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 \(\frac{1}{5} \lt \frac{2}{7}\).
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.
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 \(\frac{29}{60}\).
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 \(\frac{19}{45}\). 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.
Another good choice for a benchmark comparison is the fraction \(\frac25\).
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 $$\frac{8}{25} \lt \frac25 \lt \frac{19}{45}.$$
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. However, there are no fifths between these \(\frac{8}{25}\) and \(\frac{14}{39}\), and consequently students might spend a lot of time spinning their wheels trying to make one of those comparisons work. Both fractions are less than \(\frac{1}{2}\), so identifying \(\frac{1}{3}\) as a possibility for comparison hopefully will come from the students but may need to be suggested if they struggle.