Equivalent Fractions and Simplifying: Free Response

Follow the two counts separately, because one cutting acts on both of them.

The strip was in $7$ equal parts and every part became $3$ pieces, so the whole strip now holds

$7 \times 3 = 21 \text{ pieces.}$

The $2$ coloured parts went under the same knife, so each of them became $3$ pieces too:

$2 \times 3 = 6 \text{ coloured pieces.}$

Six of the $21$ pieces are coloured, so in terms of the new pieces the coloured share is

$\frac{6}{21}.$

No card was added and none was taken away; only the cut lines changed. So $\frac{6}{21}$ and $\frac{2}{7}$ are two descriptions of one and the same coloured amount.

Take the classmate's line at its word before judging it. A denominator counts the pieces the whole is in, so $\frac{8}{9}$ claims the strip is in $9$ pieces with $8$ of them coloured.

That strip is not in $9$ pieces any more. Every one of its parts was cut into $2$, so

$9 \times 2 = 18$

pieces make up the strip now, and the $4$ coloured parts became

$4 \times 2 = 8$

coloured pieces. The classmate scaled the coloured count and left the piece count exactly as it was, which describes a strip whose parts were never cut at all.

Read the two lines back as amounts of colour. $\frac{4}{9}$ colours $4$ of the $9$ parts, under half the strip; $\frac{8}{9}$ colours $8$ of them, nearly all of it. Drawing cut lines cannot colour in more card, so the classmate's line cannot be a record of this cutting.

With both counts scaled by the same $2$, the record is

$\frac{4}{9} = \frac{8}{18}.$

The two numbers in a fraction count different things, and one act of cutting changes both of them at once.

Start from what the fraction says: the whole is in $b$ equal parts and $a$ of them are coloured. Cut every part into $c$ equal pieces. The whole was in $b$ parts and each became $c$ pieces, so the whole is now in $b \times c$ pieces. The coloured parts met the same knife, so the $a$ coloured parts became $a \times c$ coloured pieces. Neither count could change without the other, because a single cutting produced both:

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

No card was added and none removed, so the coloured amount is exactly what it was, described in smaller pieces.

Now suppose only one of the two numbers is scaled, on a strip that has some colour on it. Scaling the numerator alone says more pieces are coloured while the whole is still in its original parts, which is somebody colouring in extra card. Scaling the denominator alone says the whole was cut finer while the coloured count stood still, which is somebody rubbing colour out. Either way the amount moves, and the two fractions no longer name the same number.

The words about colour are doing real work there. A strip with no colour at all is the one case where a lopsided change does no damage: $\frac{0}{7}$ and $\frac{0}{14}$ both name $0$, because no colour is no colour however finely the card is cut. That is not a loophole in the rule. The building rule scales both counts and gets the value right in every case, this one included; what the exception shows is that scaling both is what always works, not merely the only thing that ever can.

The multiplier can never be $0$. Cutting each part into $0$ pieces leaves no pieces to count, and both counts collapse at once, since the fraction it would produce is

$\frac{a \times 0}{b \times 0} = \frac{0}{0}.$

A denominator of $0$ names no number, because it asks for a division by $0$, and no such division picks out one number: for $\frac{6}{0}$ nothing multiplied by $0$ gives $6$, while for $\frac{0}{0}$ every number would qualify.

The plan is a fraction of a bed and the bed is a number of plots, so rewrite $\frac{5}{8}$ with $24$ underneath it.

The multiplier comes from the denominators. The plan counts eighths and this bed is in $24$ plots, and

$8 \times 3 = 24,$

so the multiplier is $3$. In plain terms, one eighth of this bed is $3$ plots. The building rule sends that same $3$ to the numerator:

$\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}.$

So the herbs take $15$ plots. The unit is worth keeping straight: $15$ is a count of plots, while $\frac{15}{24}$ is the share of the bed those plots make up, and the two are the same fact stated in different currencies.

The planted share is $18$ plots out of the $30$ the bed holds, which is $\frac{18}{30}$ before any simplifying. Reach lowest terms in one division by dividing by the greatest common factor, so find that first:

$18 = 2 \times 3 \times 3, \qquad 30 = 2 \times 3 \times 5.$

The primes they share are one $2$ and one $3$, so $\operatorname{GCF}(18, 30) = 6$. Divide both terms by it:

$\frac{18}{30} = \frac{18 \div 6}{30 \div 6} = \frac{3}{5}.$

Now check that the job is finished. The only factor $3$ and $5$ share is $1$, so there is nothing left to divide out and $\frac{3}{5}$ is in lowest terms.

Read it back at the bed. Grouping the $30$ plots into $5$ blocks of $6$, the vegetables fill $3$ of those blocks, which is the same land the $18$ plots covered.

The question is whether the plan's fraction can be written with $20$ underneath it using a whole-number multiplier.

That multiplier would have to carry $8$ to $20$:

$8 \times c = 20.$

The multiples of $8$ run $8, 16, 24, 32, \ldots$, and $20$ is not among them, so no whole number $c$ does it. Marking $\frac{5}{8}$ of a $20$-plot bed would need part of a plot, and a plot is the smallest piece this bed has.

The same argument says exactly which beds do work. A bed of $n$ plots serves the plan when $8 \times c = n$ for some whole number $c$, and that is precisely the statement that $n$ is a multiple of $8$:

$n = 8, \; 16, \; 24, \; 32, \; 40, \; \ldots$

Both directions hold here: every such bed can be marked out, because $c$ plots then make one eighth and $5 \times c$ plots make the herb share, and no other bed can, because the multiplier would not be whole. The first bed in this question has $24$ plots, which is on the list, and $20$ is not.

The supplier's bed is not useless, of course. It can be marked out if the plots themselves are subdivided, but then the bed is no longer in $20$ pieces, which is a different bed from the one that was bought.

One division is enough only if it divides out everything the two numbers share, so find the greatest common factor first. Write each as a product of primes:

$32 = 2 \times 2 \times 2 \times 2 \times 2, \qquad 56 = 2 \times 2 \times 2 \times 7.$

The copies of $2$ they share number three, and there is no other shared prime, so

$\operatorname{GCF}(32, 56) = 2 \times 2 \times 2 = 8.$

Divide the numerator and the denominator by that one number:

$\frac{32}{56} = \frac{32 \div 8}{56 \div 8} = \frac{4}{7}.$

Both terms were divided by the same $8$, so the result names the same amount as the fraction it came from. And $4$ and $7$ share no factor above $1$, so there is nothing further to take out.

Halving acts on both terms at once, so it is the simplifying rule with a common factor of $2$, applied over and over:

$\frac{32}{56} = \frac{16}{28} = \frac{8}{14} = \frac{4}{7}.$

At each of those steps both numbers were even, so both halved to whole numbers. The route stops at $\frac{4}{7}$ because halving again would ask for half of $7$, which is not a whole number.

Now set the two routes side by side. The halving route took three divisions; the single-division route took one, and that is not a coincidence:

$2 \times 2 \times 2 = 8,$

so the three halvings between them divided out exactly the number the one division divided out at a stroke.

Both routes end at the same fraction here, and that is not luck, though it is not automatic either. Every step of either route divides both terms by a common factor, so neither route moves the value. The three halvings between them removed everything $32$ and $56$ share, so the halving route lands in lowest terms as well, and a fraction has only one lowest-terms form for the two of them to land on.

That second condition is the one to watch, because it can fail. Halving stops as soon as one of the terms is odd, which need not be lowest terms at all: on a fraction whose numerator and denominator share an odd factor above $1$, the halving route runs out early and leaves that factor sitting there. It finished the job here only because everything $32$ and $56$ share is built out of $2$s.

The difference between the routes is therefore bookkeeping and risk. The halving route needs a check after every step, and a check again when it stops, to see whether anything is left.

A claim that says always is settled by one example, and there is one already on the page.

The first halving divides by the common factor $2$, which is certainly bigger than $1$:

$\frac{32}{56} = \frac{16}{28}.$

By the claim, $\frac{16}{28}$ should now be in lowest terms. It is not: $16$ and $28$ are both even, so $2$ divides both of them again, and in fact $4$ does. One counterexample is enough, so the claim does not hold.

What went wrong is that $2$ is a common factor but not the greatest one. Dividing by a common factor removes that factor and nothing else, so whatever else the two numbers share is still sitting there afterwards. The greatest common factor is the one number that carries everything they share, which is why dividing by it finishes the job at a stroke.

The test for a finished fraction says nothing about the route taken to it: a fraction is in lowest terms when its numerator and denominator share no factor above $1$. Applied to $\frac{16}{28}$ that test says keep going. Applied to $\frac{4}{7}$ it says stop, since the only factor $4$ and $7$ share is $1$.

Take each fraction on its own and divide out everything its two terms share.

For the first, $21 = 3 \times 7$ and $49 = 7 \times 7$, so the only prime they share is one $7$ and $\operatorname{GCF}(21, 49) = 7$:

$\frac{21}{49} = \frac{21 \div 7}{49 \div 7} = \frac{3}{7}.$

For the second, $27 = 3 \times 3 \times 3$ and $63 = 3 \times 3 \times 7$, so they share two $3$s and $\operatorname{GCF}(27, 63) = 9$:

$\frac{27}{63} = \frac{27 \div 9}{63 \div 9} = \frac{3}{7}.$

Both land on $\frac{3}{7}$. Simplifying never moves a value, so each fraction names whatever $\frac{3}{7}$ names, and the two therefore name the same number.

Notice that the two multipliers were different, $7$ in one case and $9$ in the other. Fractions built from one starting fraction by different multipliers still name that same number, which is precisely why matching lowest-terms forms is what the test looks at rather than matching numerators or denominators.

The cross-product test pairs each numerator with the other fraction's denominator:

$14 \times 35 = 490,$

$24 \times 21 = 504.$

The two products are $490$ and $504$, which are different numbers, so the fractions are not equal.

It is worth seeing why those products decide anything. Write both fractions over the shared denominator $24 \times 35$:

$\frac{14}{24} = \frac{14 \times 35}{24 \times 35}, \qquad \frac{21}{35} = \frac{21 \times 24}{35 \times 24}.$

Each rewriting multiplies a numerator and a denominator by the same number, so neither has moved its fraction. With one and the same denominator underneath, the two can name the same amount only when the numerators agree, and those numerators are exactly the two cross products.

Simplifying agrees with the verdict, at the cost of two greatest common factors: $\frac{14}{24} = \frac{7}{12}$ and $\frac{21}{35} = \frac{3}{5}$, which are different lowest-terms forms.

Answer the two questions the pair raises, in order.

Are they equal? Simplify each:

$\frac{6}{10} = \frac{6 \div 2}{10 \div 2} = \frac{3}{5}, \qquad \frac{9}{15} = \frac{9 \div 3}{15 \div 3} = \frac{3}{5}.$

Same lowest-terms form, so the two name the same number.

Can either be built from the other by a nonzero whole-number multiplier? Building the second from the first needs a whole number $c$ with $10 \times c = 15$; the multiples of $10$ run $10, 20, 30, \ldots$, so there is none. Building the first from the second needs $15 \times c = 10$, and multiplying $15$ by any whole number of $1$ or more never lands below $15$, so there is none there either.

So here is a pair that is equal and in which neither fraction is a build of the other. The proposal said equality happens exactly in that case, and this pair is outside the case while still being equal, so the proposal is not a correct test as stated.

Be careful to take only what the counterexample gives. The proposal has two halves and they fare differently. The half that survives is the building rule itself: if one fraction is built from the other by multiplying both terms by the same nonzero whole number, the two certainly are equal. The word nonzero is not decoration there, and part of this set is about why: a multiplier of $0$ would produce $\frac{0}{0}$, which is no fraction at all. The half that breaks is the other direction, the claim that equal fractions must be related that way.

The repair is to go through lowest terms rather than directly between the two fractions. Both of these are built from $\frac{3}{5}$, by $2$ in one case and by $3$ in the other, and that shared starting fraction is what the proposal was missing. Two fractions are equal exactly when they have the same lowest-terms form, and that test does decide this pair correctly.

This job fixes the numerator, so read the multiplier off the numerators rather than the denominators:

$7 \times 5 = 35,$

so the multiplier is $5$. The building rule requires the denominator to take that same multiplier:

$\frac{7}{12} = \frac{7 \times 5}{12 \times 5} = \frac{35}{60}.$

The method is the familiar one read from the other row: find what the term you were told about was multiplied by, then do the same to the term you were not told about. Multiplying only the numerator would give $\frac{35}{12}$, which is a different number entirely, since only one of the two counts would have been scaled.

Each fraction needs its own multiplier, because each starts from a different denominator.

For the first, $10 \times 3 = 30$, so the multiplier is $3$:

$\frac{3}{10} = \frac{3 \times 3}{10 \times 3} = \frac{9}{30}.$

For the second, $15 \times 2 = 30$, so the multiplier is $2$:

$\frac{4}{15} = \frac{4 \times 2}{15 \times 2} = \frac{8}{30}.$

Neither rewriting moved a value, since each multiplied a numerator and a denominator by the same number. What has changed is the size of piece being counted: before, one fraction counted tenths and the other counted fifteenths, and now both count thirtieths, which is what having a shared denominator means.

A denominator can serve a fraction only if a whole-number multiplier reaches it from that fraction's own denominator, since a multiplier that is not whole would not leave whole numbers to count with.

For $\frac{3}{10}$ the new denominator must be $10 \times c$ for some whole number $c$, which says it is a multiple of $10$. For $\frac{4}{15}$ it must be $15 \times d$ for some whole number $d$, which says it is a multiple of $15$. To serve both at once it has to be a common multiple of $10$ and $15$.

Now pin that family down with prime factorizations. Being a multiple of $10 = 2 \times 5$ demands at least one $2$ and one $5$; being a multiple of $15 = 3 \times 5$ demands at least one $3$ and one $5$. Put the two demands together and the number must carry a $2$, a $3$ and a $5$, so it is a multiple of

$2 \times 3 \times 5 = 30.$

The other direction holds as well, which is what makes the description exact rather than merely necessary. Since $30 = 10 \times 3$ and $30 = 15 \times 2$, the denominator $30 \times k$ is reached from $10$ by the whole-number multiplier $3 \times k$ and from $15$ by $2 \times k$, so every multiple of $30$ really does serve both fractions. The family is therefore exactly

$30, \; 60, \; 90, \; 120, \; \ldots$

and $30$, the least common multiple of $10$ and $15$, is the smallest member. Any of the others would have done the job in part B; they describe the same two amounts in smaller pieces, so $\frac{3}{10} = \frac{18}{60}$ and $\frac{4}{15} = \frac{16}{60}$ carry exactly the same information as the thirtieths did.