Chapter 7 · Add and subtract fractions

Compare Fraction Sums and Differences and Fix Mistakes

Use benchmarks to compare sums and differences of fractions, and compute when benchmarks cannot decide. Then find, prove wrong, and fix common fraction mistakes.

12 min read

You can already add and subtract fractions such as $\frac{1}{2} + \frac{1}{3}$ , where the denominators are different. The denominator is the bottom number. It tells how many equal parts make the whole. The numerator is the top number. It counts how many of those parts you have.

In this lesson you use that skill in two new ways.

Compare results. A sum is the answer when you add. A difference is the answer when you subtract. You will decide which of two sums or differences is greater.
Catch mistakes. You will spot a wrong answer, show that it cannot be right, and fix it. Often you can tell that an answer is wrong before you compute anything.

One condition holds all through the lesson. You can add, subtract or compare fractions as amounts only when they are parts of the same whole, or of wholes of equal size. Half of a small pizza is not the same amount as half of a large pizza. In every problem in this lesson, all the fractions are parts of the same whole.

A benchmark is a number that is easy to picture. You use it to judge the size of other numbers. The benchmarks in this lesson are $0$ , $\frac{1}{2}$ and $1$ .

To compare a fraction with $\frac{1}{2}$ , look at half of its denominator.

$\frac{5}{8}$ : half of $8$ is $4$ , and $5$ is more than $4$ . So $\frac{5}{8}$ is greater than $\frac{1}{2}$ .
$\frac{3}{8}$ : $3$ is less than $4$ . So $\frac{3}{8}$ is less than $\frac{1}{2}$ .
$\frac{4}{8}$ : $4$ is exactly half of $8$ . So $\frac{4}{8}$ equals $\frac{1}{2}$ .

This works because half of the equal parts make half of the whole. Eight eighths make the whole, so four eighths make half of it.

To compare a fraction with $1$ , look at its numerator and its denominator. $\frac{7}{8}$ is less than $1$ because $7$ is less than $8$ . $\frac{11}{8}$ is greater than $1$ because $11$ is greater than $8$ .

Each check compares a fraction with $\frac{1}{2}$ or $1$ of the same whole. The table shows how benchmarks help you judge the size of a sum or a difference before you compute it.

Judging the size of a sum or a difference when all the fractions are parts of the same whole

Sum or difference	What you notice	What the benchmark tells you
$\frac{2}{3} + \frac{5}{6}$	Each fraction is greater than $\frac{1}{2}$ .	The sum is greater than $\frac{1}{2} + \frac{1}{2}$ , so it is greater than $1$ .
$\frac{1}{4} + \frac{3}{8}$	Each fraction is less than $\frac{1}{2}$ .	The sum is less than $\frac{1}{2} + \frac{1}{2}$ , so it is less than $1$ .
$\frac{1}{2} + \frac{1}{5}$	You start at $\frac{1}{2}$ and add an amount greater than $0$ .	The sum is greater than $\frac{1}{2}$ .
$\frac{5}{6} - \frac{1}{2}$	You start below $1$ and take away $\frac{1}{2}$ .	The difference is less than $1 - \frac{1}{2}$ , so it is less than $\frac{1}{2}$ .
$\frac{2}{4} - \frac{1}{2}$	Both fractions name the same amount.	The difference is exactly $0$ .

Worked example

Compare with benchmarks only

On Saturday Ana walks $\frac{5}{8}$ mile and then $\frac{7}{10}$ mile. On Sunday she walks $\frac{3}{8}$ mile and then $\frac{2}{5}$ mile. Every fraction is a part of the same whole, $1$ mile. On which day does she walk farther?

Write each day's distance as a sum, because walking one part and then another joins the two parts. Every fraction measures part of the same mile, so you can compare the two sums.
$\text{Saturday: } \frac{5}{8} + \frac{7}{10} \qquad \text{Sunday: } \frac{3}{8} + \frac{2}{5}$
Judge Saturday's sum. Half of 8 is 4, and 5 is more than 4. Half of 10 is 5, and 7 is more than 5. So each part is greater than one half of a mile, and together they are greater than two halves, which is 1 mile.
$\frac{5}{8} + \frac{7}{10} > \frac{1}{2} + \frac{1}{2} = 1$
Judge Sunday's sum. Half of 8 is 4, and 3 is less than 4. Half of 5 is two and a half, and 2 is less than that. So each part is less than one half of a mile, and together they are less than two halves, which is 1 mile.
$\frac{3}{8} + \frac{2}{5} < \frac{1}{2} + \frac{1}{2} = 1$
Compare the two days. Saturday's sum is greater than 1 mile, and Sunday's sum is less than 1 mile. An amount above 1 mile is greater than an amount below 1 mile, so the benchmark decides it. You do not need to add.
$\frac{5}{8} + \frac{7}{10} > 1 > \frac{3}{8} + \frac{2}{5}$

Answer

Ana walks farther on Saturday. Her Saturday distance, $\frac{5}{8} + \frac{7}{10}$ , is greater than $1$ mile, and her Sunday distance, $\frac{3}{8} + \frac{2}{5}$ , is less than $1$ mile.

Benchmarks do not always decide a comparison. Sometimes both results land on the same side of a benchmark. Both might be less than $1$ , for example. Then the benchmark does not tell you which one is greater.

A benchmark can also be unclear about a single result. If one fraction is greater than $\frac{1}{2}$ and the other is less, the benchmark $\frac{1}{2}$ alone does not tell you whether their sum is above or below $1$ .

When benchmarks cannot decide, compute. Find each result, then compare the results. They are easy to compare when they share a common denominator, which is one denominator that both fractions can use. If they do not share one, rename them. To rename a fraction is to write the same amount with a different denominator, as in $\frac{1}{2} = \frac{4}{8}$ . Once the parts are the same size, the result with more parts is greater.

Do not guess. A difference can turn out greater than a sum, as the next example shows. The fractions must still be parts of the same whole, or of wholes of equal size.

Worked example

Compute when benchmarks cannot decide

Which is greater, the sum $\frac{1}{3} + \frac{1}{4}$ or the difference $\frac{11}{12} - \frac{1}{6}$ ? All four fractions are parts of the same whole.

Try benchmarks first. One third and one fourth are each less than one half, because 1 is less than half of 3 and less than half of 4. So the sum is less than 1. Eleven twelfths is less than 1, and you take away an amount greater than 0, so the difference is less than 1 too. Both results are below 1, so the benchmark cannot decide.
$\frac{1}{3} + \frac{1}{4} < 1 \qquad \frac{11}{12} - \frac{1}{6} < 1$
Compute the sum. Thirds and fourths can both be cut into twelfths, so rename both fractions as twelfths. One third is four twelfths, and one fourth is three twelfths. Now the parts are the same size, so you add the numerators.
$\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$
Compute the difference. Each sixth is the same amount as two twelfths, so rename one sixth as two twelfths. Now both fractions count twelfths, so you subtract the numerators.
$\frac{11}{12} - \frac{1}{6} = \frac{11}{12} - \frac{2}{12} = \frac{9}{12}$
Compare the results. Both count twelfths of the same whole, so the parts are the same size. Nine parts are more than seven parts, so nine twelfths is the greater amount.
$\frac{9}{12} > \frac{7}{12}$

Answer

The difference $\frac{11}{12} - \frac{1}{6}$ is greater than the sum $\frac{1}{3} + \frac{1}{4}$ , because $\frac{9}{12}$ is greater than $\frac{7}{12}$ .

Key concept

Benchmarks first, then compute if you need to

These ideas are for fractions that are parts of the same whole, or of wholes of equal size.

To compare two sums or differences, first test each one against a benchmark: $0$ , $\frac{1}{2}$ or $1$ . If one is greater than a benchmark and the other is less than it, you know which is greater. If the benchmarks cannot decide, compute both and compare.
To check an answer, ask where the benchmarks say it must be. An answer that is not there is impossible.

A benchmark can prove that an answer is wrong. It cannot prove that an answer is right. Only correct computing does that.

Now put the checking idea to work. A benchmark can tell you that an answer is wrong. It does not tell you where the work went wrong. To find the slip, remember what the two numbers in a fraction mean. The denominator names the size of the parts. The numerator counts the parts.

Two common slips forget this:

adding the numerators and the denominators
renaming a denominator without renaming its numerator

So catching a mistake has three parts. Test the answer against a benchmark. Find the slip. Then fix the work and test the new answer. As before, the fractions must be parts of the same whole, or of wholes of equal size.

Mistake: add the numerators and add the denominators. Suppose someone adds two parts of the same whole like this: $\frac{3}{4} + \frac{2}{3} = \frac{5}{7}$ . They added $3 + 2$ and $4 + 3$ .

Why it fails. Fourths and thirds are parts of different sizes, so $3 + 2$ counts unlike parts as if they were alike. And $4 + 3 = 7$ makes sevenths, a part size that neither fraction has.

A benchmark proves it. Both $\frac{3}{4}$ and $\frac{2}{3}$ are greater than $\frac{1}{2}$ , so the sum must be greater than $\frac{1}{2} + \frac{1}{2} = 1$ . But $\frac{5}{7}$ is less than $1$ . That answer is impossible.

The fix. Rename both fractions as twelfths: $\frac{9}{12} + \frac{8}{12} = \frac{17}{12}$ . This is greater than $1$ , as it must be.

Worked example

Prove an answer wrong, then fix it

Kai subtracts two parts of the same whole. His work is $\frac{7}{8} - \frac{1}{2} = \frac{7}{8} - \frac{1}{8} = \frac{6}{8}$ , and it has a mistake. Show that his answer is impossible, find his slip, and correct the work.

Ask the benchmark where the answer must be. If you took one half away from 1, exactly one half would be left. Seven eighths is less than 1, so when you take one half away from it, less than one half is left.
$\frac{7}{8} - \frac{1}{2} < 1 - \frac{1}{2} = \frac{1}{2}$
Test Kai's answer. Half of 8 is 4, and 6 is more than 4, so six eighths is greater than one half. The difference must be less than one half, so six eighths is impossible.
$\frac{6}{8} > \frac{1}{2}$
Find the slip. Kai wrote one half as one eighth. He changed the denominator from 2 to 8 but left the numerator at 1. A whole cut into 8 equal parts has smaller parts than the same whole cut into 2, so one eighth is less than one half. His renamed fraction is not the same amount.
$\frac{1}{8} < \frac{1}{2}$
Correct the renaming. Each half is the same amount as 4 eighths, so multiply the numerator and the denominator of one half by 4. Then both fractions count eighths, and you subtract the numerators.
$\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8} \qquad \frac{7}{8} - \frac{4}{8} = \frac{3}{8}$
Test the new answer. Half of 8 is 4, and 3 is less than 4, so three eighths is less than one half. That is where the benchmark said the answer must be.
$\frac{3}{8} < \frac{1}{2}$

Answer

Kai renamed the denominator of $\frac{1}{2}$ but not its numerator, so $\frac{6}{8}$ is wrong. The correct difference is $\frac{7}{8} - \frac{1}{2} = \frac{3}{8}$ .

Here is what to take with you. For fractions that are parts of the same whole, or of wholes of equal size, a benchmark such as $0$ , $\frac{1}{2}$ or $1$ gives you a quick sense of how large a sum or a difference must be. That sense does two jobs. It can show which of two results is greater, and when it cannot, you compute and compare with a common denominator. It can also show that an answer is impossible, which sends you back to find the slip and fix it. Make a benchmark check your habit each time you add or subtract fractions.

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