Understanding Fractions: Free Response

The wall is one whole, and it was marked into $9$ equal panels, so every fraction in this question counts ninths.

Four of those nine panels carry paint, so the painted amount is four ninths:

$\frac{4}{9}$

The blank panels are the ones left over, and there are $9 - 4 = 5$ of them, each one a ninth of the wall:

$\frac{5}{9}$

The two numbers are doing different jobs, and that is why one of them moved and the other did not. The $9$ underneath is a report about the cutting: the wall was divided into $9$ equal panels, so a ninth is the size of one panel. The number on top is a plain count of those panels, in the same way that $4$ metres is a count of metres. Painting more panels changes the count, not the size of a panel, so the denominator stays at $9$ in both fractions while the numerator changes.

One panel is one of the equal parts its wall was marked into, so each of these fractions has numerator $1$.

The first wall carries $9$ equal panels, so one of them is

$\frac{1}{9}$

of that wall. The second wall carries $6$ equal panels, so one of them is

$\frac{1}{6}$

of that wall.

The two walls are the same size, so the two panels can be set side by side and compared. Sharing one wall among $9$ panels leaves each panel less wall than sharing the same wall among $6$ panels does, because more cuts leave smaller pieces:

$\frac{1}{9} < \frac{1}{6}$

So a single panel of the second wall covers more, even though $9$ is the larger number.

One detail is worth keeping straight. The inequality $\frac{1}{9} < \frac{1}{6}$ is a statement about two numbers, and it holds whatever the walls happen to look like. What the walls being the same size buys is the step from that inequality to a claim about the panels themselves: a fraction measures against the whole it came from, so if the two walls differed in size, knowing $\frac{1}{9} < \frac{1}{6}$ would not on its own settle which panel covers more.

The visitor's arithmetic is not in question. It is perfectly true that $12$ is bigger than $6$. The failure is in what that comparison is taken to be about.

A denominator does not measure a panel. It counts how many equal panels one whole wall was divided into, and the wall being divided is the same size in both cases. A fixed amount of wall shared among more panels leaves less wall for each of them, so pushing the count of panels up forces the size of one panel down:

$\frac{1}{12} < \frac{1}{6}$

The corrected claim keeps the visitor's observation and reverses what it is taken to show: the third wall was marked into more panels, so each of its panels is the smaller one.

One test keeps this straight in future. Push the count as far as you like: mark a wall into $100$ equal panels and one panel is a thin strip. Nobody expects a hundred panels to make each panel large, and the same reasoning settles $12$ against $6$. The visitor read the denominator as an amount of wall per panel, when it is a count of panels per wall.

Sharing equally is a division, and the two numbers arrive in a fixed order: the amount being shared goes first, the number of people it is shared among goes second.

$5 \div 6$

The whole numbers have no answer to that division, but the fraction bar does, because the bar is that division written another way:

$5 \div 6 = \frac{5}{6}$

It is worth seeing why five sixths is believable and not merely formal. Cut each of the $5$ loaves into $6$ equal pieces, which gives $30$ pieces, every one of them a sixth of a loaf. Deal those pieces out to the $6$ people:

$30 \div 6 = 5 \text{ pieces each}$

Five pieces, each a sixth, is five sixths of a loaf.

One whole loaf would be six sixths, and each share is one sixth short of that, so nobody receives a whole loaf.

Read each bar as a division and ask whether that division comes out even.

$\frac{27}{9} = 27 \div 9 = 3$

$\frac{13}{13} = 13 \div 13 = 1$

The second needs no arithmetic at all: a numerator equal to its denominator says the whole was cut into $13$ equal parts and all $13$ were taken back, which is the whole itself.

$\frac{15}{1} = 15 \div 1 = 15$

A denominator of $1$ says the whole was never cut, so each part is a whole and taking $15$ of them gives $15$.

$\frac{0}{7} = 0 \div 7 = 0$

Here the whole was cut into $7$ equal parts and none was taken, which is a perfectly good amount: none.

That leaves $\frac{22}{9}$. Nine goes into $22$ twice with $4$ left over, so that division does not come out even and no whole number equals it. This is not a gap in the answer. The fraction $\frac{22}{9}$ is itself the answer to $22 \div 9$, and it is a complete one.

A division always asks the same shape of question: $a \div b$ asks which number, multiplied by $b$, gives back $a$. Put both fractions to that test.

First $\frac{0}{7}$, which is $0 \div 7$. The question is which number multiplied by $7$ gives $0$, and $0$ answers it:

$0 \times 7 = 0$

so $\frac{0}{7} = 0$. The parts reading agrees. The whole was cut into $7$ equal parts and none of them was taken, and taking none of something is a real amount, namely nothing at all.

Now $\frac{7}{0}$, which is $7 \div 0$. The question is which number multiplied by $0$ gives $7$. Every number multiplied by $0$ gives $0$, never $7$:

$n \times 0 = 0 \text{ for every number } n$

So nothing answers the question, and the fraction names no number. We say it is undefined.

The parts reading breaks in the same place. A denominator of $0$ claims the whole was cut into no parts at all, and then there are no parts for a numerator to count.

The asymmetry is the thing to hold on to. The two positions in a fraction are not interchangeable: a numerator of $0$ is an ordinary count that happens to be none, while a denominator of $0$ removes the very thing a fraction is built out of.

The denominator and the numerator give two instructions, and the denominator moves first.

The denominator $7$ says to split the choir into $7$ equal groups, so begin by sizing one group:

$35 \div 7 = 5$

Each group holds $5$ singers, so one seventh of the choir is $5$ singers. The numerator $2$ says to take two of those groups:

$2 \times 5 = 10$

So $10$ singers take the low harmony line.

The order of the two steps is not a matter of taste. Divide by the denominator to find the size of one part, then multiply by the numerator to count how many of those parts you want. And the split lands on whole singers only because $7$ divides $35$ evenly, which is what let the first step come out as a whole number.

The choir is split into $7$ equal groups and $2$ of them take the harmony, so the groups left for the melody number

$7 - 2 = 5$

out of the seven. That is five sevenths of the choir.

Now size the groups and count them, exactly as before. One group holds

$35 \div 7 = 5 \text{ singers}$

and five groups hold

$5 \times 5 = 25 \text{ singers}$

so $25$ singers take the melody.

All seven groups together are $\frac{7}{7}$ of the choir. A fraction whose numerator equals its denominator is one whole, because taking all $7$ of the $7$ equal groups is taking the choir back entire, with nobody left out and nobody counted twice. That is the check on the whole arrangement: the two lines between them use up every group, and every group holds $5$ singers.

The method starts the same way. Split the new choir into $7$ equal groups:

$38 \div 7$

Seven goes into $38$ five times with $3$ left over, so this division does not come out even and no whole number answers it. The fraction bar does answer it:

$38 \div 7 = \frac{38}{7}$

So one seventh of $38$ singers is exactly $\frac{38}{7}$ singers, a quantity larger than $5$ and smaller than $6$.

That is a perfectly good number, and for a length of rope or a weight of flour it would end the question. Singers are different: a group cannot hold a fraction of a person. So the arithmetic is reporting something real about the situation rather than failing at it. A choir of $38$ cannot be divided into seven equal groups of whole singers at all, and no clever arrangement of the rehearsal will change that.

What the conductor has to change is something else: the total, by recruiting or resting singers until it is a multiple of $7$, or the plan, by giving up on groups of exactly equal size.

The season of $35$ worked only because $7$ divides $35$ evenly. The fraction $\frac{2}{7}$ was never the difficulty. What matters is whether the denominator divides the number being split.

Each step is one of $5$ equal parts of a unit, so every step is $\frac{1}{5}$ and the marks count fifths. Seven steps from $0$ is seven fifths:

$\frac{7}{5}$

Its numerator is larger than its denominator, so it is an improper fraction, and an improper fraction has reached or passed one whole.

Where it lands can be counted rather than guessed. Five steps carry you to

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

and the two steps after that are still short of the next whole, which would need five more. So $\frac{7}{5}$ sits between $1$ and $2$, two fifths of the way along that second unit.

Nothing about the counting changed when the marks passed $1$. The line simply kept going, the steps stayed the same size, and the numerator went on counting them.

Counting in fifths, five steps make one whole:

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

because taking all $5$ of the $5$ equal parts of a unit is taking that unit back. Ten steps cross a second unit in the same way:

$\frac{10}{5} = 10 \div 5 = 2$

The test that needs no picture comes from reading the bar as a division. A fraction names a whole number exactly when the division it stands for comes out even, that is, when the denominator divides the numerator with nothing left over. It works in both directions: if the denominator divides the numerator, the division has a whole-number answer, and if the fraction equals a whole number, then that whole number times the denominator is the numerator, so the denominator divides it.

Here $5$ divides $5$ and $5$ divides $10$, so both of those marks land on whole numbers, while $5$ does not divide $7$, which is why the seventh mark landed strictly between two of them. The whole-number marks on this line are exactly the ones whose numerator is a multiple of $5$: $\frac{5}{5}$ at $1$, $\frac{10}{5}$ at $2$, and on a longer line $\frac{15}{5}$ at $3$.

One mark from this very line settles the claim.

Placing a fraction has two stages: the denominator fixes the size of a step, and the numerator says how many steps to count from $0$. Nothing in the second stage says to stop at the $1$ mark. The line continues, the steps continue, and the count runs as far as the numerator says:

$\frac{7}{5} > 1$

because seven steps of a fifth carry you past the five steps that make one whole. A single case like that is enough to sink a claim about every fraction.

What survives is a sharper statement, and the line shows all three cases at once. When the numerator is smaller than the denominator, the count runs out before it has taken the number of steps a whole requires, so the mark lands short of $1$: those are the proper fractions. When the numerator equals the denominator, the count lands exactly on $1$. When the numerator is larger, the count carries past it, which is the improper case.

The phrase that misled the classmate is "a part of a whole". A fraction is a count of equal parts, and nothing forbids counting out more parts than a single whole contains, provided there is more than one whole to take them from.

The first pair counts the same size of part. Both fractions are built from sixteenths, so the only thing that differs is how many sixteenths each one holds, and eleven of a thing beats seven of the very same thing:

$\frac{7}{16} < \frac{11}{16}$

The second pair counts the same number of parts, six each, but the parts are not the same size. One whole cut into $7$ pieces gives larger pieces than the same whole cut into $13$, so six of the larger pieces outweigh six of the smaller ones:

$\frac{6}{7} > \frac{6}{13}$

The two shortcuts read the numbers in opposite directions, and it is worth saying why. With the denominators matching, the larger numerator wins, because it counts more of an identical part. With the numerators matching, and that shared count not zero, the smaller denominator wins, because fewer cuts leave bigger parts.

That second condition earns its place. Taking none of a part is nothing at all, whatever the parts are like, so a shared numerator of $0$ leaves two fractions equal rather than one of them ahead:

$\frac{0}{3} = 0 = \frac{0}{7}$

With a genuine count of parts the rule holds, and it only feels backwards if the denominator is read as a size; it is a count of how many parts the whole was cut into.

Both requests come down to choosing which of the two numbers to hold fixed.

For the first, hold the denominator at $11$ so that everything counts elevenths. The order of the fractions is then the order of their numerators, and any numerator strictly between $3$ and $8$ does the job:

$\frac{3}{11} < \frac{5}{11} < \frac{8}{11}$

The numerators $4$, $6$ and $7$ serve just as well. There are exactly four fractions with denominator $11$ that qualify, because there are exactly four whole numbers strictly between $3$ and $8$.

For the second, hold the numerator at $3$ so that both fractions take three parts, and then make the parts smaller by cutting the whole into more of them. Any denominator larger than $11$ works:

$\frac{3}{14} < \frac{3}{11}$

Three fourteenths is three of a smaller piece than three elevenths, so it is the smaller amount. A denominator of $12$, of $20$ or of $100$ would do equally well.

The two requests are mirror images of one another. The first freezes the size of the part and varies the count; the second freezes the count and varies the size. Each shortcut exists because exactly one thing is being allowed to move.

Test the claim against the pair rather than arguing about it in general.

The first shortcut needs the two fractions to be built from the same size of part, which is to say it needs them to share a denominator. Here the denominators are $7$ and $9$, so it does not apply.

The second shortcut needs the two fractions to take the same number of parts, which is to say it needs them to share a numerator, and that shared count must not be zero. Here the numerators are $4$ and $5$, so it does not apply either.

The extra condition on the second one is worth stating rather than assuming. If the shared numerator were $0$, both fractions would be $0$ however their parts were sized:

$\frac{0}{3} = 0 = \frac{0}{7}$

and the smaller denominator would win nothing, because the two amounts are equal. A shortcut is only as good as the conditions attached to it.

Neither shortcut reaches the pair, so the classmate's claim is too strong. Now be exact about what that does and does not show. The two fractions certainly have a definite order: each names a single point on the number line, and one of those two points lies to the left of the other. What is missing is a method, and both shortcuts are missing it for the same reason. A shortcut works by freezing one of the two numbers so that only one thing varies, and

$\frac{4}{7} \quad \text{against} \quad \frac{5}{9}$

lets the count of parts and the size of the parts change at once. Comparing $4$ against $5$ ignores that the parts differ; comparing $7$ against $9$ ignores that the counts differ. Neither number can be read on its own.

The repair is to rewrite one or both fractions so that they do share a denominator, which is the business of the next lesson. Until then, the honest report on this pair is that these two shortcuts do not decide it.