4.NF Using Benchmarks to Compare Fractions


Alignments to Content Standards: 4.NF.A.2

Task

Melissa gives her classmates the following explanation for why $\frac{1}{5} \lt \frac{2}{7}$:

I can compare both $\frac{1}{5}$ and $\frac{2}{7}$ to $\frac{1}{4}$.
Since $\frac{1}{5}$ and $\frac{1}{4}$ are unit fractions and fifths are smaller than fourths, I know that $\frac{1}{5} \lt \frac{1}{4}$.
I also know that $\frac{1}{4}$ is the same as $\frac{2}{8}$, so $\frac{2}{7}$ is bigger than $\frac{1}{4}$.
Therefore $\frac{1}{5} \lt \frac{2}{7}$.

 

  1. Explain each step in Melissa's reasoning. Is she correct?
  2. Use Melissa's strategy to compare $\frac{29}{60}$ and $\frac{45}{88}$, this time comparing both fractions with $\frac{1}{2}$.
  3. Use Melissa's strategy to compare $\frac{8}{25}$ and $\frac{19}{45}$. Explain which fraction you chose for comparison and why.

 

IM Commentary

This task is intended primarily for instruction. The goal is to provide examples for comparing two fractions, $\frac{1}{5}$ and $\frac{2}{7}$ in this case, by finding a benchmark fraction which lies in between the two. In Melissa's example, she chooses $\frac{1}{4}$ as being larger than $\frac{1}{5}$ and smaller than $\frac{2}{7}$.

This is an important method for comparing fractions and one which requires a strong number sense and ability to make mental calculations. It is, however, a difficult ability to assess because the method is only appropriate when there is a clear benchmark fraction to be used. In part (c) of the problem, for example, students may see the denominator of $25$ and think that $\frac{1}{5}$ or $\frac25$ would be potential fractions to use for comparison. In this case, it turns out that $\frac{2}{5}$ is an excellent choice which works well. However, if the numbers were different (for example $\frac{8}{25}$ and $\frac{14}{39}$) then there may be no fifths between them and students might spend a lot of time spinning their wheels trying to make $\frac{1}{5}$ or $\frac{2}{5}$ work. In addition to $\frac{2}{5}$, suggested by the denominator 25, both fractions are less than $\frac{1}{2}$, so identifying $\frac{1}{3}$ as a possibility for comparison might also come from the students and could be suggested if they struggle.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This particular task is linked very intentionally to Mathematical Practice Standard 3, critique the reasoning of others. Students are asked to explain and critique the reasoning of their classmate, Melissa.  This type of task provides students with an opportunity to distinguish a reasonable explanation from that which is flawed.  If there is a flaw in the argument they can further explain why it is flawed. Learning how to argue whether a claim is true or false concisely and precisely becomes a routine part of a student’s mathematical work. This task is further extended by directing students to explore Melissa’s strategy with 2 additional examples. Their exploration may spark a conversation about when this strategy is most effective and what other strategies may be more effective and why. (MP.5)

Solution

  1. Melissa's reasoning is correct. For the first step $\frac{1}{5}$ represents one of five equal pieces that make up a whole. $\frac{1}{4}$ represents one of four equal pieces making up the same whole. Since there are fewer of the equal pieces of size $\frac{1}{4}$ making up the same whole, $\frac{1}{5} \lt \frac{1}{4}$.

    Next, Melissa argues that $\frac{1}{4} \lt \frac{2}{7}$. To compare these two fractions, she first changes the denominator of $\frac{1}{4}$ from $4$ to $8$. To write $\frac{1}{4}$ as a fraction with $8$ in the denominator means that the denominator is multiplied by $2$. Multiplying the numerator by $2$ also gives $$ \frac{1}{4} = \frac{2 \times 1}{2 \times 4} = \frac{2}{8}. $$ Now $\frac{2}{8} \lt \frac{2}{7}$ because $\frac{2}{8}$ represents two of eight equal pieces which make up a whole while $\frac{2}{7}$ represents two of seven equal pieces that make up the same whole. Since there are fewer of the equal pieces of size $\frac{1}{7}$ making up the same whole, $\frac{2}{8} \lt \frac{2}{7}$.

    Combining the work from the first two paragraphs gives $$ \frac15 \lt \frac14 \lt \frac{2}{7} $$ and so $\frac15 \lt \frac{2}{7}$. Melissa's reasoning is involved but correct.
  2. Using Melissa's strategy, the goal is to compare $\frac{29}{60}$ to $\frac{1}{2}$ and then to compare $\frac{45}{88}$ to $\frac{1}{2}$. For $\frac{29}{60}$ and $\frac{1}{2}$ we can compare these fractions by finding a common denominator. Since $2$ is a factor of $60$ we can use $60$ as a common denominator. To write $\frac{1}{2}$ with a denominator of $60$ we need to multiply the denominator (and numerator) by $30$: $$ \frac{1}{2} = \frac{30 \times 1}{30 \times 2} = \frac{30}{60}. $$ Now we can see that $\frac{29}{60} \lt \frac{30}{60}$ since we are comparing $29$ pieces to $30$ pieces where these pieces all have the same size. So we find $$ \frac{29}{60} \lt \frac{1}{2}. $$

    Next, to compare $\frac{1}{2}$ to $\frac{45}{88}$ we can write $\frac{1}{2}$ with a denominator of $88$, multiplying numerator and denominator by $44$ this time: $$ \frac{1}{2} = \frac{44 \times 1}{44 \times 2} = \frac{44}{88}. $$ We know that $\frac{44}{88} \lt \frac{45}{88}$ because $44$ pieces is less than $45$ pieces and the pieces all have the same size. So we see that $$ \frac{1}{2} \lt \frac{45}{88}. $$

    Combining the reasoning of the two paragraphs above gives $$ \frac{29}{60} \lt \frac{1}{2} \lt \frac{45}{88} $$ and so $\frac{45}{88}$ is greater than $\frac{29}{60}$.

  3. The reasoning here will be like that of parts (a) and (b) if we can identify the benchmark fraction to compare with $\frac{8}{25}$ and $\frac{19}{45}$. One possible choice for a benchmark comparison is the fraction $\frac25$, convenient because one of our fractions has 25 as a denominator. Since $25=5 \times 5$, we can convert the fraction $\frac25$ to twenty-fifths: $$\frac25=\frac{5 \times 2}{5 \times 5}=\frac{10}{25}.$$ Now $\frac{8}{25} \lt \frac{10}{25}$ because $8$ is less than $10$ and both fractions have a denominator of $25$. So we have found that $$\frac{8}{25} \lt \frac25.$$ Since we used $\frac25$ for comparison with $\frac{8}{25}$, we should also use $\frac25$ for comparison with $\frac{19}{45}$. Since $45=9 \times 5$, we can convert the fraction $\frac25$ to forty-fifths: $$\frac25=\frac{9 \times 2}{9 \times 5}=\frac{18}{45}.$$ Now $\frac{18}{45} \lt \frac{19}{45}$ because $18$ is less than $19$ and both fractions have a denominator of $45$. So we have found that $$\frac25 \lt \frac{19}{45}.$$ Combining the previous work, we see that $$\frac{8}{25} \lt \frac25 \lt \frac{19}{45}.$$

    Since $8 \times 3 = 24$, we have $$ \frac{1}{3} = \frac{8 \times 1}{8 \times 3} = \frac{8}{24}. $$ This is close to $\frac{8}{25}$ and this was what motivated the choice of $\frac{1}{3}$ (we will see below that $\frac{19}{45}$ is also close to $\frac{1}{3}$, making $\frac{1}{3}$ an appropriate fraction for comparison). To see which is larger, $\frac{1}{3}$ or $\frac{8}{25}$, note that $\frac{1}{25} \lt \frac{1}{24}$ because if a whole is broken into $24$ equal sized pieces these pieces will be larger than if the same whole is broken into $25$ equal sized pieces. So we can conclude that $\frac{8}{25} \lt \frac{8}{24}$ giving $$\frac{8}{25} \lt \frac{1}{3}.$$

    Since we used $\frac{1}{3}$ for comparison with $\frac{8}{25}$ we should also use $\frac{1}{3}$ for comparison with $\frac{19}{45}$. Since $45 = 15 \times 3$, we can convert the fraction $\frac{1}{3}$ to forty-fifths: $$ \frac{1}{3} = \frac{15 \times 1}{15 \times 3} = \frac{15}{45}. $$ Now $\frac{15}{45} \lt \frac{19}{45}$ because $15$ is less than $19 $ and both fractions have a denominator of $45$. So we have found that $$ \frac{1}{3} \lt \frac{19}{45}. $$ Combining the work of the previous two paragraphs we see that $$ \frac{8}{25} \lt \frac{1}{3} \lt \frac{19}{45}. $$

    The key to using this method for comparing fractions is identifying a benchmark fraction for comparison. This requires either a good number sense or a lot of experience.

Anthony says:

over 1 year

Finding benchmark fractions that one can compare to unit or common fractions is a great strategy in dealing with fractions that are unfamiliar or not used often by students. We all know unfamiliarity causes much stress in students. Students need more examples like this in which they can manipulate numbers to come up with a solution. I often use words like since..... then...... why use sevenths when using eighths is come much easier. Make the numbers easier to use.

Michael Nakamaye says:

almost 2 years

You are exactly right about what happened with changing fractions in the problem and then forgetting to change the commentary. This has been updated now and hopefully is clearer. Thank you for catching this!

Meredith Hegg says:

about 2 years

I assume part c was changed from the original (which presumably used 8/25 and 14/39). The solutions reflect this but the commentary does not, i.e., paragraph two needs to be rewritten. Furthermore, with the new change, 2/5 is not only a reasonable choice for benchmarking but is, I believe, the likelier choice. In the solution, I would probably present 2/5 before 1/3.

BArbara says:

over 2 years

Why wasn't the common denominator (35) used in finding a solution?

Cam says:

over 2 years

You'd have to ask Melissa! But seriously, certainly a common denominator approach is a perfectly valid one. It is worth noting, however, that this comparison can be made without referencing denominators larger than 10. Part of the point of this exercise is to encourage thinking about an alternative approach to fraction comparison.

Ruth Balf says:

almost 4 years

I am unclear about the appropriateness of fractions to use with 4th graders. The Common Core has a footnote that says denominators for Grade 4 should be limited to 2, 3, 4, 6, 8, 10, 12 and 100, so I am wondering how this task fits.

Mike says:

almost 4 years

I agree, we worked on the denominators you mentioned and were tested on our summative with higher denominators. Of course most of the kids were lost and were unable to show what they really knew.

Kristin says:

almost 4 years

The limits on the denominators are actually supposed to be for summative assessment purposes, not instruction. So the task would be appropriate for an instructional context but not a test. It's unfortunate that the summative assessment your students took did not reflect this.

Ken Mullen says:

about 4 years

This is an important method for comparing fractions, but the example comparison to 1/4 is somewhat weak in terms of showing the importance. It would have been quicker and clearer for Melissa to have rewritten 1/5 as 8/40 and shown the comparison in one step. For the method given, she had to do the same kind of rewriting plus make a side argument about 1/5 and 1/4.

Michael Nakamaye says:

about 4 years

Agreed! This is a good observation. We could change the $\frac{11}{40}$ to something simpler such as $\frac{2}{7}$ where the method works quite well or go for a fraction with a little larger denominator such as $\frac{4}{15}$. I agree that it should be changed but wanted to check to see if you have a preference for what type of number we put in place of $\frac{11}{40}$. Thanks!