Adding and Subtracting Fractions: Free Response

Both calculations sit over the same denominator, so the pieces are already one size and only the counts move.

For the total, add the counts of fifteenths and keep fifteenths as the size of the piece:

$\frac{4}{15} + \frac{6}{15} = \frac{4 + 6}{15} = \frac{10}{15}.$

The greatest common factor of $10$ and $15$ is $5$, so the total in lowest terms is

$\frac{10}{15} = \frac{10 \div 5}{15 \div 5} = \frac{2}{3}.$

For the piece that is left, take one count away from the other:

$\frac{13}{15} - \frac{4}{15} = \frac{13 - 4}{15} = \frac{9}{15}.$

Here the greatest common factor is $3$:

$\frac{9}{15} = \frac{9 \div 3}{15 \div 3} = \frac{3}{5}.$

So the two projects use $\frac{2}{3}$ of a roll and $\frac{3}{5}$ of a roll is left. Each answer is a fraction of one roll, which is the only reason the numbers were comparable in the first place.

Put the printed value into fifteenths first, using the building rule in reverse. The denominator $30$ becomes $15$ when it is divided by $2$, so the numerator is divided by $2$ as well:

$\frac{10}{30} = \frac{10 \div 2}{30 \div 2} = \frac{5}{15}.$

Now the printed total and the amounts it came from are all counted in fifteenths, so they can be compared by their counts alone:

$\frac{5}{15} < \frac{6}{15}.$

The printed total is smaller than the amount the second project uses on its own. Putting the first project's cord together with it cannot possibly leave less than the second project alone, so this is not a small slip in the arithmetic. The line is not a total at all.

What produced it is visible in the numbers: $4 + 6 = 10$ on top, and $15 + 15 = 30$ underneath. The denominators were added as though they were counts. They are not: $15$ says how big one piece is, and putting two groups of fifteenths together leaves every piece exactly as big as it was.

A fraction such as $\frac{4}{15}$ makes two different statements with its two numbers.

The denominator $15$ says how big one piece is: the roll has been cut into $15$ equal pieces, so one piece is a fifteenth of a roll. That is a description of a piece, and it tallies nothing.

The numerator $4$ says how many of those pieces are being held. That is a count, and counting is exactly what addition and subtraction do.

So when two amounts made of fifteenths are put together, the only thing that can move is the count:

$\frac{4}{15} + \frac{6}{15} = \frac{4 + 6}{15} = \frac{10}{15}.$

Nothing in that step made a piece bigger or smaller. Four fifteenth-sized pieces together with six more of the same size are ten pieces of that same size, so the denominator comes through untouched while the top becomes a number that appeared in neither fraction to begin with.

This is also why unlike denominators need work before anything else. If one amount were measured in fifteenths and the other in eighths, there would be no single piece being counted, and the two counts could not be totalled until both amounts were described in one size.

The least common denominator is the least common multiple of the two denominators, so work on $9$ and $12$ first.

By prime factorization, $9 = 3 \times 3$ and $12 = 2 \times 2 \times 3$. A common multiple has to carry the two $2$s that $12$ needs and the two $3$s that $9$ needs, and the smallest number that does is

$\operatorname{lcm}(9, 12) = 2 \times 2 \times 3 \times 3 = 36.$

Listing multiples gives the same answer: $12, 24, 36$, and $36$ is the first of them that $9$ divides.

Now rebuild each fraction over $36$. The denominator $9$ reaches $36$ when it is multiplied by $4$, so the numerator is multiplied by $4$ as well:

$\frac{4}{9} = \frac{4 \times 4}{9 \times 4} = \frac{16}{36}.$

The denominator $12$ reaches $36$ when it is multiplied by $3$:

$\frac{5}{12} = \frac{5 \times 3}{12 \times 3} = \frac{15}{36}.$

Both amounts are now counted in thirty-sixths, which is what makes the next step possible.

Both fractions are now counted in one size of piece, so the like-denominator rule finishes the job: add the counts and keep the size.

$\frac{4}{9} + \frac{5}{12} = \frac{16}{36} + \frac{15}{36} = \frac{31}{36}.$

The step in the middle is where the counting happens:

$\frac{16 + 15}{36} = \frac{31}{36}.$

Read the answer back: it counts $31$ pieces, each one a thirty-sixth of one whole.

Check lowest terms before stopping. $31$ is prime, and $36$ is not a multiple of $31$, so the two share no factor above $1$ and $\frac{31}{36}$ is already as simple as it can be.

Start from what the rebuild does to the pieces, not to the symbols.

Taking $\frac{4}{9}$ to thirty-sixths multiplies the denominator by $4$, which is the same as cutting each ninth into $4$ equal pieces. Every ninth becomes $4$ smaller pieces, so the $4$ ninths being held become $16$ of those smaller pieces, and that is why the numerator is multiplied by $4$ too:

$\frac{4}{9} = \frac{4 \times 4}{9 \times 4} = \frac{16}{36}.$

Nothing was added and nothing was taken away; the same amount is simply being described in smaller pieces. So $\frac{16}{36}$ and $\frac{4}{9}$ are equal, not merely close, and replacing one by the other cannot change what the sum comes to. The same holds for the other fraction, so the total in part B is a total of the two amounts you were actually given.

Multiplying only the denominator is a different thing entirely. Writing

$\frac{4}{36}$

keeps the count at $4$ while every piece has shrunk to a quarter of its old size, so it names a quarter of the amount you started with. Both numbers have to move, and by the same factor, precisely because one of them names the size of a piece and the other counts pieces.

Look at the denominators before anything else. They are $10$ and $2$, and $10$ is itself a multiple of $2$, since $2 \times 5 = 10$. So the least common multiple of $10$ and $2$ is $10$: tenths are the shared size, and $\frac{3}{10}$ is already written in them.

Only the half needs rebuilding, and $2$ reaches $10$ when it is multiplied by $5$:

$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}.$

Now add the counts and keep the size:

$\frac{3}{10} + \frac{1}{2} = \frac{3}{10} + \frac{5}{10} = \frac{8}{10}.$

Finish in lowest terms. The greatest common factor of $8$ and $10$ is $2$:

$\frac{8}{10} = \frac{8 \div 2}{10 \div 2} = \frac{4}{5}.$

The crew used $\frac{4}{5}$ of a can of blue over the two days.

Check the denominators the same way. Neither divides the other: $8$ is not a multiple of $3$, and $3$ is not a multiple of $8$. So the shared size is a new one that neither fraction is written over yet.

Since $3$ and $8$ have no factor above $1$ in common, their least common multiple is their product:

$\operatorname{lcm}(3, 8) = 3 \times 8 = 24.$

Both fractions have to be rebuilt over $24$:

$\frac{1}{3} = \frac{1 \times 8}{3 \times 8} = \frac{8}{24}.$

$\frac{3}{8} = \frac{3 \times 3}{8 \times 3} = \frac{9}{24}.$

Now add the counts:

$\frac{1}{3} + \frac{3}{8} = \frac{8}{24} + \frac{9}{24} = \frac{17}{24}.$

Since $17$ is prime and $24$ is not a multiple of $17$, the total is already in lowest terms: the crew used $\frac{17}{24}$ of a can of yellow. This colour took twice the rebuilding that the blue took, and the reason is visible in the denominators rather than anywhere in the fractions themselves.

Compare what the two parts above had in common and where they parted.

In part A the denominators were $10$ and $2$, and $2 \times 5 = 10$, so $10$ is a multiple of both of them. Every common multiple of $2$ and $10$ has to be a multiple of $10$, so none of them is smaller than $10$, which makes $10$ the least common denominator. The fraction already written over $10$ was therefore already in the shared size, and only the other one had to move.

In part B the denominators were $3$ and $8$, and neither is a multiple of the other. The shared size, $24$, is a denominator that neither fraction carried, so both had to be rebuilt.

The condition, then, is that the two denominators differ and the larger is a multiple of the smaller. When it holds, the larger denominator is itself the least common denominator, and the fraction carrying it is left exactly as it is. When neither denominator is a multiple of the other, the least common denominator is a new number and both fractions have to be rebuilt. (Denominators that are already equal, as in the first question of this set, need no rebuilding at all.)

Now test the given pair. Divide the larger denominator by the smaller and see whether it comes out whole:

$18 = 6 \times 3.$

It does, so the condition holds. The least common denominator is $18$, the fraction $\frac{7}{18}$ stays as it is, and only the other one is rebuilt:

$\frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18}.$

Carry each method out in full, so that the comparison is between two finished calculations rather than between one calculation and an opinion.

The student's rule adds the two tops and the two bottoms:

$\frac{1 + 1}{2 + 2} = \frac{2}{4} = \frac{1}{2}.$

The like-denominator method leaves the size of the piece alone and adds only the counts. Both fractions are already halves, so

$\frac{1}{2} + \frac{1}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1.$

The two methods report $\frac{1}{2}$ and $1$. This particular sum is one that can be settled without either rule: two halves of one thing are that whole thing, so $1$ is right and the student's rule has landed somewhere else.

Read the output as a fraction. The value $\frac{2}{4}$ says: two pieces, each one a quarter of the whole.

That is not what was in front of the student. They started with two pieces, each one a half of the whole. The count of pieces, $2$, came through correctly. What changed is the description of the pieces: halves went in, and quarters came out.

$\frac{1}{2} + \frac{1}{2} \longrightarrow \frac{2}{4}.$

Nothing in the situation cut anything smaller. The rule did that by itself, by working out $2 + 2$ underneath as though the bottom number were a tally of something. It is not a tally: the bottom number says how many equal pieces the whole was cut into, which is a statement about the size of one piece. Two groups of half-sized pieces are still made of half-sized pieces.

That is exactly why the total came out too small. Each of the two pieces was reported at half its real size, so the total was reported at half its real size too, which is the gap between $\frac{1}{2}$ and $1$.

Test it rather than argue about it. The rule gives

$\frac{1 + 1}{3 + 6} = \frac{2}{9}.$

A correct total needs one size of piece first. Since $6$ is a multiple of $3$, sixths will serve, and only the first fraction is rebuilt:

$\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}.$

So the rule fails on unlike denominators as well. It is worth seeing how it fails, because this answer can be rejected without knowing the right one. Put $\frac{2}{9}$ and $\frac{1}{3}$ over ninths:

$\frac{1}{3} = \frac{3}{9}, \qquad \frac{2}{9} < \frac{3}{9}.$

The rule's total is smaller than one of the two amounts being added, and adding one positive amount to another cannot leave less than either of them. Any answer that does that is wrong, whatever produced it.

Both cases have the same cause. The rule treats the bottom number as something to be totalled, as though it counted pieces. It does not: it says how large one piece is. Adding two such descriptions together invents a smaller piece that nobody asked for, and reporting the same count in smaller pieces reports less than there was. The correct method matches the sizes first and then adds only the counts, which is the one number in a fraction that counts anything.

A whole number is a fraction with denominator $1$, so it can be rebuilt over whatever denominator you need, exactly as any other fraction can.

In the first calculation the fraction is in sevenths, so write $3$ in sevenths. Each whole is $7$ sevenths, so three wholes are $21$ of them:

$3 = \frac{3}{1} = \frac{3 \times 7}{1 \times 7} = \frac{21}{7}.$

Both amounts are now counted in sevenths, so add the counts:

$3 + \frac{4}{7} = \frac{21}{7} + \frac{4}{7} = \frac{25}{7}.$

In the second the fraction is in eighths, and each whole is $8$ eighths:

$2 = \frac{2 \times 8}{1 \times 8} = \frac{16}{8}.$

Subtract the counts:

$2 - \frac{5}{8} = \frac{16}{8} - \frac{5}{8} = \frac{11}{8}.$

Check lowest terms on each: $25$ and $7$ share no factor above $1$, and neither do $11$ and $8$. Each answer is a single fraction counting pieces of the size its denominator names, twenty-five sevenths and eleven eighths, and each is a complete answer as it stands.

Take the student's rewrite on its own terms first. The fraction $\frac{3}{7}$ counts three pieces, each a seventh of one whole, which does not even reach a single whole, while $3$ is three wholes. So the rewrite is not a description of the same amount, and everything after it is working with the wrong quantity.

The step that went wrong is the rebuild. Changing the denominator from $1$ to $7$ cuts each whole into seven pieces, so the count has to be multiplied by $7$ as well:

$3 = \frac{3}{1} = \frac{3 \times 7}{1 \times 7} = \frac{21}{7}.$

The student multiplied the bottom by $7$ and left the top alone. It is the same slip as writing $\frac{1}{2}$ in sixths as $\frac{1}{6}$ instead of $\frac{3}{6}$.

The reported total can be read back as a second check:

$\frac{7}{7} = 1.$

A calculation that starts with three wholes and then adds to them cannot come out at one whole. Repairing the rebuild repairs the total:

$3 + \frac{4}{7} = \frac{21}{7} + \frac{4}{7} = \frac{25}{7}.$

Test the claim on one of the cases offered. Cut a whole into $5$ equal pieces. Each piece is $\frac{1}{5}$, so taking all five of them puts the whole back together:

$\frac{5}{5} = 1.$

Nothing in that argument depended on the number $5$. Cut the whole into $12$ equal pieces and take all $12$, or into $100$ and take all $100$: each time you are holding the whole thing again, because the denominator says how many pieces the whole was cut into and the numerator says you have taken that many. So

$1 = \frac{c}{c}$

for any whole number $c$ that is not zero, and the classmate is right. The one value ruled out is $c = 0$, since a whole cannot be cut into no pieces at all.

The same freedom belongs to every whole number, not only to $1$: three wholes written in sevenths are $\frac{21}{7}$, and written in halves they are $\frac{6}{2}$.

Which denominator to choose is settled by the fraction standing beside it. To add $1$ to $\frac{4}{9}$, write $1 = \frac{9}{9}$, because ninths are already on the page and choosing them leaves nothing else to rebuild:

$1 + \frac{4}{9} = \frac{9}{9} + \frac{4}{9} = \frac{13}{9}.$

Choosing $\frac{5}{5}$ instead would not be wrong, but it would leave two different sizes of piece and a second rebuild to do before anything could be added.