides by \displaystyle \frac{4}{3}, the reciprocal of \displaystyle \frac{3}{4}, leaves n alone, which is one of the properties of equality at work.
Boundaries
The divisor cannot be 0. The rule needs the reciprocal of the divisor, and 0 has none, so \displaystyle \frac{2}{3} \div 0 has no value. The number being divided may be 0: \displaystyle 0 \div \frac{3}{4} = 0 \times \frac{4}{3} = 0, since an empty jug fills no bottles.
Only the divisor flips. Division is not commutative: \displaystyle \frac{2}{5} \div 4 and \displaystyle 4 \div \frac{2}{5} are different numbers, reciprocals of each other, as Example 3 found. Flipping the first fraction instead of the second produces the reciprocal of the right answer.
Simplify after flipping. Dividing a top number and a bottom number by a common factor is allowed in a product. Across a division sign, the top of the first fraction and the bottom of the divisor are not such a pair: dividing both by the same number changes the answer, as Exercise 23 shows. That is why this lesson always flips first.
Fractions and whole numbers here. A mixed number has to become an improper fraction before it can be flipped, which is dividing mixed numbers. With negative numbers the same rule holds, and the sign of the answer follows the rules in multiplying and dividing rational numbers.
The size predictions are for positive numbers. “Dividing by less than 1 makes it larger” is true for positive amounts, which is all this lesson uses.
The answer is a number; the story decides what it means. \displaystyle \frac{12}{5} bottles is two full bottles and part of a third. Whether the part counts depends on the question, and choosing the right division for a situation is the work of word problems that divide fractions.
Where this goes next
Mixed numbers come next: dividing mixed numbers rewrites them as improper fractions and then uses exactly this move. The same division finds a whole from a part: if \displaystyle \frac{3}{4} of a jug is 6 liters, the full jug is \displaystyle 6 \div \frac{3}{4} = 8 liters, which is finding the whole from a fraction. Deciding which number to divide by which, in a story, is word problems that divide fractions.
Further on, a rate such as \displaystyle \frac{1}{2} liter in \displaystyle \frac{3}{4} of a minute becomes a unit rate by this division: \displaystyle \frac{1}{2} \div \frac{3}{4} = \frac{2}{3} of a liter per minute. And “multiply by the reciprocal” is how equations such as \displaystyle \frac{2}{3}x = 10 are solved, by multiplying both sides by \displaystyle \frac{3}{2}.
Exercises
Day 1: reciprocals and the rule
- Write the reciprocal of \displaystyle \frac{3}{7}, of 9, of \displaystyle \frac{1}{12} and of \displaystyle \frac{11}{4}.
- \displaystyle \frac{1}{2} \div \frac{3}{5}
- \displaystyle \frac{3}{4} \div \frac{2}{7}
- \displaystyle \frac{4}{5} \div \frac{7}{10}
- \displaystyle \frac{2}{9} \div \frac{5}{6}
Day 1: whole numbers
- \displaystyle 8 \div \frac{2}{3}
- \displaystyle \frac{5}{7} \div 3
- \displaystyle \frac{6}{11} \div 6
- Valentina has a 5-liter jug and bottles that hold \displaystyle \frac{5}{8} of a liter. How many bottles can she fill?
- Esi shares \displaystyle \frac{2}{3} of a liter of syrup equally among 4 bottles. How much syrup goes into each bottle?
Day 2: simplifying first
Simplify before you multiply.
- \displaystyle \frac{10}{21} \div \frac{5}{7}
- \displaystyle \frac{9}{16} \div \frac{3}{8}
- \displaystyle \frac{14}{15} \div \frac{7}{20}
- \displaystyle 12 \div \frac{9}{10}
Day 2: estimating and checking
- Without working them out, say whether each quotient is larger than, smaller than or equal to \displaystyle \frac{3}{5}. Then find each one. (a) \displaystyle \frac{3}{5} \div \frac{1}{4} (b) \displaystyle \frac{3}{5} \div \frac{4}{3} (c) \displaystyle \frac{3}{5} \div 1
- Without working them out, put these in order from smallest to largest, then find each one to confirm: \displaystyle \frac{5}{6} \div \frac{1}{2}, \displaystyle \frac{5}{6} \div 2, \displaystyle \frac{5}{6} \div \frac{2}{3}.
- Estimate \displaystyle \frac{11}{12} \div \frac{1}{5} by rounding, then find it exactly.
- Pavel says \displaystyle \frac{5}{8} \div \frac{2}{3} = \frac{5}{12}. Check his answer by multiplying. If it is wrong, find the right one and check that.
- Callum has \displaystyle \frac{7}{8} of a liter left in his jug and bottles that hold \displaystyle \frac{1}{4} of a liter. How many full bottles can he fill, and how full is the last one?
- Find the number that, multiplied by \displaystyle \frac{5}{6}, gives \displaystyle \frac{10}{9}.
Reasoning
- \displaystyle \frac{3}{4} \div 6 = \frac{1}{8} and \displaystyle 6 \div \frac{3}{4} = 8. Explain why swapping the two numbers in a division always turns the answer into its reciprocal.
- Callum works out \displaystyle \frac{2}{5} \div \frac{3}{4} by flipping the first fraction: \displaystyle \frac{5}{2} \times \frac{3}{4} = \frac{15}{8}. Show by multiplying back that this is wrong, find the right answer, and say how his answer is related to it.
- Esi works out \displaystyle \frac{3}{5} \div \frac{2}{9} by dividing the 3 and the 9 by 3 before flipping, which gives \displaystyle \frac{1}{5} \div \frac{2}{3} = \frac{3}{10}. Find the right answer and explain what her early step changed.
- Explain why dividing a positive number by a fraction less than 1 always gives an answer larger than the number you started with.
Answers
- \displaystyle \frac{7}{3}, \displaystyle \frac{1}{9}, 12 and \displaystyle \frac{4}{11}. Each times its number is 1, for example \displaystyle \frac{11}{4} \times \frac{4}{11} = \frac{44}{44} = 1.
- \displaystyle \frac{5}{6}. \displaystyle \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}.
- \displaystyle \frac{21}{8}. \displaystyle \frac{3}{4} \times \frac{7}{2} = \frac{21}{8}.
- \displaystyle \frac{8}{7}. \displaystyle \frac{4}{5} \times \frac{10}{7} = \frac{40}{35}, and dividing top and bottom by 5 gives \displaystyle \frac{8}{7}.
- \displaystyle \frac{4}{15}. \displaystyle \frac{2}{9} \times \frac{6}{5} = \frac{12}{45}, and dividing by 3 gives \displaystyle \frac{4}{15}.
- 12. \displaystyle \frac{8}{1} \times \frac{3}{2} = \frac{24}{2} = 12.
- \displaystyle \frac{5}{21}. \displaystyle \frac{5}{7} \times \frac{1}{3} = \frac{5}{21}.
- \displaystyle \frac{1}{11}. \displaystyle \frac{6}{11} \times \frac{1}{6} = \frac{6}{66} = \frac{1}{11}.
- 8 bottles. \displaystyle 5 \div \frac{5}{8} = \frac{5}{1} \times \frac{8}{5} = \frac{40}{5} = 8.
- \displaystyle \frac{1}{6} of a liter. \displaystyle \frac{2}{3} \div 4 = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}.
- \displaystyle \frac{2}{3}. \displaystyle \frac{10}{21} \times \frac{7}{5}: 5 goes into 10 and 5, 7 into 7 and 21, leaving \displaystyle \frac{2 \times 1}{3 \times 1}.
- \displaystyle \frac{3}{2}. \displaystyle \frac{9}{16} \times \frac{8}{3}: 3 goes into 9 and 3, 8 into 8 and 16, leaving \displaystyle \frac{3 \times 1}{2 \times 1}.
- \displaystyle \frac{8}{3}. \displaystyle \frac{14}{15} \times \frac{20}{7}: 7 goes into 14 and 7, 5 into 20 and 15, leaving \displaystyle \frac{2 \times 4}{3 \times 1}.
- \displaystyle \frac{40}{3}. \displaystyle \frac{12}{1} \times \frac{10}{9}: 3 goes into 12 and 9, leaving \displaystyle \frac{4 \times 10}{1 \times 3}.
- (a) Larger, since \displaystyle \frac{1}{4} < 1: \displaystyle \frac{3}{5} \times 4 = \frac{12}{5}. (b) Smaller, since \displaystyle \frac{4}{3} > 1: \displaystyle \frac{3}{5} \times \frac{3}{4} = \frac{9}{20}. (c) Equal: dividing by 1 gives \displaystyle \frac{3}{5}.
- \displaystyle \frac{5}{6} \div 2 = \frac{5}{12}, then \displaystyle \frac{5}{6} \div \frac{2}{3} = \frac{5}{4}, then \displaystyle \frac{5}{6} \div \frac{1}{2} = \frac{5}{3}. The amount is the same each time, so the smaller the divisor, the more times it fits, and \displaystyle \frac{1}{2} < \frac{2}{3} < 2.
- About 5, since \displaystyle \frac{11}{12} is close to 1 and \displaystyle 1 \div \frac{1}{5} = 5. Exactly, \displaystyle \frac{11}{12} \times 5 = \frac{55}{12}, a little under 5 as expected.
- Wrong: \displaystyle \frac{5}{12} \times \frac{2}{3} = \frac{10}{36} = \frac{5}{18}, not \displaystyle \frac{5}{8}. The right answer is \displaystyle \frac{15}{16}: \displaystyle \frac{5}{8} \times \frac{3}{2} = \frac{15}{16}, and \displaystyle \frac{15}{16} \times \frac{2}{3} = \frac{30}{48} = \frac{5}{8}. Pavel multiplied by \displaystyle \frac{2}{3} instead of dividing; his answer is smaller than \displaystyle \frac{5}{8}, which dividing by a number less than 1 can never give.
- 3 full bottles, and a fourth one half full. \displaystyle \frac{7}{8} \div \frac{1}{4} = \frac{7}{8} \times 4 = \frac{28}{8} = \frac{7}{2}, and \displaystyle \frac{7}{2} = 3 + \frac{1}{2} = 3\frac{1}{2}.
- \displaystyle \frac{4}{3}. The number is \displaystyle \frac{10}{9} \div \frac{5}{6} = \frac{10}{9} \times \frac{6}{5}; 5 goes into 10 and 5, 3 into 6 and 9, leaving \displaystyle \frac{2 \times 2}{3 \times 1}. Check: \displaystyle \frac{4}{3} \times \frac{5}{6} = \frac{20}{18} = \frac{10}{9}.
- For non-zero a and b, \displaystyle a \div b = a \times \frac{1}{b} and \displaystyle b \div a = b \times \frac{1}{a}. Their product is \displaystyle \frac{a \times b}{b \times a} = 1, and two numbers whose product is 1 are reciprocals. With fractions it shows directly: \displaystyle \frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc} and \displaystyle \frac{c}{d} \div \frac{a}{b} = \frac{cb}{da}, the same fraction upside down.
- The right answer is \displaystyle \frac{8}{15}. His check fails: \displaystyle \frac{15}{8} \times \frac{3}{4} = \frac{45}{32}, not \displaystyle \frac{2}{5}. Flipping the divisor gives \displaystyle \frac{2}{5} \times \frac{4}{3} = \frac{8}{15}, and \displaystyle \frac{8}{15} \times \frac{3}{4} = \frac{24}{60} = \frac{2}{5}. His answer \displaystyle \frac{15}{8} is the reciprocal of the right one. Each of his two factors is the reciprocal of the matching factor in the right product, \displaystyle \frac{5}{2} in place of \displaystyle \frac{2}{5} and \displaystyle \frac{3}{4} in place of \displaystyle \frac{4}{3}, so his product times the right product is 1: \displaystyle \frac{15}{8} \times \frac{8}{15} = 1.
- The right answer is \displaystyle \frac{27}{10}: \displaystyle \frac{3}{5} \times \frac{9}{2} = \frac{27}{10}, and \displaystyle \frac{27}{10} \times \frac{2}{9} = \frac{54}{90} = \frac{3}{5}. Her \displaystyle \frac{3}{10} fails the check: \displaystyle \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}. Dividing the 9 by 3 before flipping changed the divisor from \displaystyle \frac{2}{9} to \displaystyle \frac{2}{3}, a bottle three times as big, and she divided the 3 by 3 as well, so her answer is 9 times too small. After the flip the 9 is a top number and the 3 is also a top number, so they cannot cancel.
- Dividing by b is multiplying by its reciprocal \displaystyle \frac{1}{b}. If b is a positive fraction less than 1, its reciprocal is greater than 1: written as \displaystyle \frac{c}{d} with c < d, the reciprocal \displaystyle \frac{d}{c} has a top larger than its bottom. Multiplying a positive number by a number greater than 1 makes it larger. In the drink station’s terms, a bottle smaller than a liter fits into each liter more than once, so the number of bottles is more than the number of liters.