Multiplying and Dividing Fractions: Free Response

The vertical cuts divide the width into $5$ equal strips and $3$ of them are yellow, which is what $\frac{3}{5}$ of the sheet means. The bands divide the height into $7$ equal rows and $2$ of them are blue, which is $\frac{2}{7}$ of the yellow region.

Carry both sets of cuts edge to edge. Every one of the $5$ columns is crossed by every one of the $7$ rows, so the sheet is cut into

$5 \times 7 = 35$

equal small rectangles, all the same size.

A rectangle carries both colours exactly when it sits in one of the yellow columns and in one of the blue rows at the same time, so the doubly shaded region is $3$ columns crossing $2$ rows:

$3 \times 2 = 6.$

Six equal pieces out of $35$ equal pieces is, by what a fraction means, $\frac{6}{35}$ of the sheet:

$\frac{2}{7} \times \frac{3}{5} = \frac{6}{35}.$

Nothing was measured. The two sets of cuts made a grid, and the grid did the counting. Since $6$ and $35$ share no factor above $1$, this is already in lowest terms.

Build the same picture with letters. Cut the sheet into $d$ equal columns and shade $c$ of them, which is $\frac{c}{d}$ of the sheet. Then cut it into $b$ equal rows and keep $a$ of them over the shaded part, which takes $\frac{a}{b}$ of what was shaded.

The two sets of cuts cross, so the whole sheet is divided into

$b \times d$

equal small rectangles. The region carrying both shadings is $a$ rows crossing $c$ columns, so it covers

$a \times c$

of those rectangles. A region of $a \times c$ equal pieces out of $b \times d$ equal pieces is

$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}.$

Part A is this statement with numbers in place of letters. The bottom product was the $5 \times 7$ that counted every rectangle on the sheet, and the top product was the $3 \times 2$ that counted the rectangles kept by both shadings.

The rule holds when one of the factors is a whole number too, since a whole number is a fraction over $1$: for instance $6 \times \frac{4}{9} = \frac{6}{1} \times \frac{4}{9} = \frac{24}{9}$, which simplifies to $\frac{8}{3}$.

The two operations are doing different jobs, and the picture makes the difference visible.

Adding starts with two amounts that already exist side by side. Thirds and fifths are pieces of different sizes, and there is no honest way to say how many pieces you hold until both amounts are re-cut into pieces of one common size. Building a common denominator is that re-cutting.

A part of a part starts with one amount. The yellow shading did not arrive beside the blue shading; the blue shading was taken from inside the yellow. So there were never two piles to reconcile. What the second set of cuts did was slice the same sheet a second way, and the moment both sets run edge to edge the sheet carries one grid of rectangles that are all the same size as each other:

$5 \times 7 = 35 \text{ equal rectangles}.$

Equal-sized pieces are exactly what addition has to work for, and here they came free, because they were produced by the cutting itself rather than assembled from two different amounts.

That is why the multiplication rule mentions no common denominator at all. The denominators are multiplied because they count cuts in two directions, and the numerators are multiplied because they count the strips kept in each direction. A common denominator would answer a question about combining, and no combining ever happened.

Look across the two fractions for a numerator and a denominator that share a factor. The $9$ on top shares $3$ with the $12$ underneath, and the $7$ on top shares $7$ with the $14$ underneath.

Divide each pair by its shared factor. The $9$ becomes $3$ and the $12$ becomes $4$; the $7$ becomes $1$ and the $14$ becomes $2$. Only small numbers are left to multiply:

$\frac{9}{14} \times \frac{7}{12} = \frac{3}{2} \times \frac{1}{4} = \frac{3 \times 1}{2 \times 4} = \frac{3}{8}$

Now the long route. Multiply straight across first:

$\frac{9}{14} \times \frac{7}{12} = \frac{9 \times 7}{14 \times 12} = \frac{63}{168}$

The greatest common factor of $63$ and $168$ is $21$, so

$\frac{63}{168} = \frac{63 \div 21}{168 \div 21} = \frac{3}{8}.$

The two routes name the same number, which they must: cancelling early and simplifying late divide the same product by the same factors, only in a different order. The difference is the size of the numbers you handle. The long route asked you to notice that $21$ divides both $63$ and $168$; the trimmed route never built a number above $8$.

Carry the student's line through to a value first, so the comparison rests on two numbers rather than on an impression:

$\frac{3}{25} \times \frac{2}{9} = \frac{3 \times 2}{25 \times 9} = \frac{6}{225} = \frac{2}{75}$

Now the product as it was written. Multiplying across gives a single fraction:

$\frac{6}{25} \times \frac{4}{9} = \frac{6 \times 4}{25 \times 9} = \frac{24}{225} = \frac{8}{75}$

Both of those last steps divide top and bottom by $3$. The two values disagree, so the opening move was not a trim, it was a change of value.

The second move is worth acquitting explicitly. Multiplying $\frac{3}{25} \times \frac{2}{9}$ across to $\frac{6}{225}$ is exactly right for the fractions it was handed; it inherits a wrong pair and processes them faultlessly. A line can be perfect arithmetic and still carry a wrong value forward, which is why the first move is the one to name.

Here is why that first move fails. Cancelling is simplifying, and simplifying divides the top and the bottom of one fraction by the same number. In this product the $6$ and the $4$ both end up on TOP, since every numerator joins the one numerator $6 \times 4$. Dividing each of them by $2$ therefore divides the numerator by $2 \times 2 = 4$ and leaves the denominator untouched, so the value falls to a quarter of what it should be:

$\frac{2}{75} \times 4 = \frac{8}{75}.$

A legitimate trim on this same product does exist, and it pairs a top with a bottom. The $6$ on top and the $9$ underneath share $3$:

$\frac{6}{25} \times \frac{4}{9} = \frac{2}{25} \times \frac{4}{3} = \frac{2 \times 4}{25 \times 3} = \frac{8}{75},$

which reaches the correct value with smaller numbers, exactly as a trim is supposed to.

Start by writing the product as the single fraction it is, which is what the multiplication rule says it comes to:

$\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$

The two starting fractions have gone. There is now one numerator, $a \times c$, and one denominator, $b \times d$.

Suppose $c$ and $b$ share a factor $k$ above $1$, so $c = k \times c'$ and $b = k \times b'$. Substituting shows $k$ standing once on top and once underneath:

$\frac{a \times c}{b \times d} = \frac{a \times k \times c'}{k \times b' \times d}$

Simplifying has one rule: dividing the top and the bottom by the same nonzero number leaves the value alone. Applying it to $k$ gives

$\frac{a \times k \times c'}{k \times b' \times d} = \frac{a \times c'}{b' \times d},$

which is exactly the product you get by replacing $c$ with $c'$ and $b$ with $b'$ before multiplying. So cancelling first is legitimate, and it produces the same number as multiplying first and simplifying afterwards.

Notice what the argument never used: which of the two starting fractions $c$ and $b$ came from. Once the product is formed, every numerator has joined the one top and every denominator has joined the one bottom, so a numerator cannot tell whether the denominator it is cancelling against was its own or its neighbour's. What the argument does use is that one number sat on top and the other underneath. Two numerators share no such relationship, which is why part B's move failed.

The recipe calls for $\frac{2}{3}$ of the oil in the jug, and the word of is the multiplication:

$\frac{2}{3} \times \frac{9}{10}$

Trim before multiplying. The $2$ on top shares $2$ with the $10$ underneath, and the $9$ on top shares $3$ with the $3$ underneath. The $2$ becomes $1$ and the $10$ becomes $5$; the $9$ becomes $3$ and the $3$ becomes $1$:

$\frac{2}{3} \times \frac{9}{10} = \frac{1}{1} \times \frac{3}{5} = \frac{3}{5}$

So the recipe calls for $\frac{3}{5}$ of a litre of oil.

The long route agrees, as it must: multiplying across gives $\frac{18}{30}$, and dividing top and bottom by their greatest common factor $6$ gives $\frac{3}{5}$ again.

The size of the answer is worth a glance. Taking a part of the oil should leave less than the jug held, and $\frac{3}{5}$ of a litre is indeed less than $\frac{9}{10}$ of a litre, since $\frac{3}{5} = \frac{6}{10}$.

Each bottle holds $\frac{3}{20}$ of a litre, so the question is how many helpings of that size fit inside the oil that was measured out. That is a division:

$\frac{3}{5} \div \frac{3}{20}$

Dividing by a fraction is multiplying by its reciprocal, and only the divisor is turned over:

$\frac{3}{5} \div \frac{3}{20} = \frac{3}{5} \times \frac{20}{3}$

Trim before multiplying. The $3$ on top cancels with the $3$ underneath, and the $20$ on top shares $5$ with the $5$ underneath:

$\frac{3}{5} \times \frac{20}{3} = \frac{1}{1} \times \frac{4}{1} = 4$

The measured oil fills $4$ bottles exactly.

The number carries no litres, and that is not an oversight. Both amounts in the division were volumes, and the question asked how many of one fit inside the other, so what comes out is a plain count of bottles. It is also a sensible size: each bottle holds less than the oil measured out, so more than one bottle must be needed.

Read the helper's line as an instruction in words. Multiplying by $\frac{3}{20}$ means taking $\frac{3}{20}$ OF the measured oil, so the answer to that instruction is a smaller amount of oil:

$\frac{3}{5} \times \frac{3}{20} = \frac{9}{100}$

That is $\frac{9}{100}$ of a litre, three twentieths of what was measured out. It is a perfectly good number and it answers a question nobody asked. What it is not is a count.

The difference is what the two operations do with the bottle size. Multiplying uses $\frac{3}{20}$ as an instruction to shrink an amount. Dividing uses $\frac{3}{20}$ as the size of one helping and asks how many helpings the amount contains. Only the second question has bottles as its answer.

There is a cheap check on the size as well. Each bottle holds less than the oil measured out, so the count has to come out above $1$. The helper's line returns $\frac{9}{100}$, which is below $1$, so whatever it is measuring, it is not a number of bottles.

The reciprocal of a fraction swaps its numerator and its denominator, so the reciprocal of $\frac{5}{9}$ is $\frac{9}{5}$. A whole number is a fraction over $1$, so $8 = \frac{8}{1}$, and flipping that gives $\frac{1}{8}$.

Multiply each number by its own reciprocal, tops together and bottoms together:

$\frac{5}{9} \times \frac{9}{5} = \frac{5 \times 9}{9 \times 5} = \frac{45}{45} = 1$

$8 \times \frac{1}{8} = \frac{8}{1} \times \frac{1}{8} = \frac{8 \times 1}{1 \times 8} = \frac{8}{8} = 1$

Neither result is a coincidence of these numbers. Flipping puts the same two numbers on the other side of the bar, so the numerator and the denominator of the product are the same product written in a different order, and a fraction whose top equals its bottom is $1$.

One number is left out of this. Flipping $\frac{0}{9}$ would put $0$ underneath, which names nothing, so $0$ has no reciprocal. Every other number here does.

Two amounts can only be counted against each other when they are cut into pieces of the same size, so re-cut the first one. Multiplying top and bottom by $2$ gives an equivalent fraction in tenths:

$\frac{4}{5} = \frac{4 \times 2}{5 \times 2} = \frac{8}{10}$

Eight tenths is eight of the pieces of size $\frac{1}{10}$, so eight of them fit. No division rule was used to get that; it was a count.

Now take the other route. The reciprocal of $\frac{1}{10}$ is $\frac{10}{1}$, and multiplying by it gives

$\frac{4}{5} \times \frac{10}{1} = \frac{4 \times 10}{5 \times 1} = \frac{40}{5} = 8.$

The count and the multiplication agree. That agreement is the evidence that the flipping rule is not an unrelated trick: on a case where the answer can be counted directly, the rule returns the counted answer. What one worked case cannot show is that the agreement holds every time, which is what part C is for.

Give the quotient a name and say what that name means. Writing $\frac{a}{b} \div \frac{c}{d} = Q$ says that $Q$ is the number which multiplies $\frac{c}{d}$ back up to $\frac{a}{b}$:

$Q \times \frac{c}{d} = \frac{a}{b}$

That step is entitled by what division means, the same sense in which $12 \div 3 = 4$ holds because $4 \times 3 = 12$. Nothing about fractions has been assumed yet.

To get $Q$ on its own, the $\frac{c}{d}$ multiplying it has to be cleared away. Multiply both sides by $\frac{d}{c}$, which is entitled by the two sides being the same number: whatever you do to one, you may do to the other.

$Q \times \frac{c}{d} \times \frac{d}{c} = \frac{a}{b} \times \frac{d}{c}$

On the left, $\frac{c}{d} \times \frac{d}{c} = 1$, which is the fact part A produced twice, and multiplying $Q$ by $1$ leaves $Q$ alone. So the left side collapses and

$Q = \frac{a}{b} \times \frac{d}{c}.$

That is the rule, and notice what the derivation never needed: no picture, no convenient common denominator, and no requirement that the count come out whole. It works for any $\frac{a}{b}$ and any $\frac{c}{d}$ that has a reciprocal.

The exclusion is the divisor $\frac{c}{d}$ being $0$, which happens exactly when $c = 0$. Flipping it would put $0$ underneath, so there is no reciprocal to multiply by. The missing-factor reading fails too, and it fails in two opposite ways depending on the dividend. If $\frac{a}{b}$ is not $0$, then no $Q$ at all satisfies $Q \times 0 = \frac{a}{b}$, because every product with $0$ is $0$. If $\frac{a}{b}$ is $0$, then every $Q$ satisfies $Q \times 0 = 0$, so there is no single number the quotient could name. Nothing to choose and everything to choose are both reasons to leave division by $0$ undefined. The rule also explains the standing warning about which fraction to turn over. It was the divisor that had to be cleared, so it is the divisor that gets flipped; turning the first fraction over instead clears nothing.

Take the two products in turn. In the first, the $2$ on top shares $2$ with the $8$ underneath:

$\frac{3}{8} \times \frac{2}{5} = \frac{3}{4} \times \frac{1}{5} = \frac{3}{20}$

In the second there is nothing to trim, since neither numerator shares a factor above $1$ with either denominator, so multiply straight across:

$\frac{3}{8} \times \frac{5}{2} = \frac{3 \times 5}{8 \times 2} = \frac{15}{16}$

Now compare each with $\frac{3}{8}$, and do it by common denominators rather than by eye.

Against the first product, twentieths and eighths both go into fortieths:

$\frac{3}{20} = \frac{6}{40}, \qquad \frac{3}{8} = \frac{15}{40}$

so $\frac{3}{20}$ is the smaller. The card's first claim survives this case.

Against the second, eighths go into sixteenths:

$\frac{3}{8} = \frac{6}{16}, \qquad \frac{15}{16} > \frac{6}{16}$

so $\frac{15}{16}$ is the larger. Multiplying has made the starting number bigger, which is not what the card said would happen. The difference between the two cases is not the operation, which was multiplication both times. It is where the second fraction sits relative to $1$.

Each division becomes a multiplication by the reciprocal of the divisor, and the reciprocals of these two divisors are each other:

$\frac{3}{8} \div \frac{2}{5} = \frac{3}{8} \times \frac{5}{2} = \frac{15}{16}$

$\frac{3}{8} \div \frac{5}{2} = \frac{3}{8} \times \frac{2}{5} = \frac{3}{20}$

So no new arithmetic was needed: these are the two products from part A, arriving in the opposite order.

The comparisons carry over with them. Since $\frac{3}{8} = \frac{6}{16}$, the value $\frac{15}{16}$ is larger; and since $\frac{3}{20} = \frac{6}{40}$ against $\frac{3}{8} = \frac{15}{40}$, the value $\frac{3}{20}$ is smaller.

That swap is the useful observation. Dividing by $\frac{2}{5}$ does exactly what multiplying by $\frac{5}{2}$ does, so any claim about what division does to size is a claim about multiplication by the flipped divisor. The card's second sentence held in one of these two cases and failed in the other, and again the deciding feature is which side of $1$ the second fraction sits on.

The four results split cleanly, and not by operation. Multiplying gave a smaller value with $\frac{2}{5}$ and a larger one with $\frac{5}{2}$; dividing gave a larger value with $\frac{2}{5}$ and a smaller one with $\frac{5}{2}$. What separates the two fractions is that $\frac{2}{5}$ is below $1$ and $\frac{5}{2}$ is above it.

So here is a version that survives all four cases. Take a positive starting number $n$. Multiplying $n$ by a fraction below $1$ leaves a smaller result, multiplying by $1$ leaves $n$ unchanged, and multiplying by a fraction above $1$ leaves a larger one.

The division half needs one extra word, and the word is positive. For a positive fraction $f$, dividing $n$ by $f$ swaps the two outer cases and leaves the middle one alone, which is no extra claim at all: dividing by $f$ is multiplying by its reciprocal, and flipping a positive fraction below $1$ produces one above $1$ and the other way round, while flipping $1$ gives $1$ back.

$\frac{3}{8} \div \frac{2}{5} = \frac{3}{8} \times \frac{5}{2}$

Drop that word and the division half is false. Dividing $\frac{3}{8}$ by $-\frac{1}{2}$ gives $-\frac{3}{4}$, which is negative and therefore smaller than $\frac{3}{8}$, even though $-\frac{1}{2}$ sits below $1$ where the sentence promises growth. A divisor of $0$ is worse still, since there is then no quotient to compare with anything. The multiplication half needs no such guard: multiplying a positive $n$ by any fraction below $1$, negative ones included, lands below $n$.

Now read each half backwards. For multiplication the three cases cover every fraction and no two of them overlap, so exactly one applies. That is what makes the backwards reading work: if a product came out smaller than the positive number it started from, the multiplier cannot have been $1$ or above $1$, because those give a result that is unchanged or larger, so it was below $1$. Division reads backwards the same way once its divisors are restricted to the positive ones, with the two outer cases exchanged. Without that restriction it does not, because a quotient smaller than $n$ could have come from a divisor above $1$ or from any negative divisor whatever.

The backwards reading is worth testing rather than assuming, because a rule and its converse are two separate claims and the second can fail while the first holds.

Which starting numbers this is about matters, and $0$ is the case that has to go. Multiplying $0$ by $\frac{2}{5}$ gives $0$, which is neither smaller nor larger, so the forwards claim fails there, and the backwards claim fails harder: every multiplier whatsoever leaves $0$ unchanged, so the result tells you nothing at all about the multiplier. Negative starting numbers would also have to be excluded from the sentence as written: multiplying $-\frac{3}{8}$ by $\frac{2}{5}$ gives $-\frac{3}{20}$, which sits to the right of $-\frac{3}{8}$ on the number line and so is larger, the opposite direction from the positive case.