## 5 Strategies for Comparing Fractions

Comparing fractions means determining which is greater and which is less. For example:

Determine which fraction represents a greater value:

\(\dfrac{7}{13}\) Vs. \(\dfrac{5}{11}\)

For many students (and adults), answering questions like this causes one of two reactions. Either they get extremely nervous or they go right to their trusty algorithm: finding common denominators. It doesn't have to be this way!

Relying on common denominators is often not the most efficient way to compare fractions. And, an over-reliance on this algorithm can obfuscate much of what makes fractions make sense to students. To help, I am sharing 5 strategies you and your students can use to compare fractions.

#### Common Denominators

Okay, let's get this one out of the way right off the bat. Finding common denominators works. It really does! And, I totally understand why so many people rely on it. You can almost turn your brain off while using this strategy because the steps are the same each time.

Let's look at the example from the top of this post. I can make common denominators by finding equivalent fractions.^{[3]}In this case, I just multiply by the opposite denominator. Since 13 and 11 are relatively prime, their product will be their LCM. Multiplying by the opposite denominator will always get a common denominator, but it won't always be as small as it can be. Often, though, most folks just multiply by the opposite denominator even if that won't necessarily be the least common denominator.

\(\dfrac{7}{13}*\dfrac{11}{11}=\dfrac{77}{143}\)

\(\dfrac{5}{11}*\dfrac{13}{13}=\dfrac{65}{143}\)

Instead of comparing the original fractions, I can compare equivalent fractions that now have the same denominator.

\(\dfrac{77}{143}>\dfrac{65}{143}\)

As promised, this strategy worked. Since both fractions now have 143 as a denominator, each fraction represents partitioning the whole into the same sized pieces. Since \(\dfrac{77}{143}\) has more of the same sized parts, it is greater.

So, that strategy was successful, but took **WAY** more calculating than was needed. Many people would not feel comfortable doing those calculations without pencil and paper or calculator. The good news is, there is another way. Making equivalent fractions with common denominators can be a great strategy for some comparisons, but I want to share some other strategies you (and hopefully your students too) can use.

#### Benchmarks

A benchmark is an idea or value you know and understand well that you can compare other things to. For example, I know that my stride is almost exactly 1 yard or 3 feet. So if I want to estimate a length, I can use my stride as a benchmark by counting my steps. Similarly, I know that I can reach about 8 feet in the air without jumping or 9 feet with a small hop. I can estimate short heights by reaching up in the air and comparing to the benchmark of my reach. This is great for estimating ceiling heights which, you know, is pretty important.

For fractions, most students can easily use 1 whole and \(\dfrac{1}{2}\) as benchmarks. Most can tell, for example, that \(\dfrac{6}{10}\) is greater than \(\dfrac{1}{2}\) because \(\dfrac{5}{10}=\dfrac{1}{2}\) and \(\dfrac{6}{10}>\dfrac{5}{10}\).

When two fractions fall on either side of a benchmark, it can be easy to tell which one is larger without manipulating them in any way. Let's return to the same example from above. How do these compare to the benchmark \(\dfrac{1}{2}\) ?

\(\dfrac{7}{13}\) is greater than \(\dfrac{1}{2}\)

\(\dfrac{5}{11}\) is less than \(\dfrac{1}{2}\)

Therefore, \(\dfrac{7}{13}>\dfrac{1}{2}>\dfrac{5}{11}\)

Comparing these fractions to \(\dfrac{1}{2}\) is more efficient because it requires less calculating. More importantly though, this strategy requires students to think about the relative sizes of fractions and their locations on the number line which helps them develop deeper understanding.

Here is a diagram of this comparison on the number line. They aren't placed exactly, but each fraction's location is correct relative to \(\dfrac{1}{2}\). Because we can locate\(\dfrac{5}{11}\) to the left of \(\dfrac{7}{13}\),\(\dfrac{7}{13}\) must be greater:

#### Common Numerators

If common denominators allow us to compare fractions just by looking at the numerators, then common numerators should allow us to compare fractions by looking only at the denominator. For this strategy to make sense, it is easiest to first compare unit fractions.^{[4]}A unit fraction is a fraction with a numerator of 1.

\(\dfrac{1}{7}\) Vs. \(\dfrac{1}{256}\)

Which of these fractions is greater? To compare these, students will often use a sharing argument. Would you rather have one piece of a cake split into 7 pieces, or one piece of a cake split into 256 pieces? The answer is obvious: \(\dfrac{1}{7}\) is larger. But, this strategy works even if the fractions are not unit fractions:

\(\dfrac{4}{19}\) Vs. \(\dfrac{4}{13}\)

One thirteenth is greater than one nineteenth, so even if you have more than 1 piece: \(\dfrac{4}{19}<\dfrac{4}{13}\)

#### More and Larger / Fewer and Smaller

The name really says it all for this strategy. Sometimes one fraction represents larger pieces while also having a greater number of those pieces. Let's look at an example:

\(\dfrac{26}{11}\) Vs. \(\dfrac{25}{12}\)

In this case, \(\dfrac{26}{11}\) has more pieces (26 > 25) while also being made up of larger pieces (\(\dfrac{1}{11} > \dfrac{1}{12}\)). Again, no calculating is necessary in order to see that: \(\dfrac{26}{11}>\dfrac{25}{12}\)

Did you notice that we were working with improper fractions in this example? It doesn't even matter! Also, aren't you glad that you did not need to find common denominators for this one?

#### Missing Parts

Okay, this is the trickiest one. Sometimes you need to compare fractions that are "missing" the same amount. For example:

\(\dfrac{7}{8}\) Vs. \(\dfrac{6}{7}\)

Both of these fractions are one part away from being equivalent to 1. So, one might say that "both fractions are missing a single piece." To tell which one is larger, I just need to see which one is missing less.

I know that \(\dfrac{1}{8} < \dfrac{1}{7}\), so that means that \(\dfrac{7}{8}\) is missing a smaller piece and is therefore closer to one.

Going back to sharing can help here too. Would you rather have \(\dfrac{1}{8}\) or \(\dfrac{1}{7}\) removed from your cake? \(\dfrac{1}{8}\) is a smaller piece, so I would much rather give that away! Therefore:

\(\dfrac{7}{8} > \dfrac{6}{7}\)

Here is a model (drawn to scale) of another comparison. Note that elevenths are larger so \(\dfrac{12}{13}\) represents a greater value.

You can even use missing parts when more than 1 part is missing. Here is an example:

\(\dfrac{93}{97}\) Vs. \(\dfrac{60}{64}\)

Using a missing parts argument, I can see that both fractions are missing 4 parts. I can compare the fractions by comparing the missing parts. I know that \(\dfrac{4}{97}\) < \(\dfrac{4}{64}\), so I am missing less from \(\dfrac{93}{97}\). Therefore, it must be greater.

#### Follow Up:

- I want to thank Ben (thanks!), who gave me a ton of feedback to help make this post better
- I often make videos for my students that can be watched for review or when students are absent. I sometimes assign them as homework to avoid wasting class time 'lecturing'. Here is a video I made on comparing fractions: