Converting Between Fractions and Decimals: Free Response

The last digit $4$ sits in the hundredths place, two digits after the point, so the denominator is a hundred and the two digits sit directly on top of it:

$0.84 = \frac{84}{100}.$

That is already correct, and it is not yet finished. Find the greatest common factor by factorizing both numbers:

$84 = 2 \times 2 \times 3 \times 7, \qquad 100 = 2 \times 2 \times 5 \times 5.$

They share exactly two factors of $2$, so the greatest common factor is $2 \times 2 = 4$. Divide the top and the bottom by it:

$\frac{84}{100} = \frac{84 \div 4}{100 \div 4} = \frac{21}{25}.$

Now check that nothing is left to cancel. Since $21 = 3 \times 7$ and $25 = 5 \times 5$ share no prime at all, $\tfrac{21}{25}$ is in lowest terms and the conversion is done.

Take them one at a time, and read the denominator off the count of digits before doing anything else.

The decimal $0.208$ has three digits after the point, so its last digit sits in the thousandths place and the denominator is a thousand:

$0.208 = \frac{208}{1000}.$

Now reduce. Factorizing gives $208 = 2 \times 2 \times 2 \times 2 \times 13$ and $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$, so the shared part is three factors of $2$ and the greatest common factor is $8$:

$\frac{208}{1000} = \frac{208 \div 8}{1000 \div 8} = \frac{26}{125}.$

Since $26 = 2 \times 13$ and $125 = 5 \times 5 \times 5$, no prime is shared and nothing is left to cancel.

The decimal $0.95$ has two digits after the point, so the denominator is a hundred, and here the greatest common factor is only $5$:

$0.95 = \frac{95}{100} = \frac{95 \div 5}{100 \div 5} = \frac{19}{20}.$

Since $19$ is prime and $20 = 2 \times 2 \times 5$, the fraction $\tfrac{19}{20}$ is in lowest terms. The two conversions needed different factors divided out, $8$ against $5$, which is why the reduction has to be worked each time rather than guessed from the last one.

Run the shortcut forwards on the answer and see whether the decimal you started from comes back. The denominator $25$ divides a hundred, since $100 = 25 \times 4$, so multiply the top and the bottom by $4$:

$\frac{21}{25} = \frac{21 \times 4}{25 \times 4} = \frac{84}{100} = 0.84.$

That is the decimal in the question, so the conversion survives the check.

Now ask what the check is actually testing. Scaling by the same number top and bottom leaves a fraction's value alone, reading a decimal off a denominator that is a $1$ followed by zeros leaves it alone, and so does the reduction those two moves undo. Every step of the round trip preserves the value, so the whole check tests one thing: whether the fraction you wrote down is the same number as the decimal you were given.

That makes it strong against any slip that changed the value. Divide the top by $4$ and the bottom by $5$, an unequal reduction, and you reach $\tfrac{21}{20}$, which scales to a hundred by $5$:

$\frac{21}{20} = \frac{105}{100} = 1.05,$

nowhere near $0.84$. Miscount the places and write $\tfrac{84}{1000}$, whose denominator is already a $1$ followed by zeros so nothing needs scaling at all, and the round trip reads back $0.084$ instead. Both slips are caught, because both changed the number.

And there is a whole class of thing the check cannot see, the one that matters here being whether the answer is in lowest terms. A fraction correct in value but only partly reduced, such as $\tfrac{42}{50}$, scales up just as happily:

$\frac{42}{50} = \frac{42 \times 2}{50 \times 2} = \frac{84}{100} = 0.84.$

The check passes it, and it is right to, because it is the same number. Being in lowest terms is a fact about the form of the answer rather than its value, and a check built entirely out of value-preserving moves cannot see the form. So the round trip is worth doing and it is not a substitute for asking whether anything is still left to cancel.

Test the three candidate denominators in turn. Dividing $10$ by $40$ does not come out at all, and $100 \div 40$ leaves a remainder of $20$, so neither works. A thousand does work:

$1000 = 40 \times 25.$

So multiply the top and the bottom by $25$. That changes nothing about the value, because multiplying a fraction's top and bottom by the same number is the equivalent-fractions move:

$\frac{17}{40} = \frac{17 \times 25}{40 \times 25} = \frac{425}{1000}.$

A denominator of a thousand means three decimal places, and the top supplies those three digits:

$\frac{17}{40} = \frac{425}{1000} = 0.425.$

A size check costs nothing and is worth having: $\tfrac{17}{40}$ is a little under $\tfrac{20}{40} = \tfrac{1}{2}$, and $0.425$ is a little under $0.5$.

For the shortcut to work, the denominator has to divide the target exactly. Divide each target by $16$ and look at what is left over:

$10 \div 16 = 0 \text{ remainder } 10, \qquad 100 \div 16 = 6 \text{ remainder } 4, \qquad 1000 \div 16 = 62 \text{ remainder } 8.$

Not one comes out exactly, so no whole-number multiplier turns $16$ into ten, a hundred or a thousand, and the shortcut has nothing to offer. Fall back on the division the fraction bar has always meant, $11 \div 16$.

Sixteen goes into $11$ zero times, so the quotient starts with $0$ and a point, the opening remainder is $11$ itself, and the division runs into $11.0000$. Sixteen into $110$ tenths goes $6$ times, since $6 \times 16 = 96$, leaving a remainder of $14$. Sixteen into $140$ hundredths goes $8$ times ($128$), leaving $12$. Sixteen into $120$ thousandths goes $7$ times ($112$), leaving $8$. Sixteen into $80$ ten-thousandths goes $5$ times ($80$), leaving nothing:

$\frac{11}{16} = 11 \div 16 = 0.6875.$

The remainder reached $0$, so the division stopped of its own accord after four places. Check the size: $\tfrac{11}{16}$ is more than $\tfrac{8}{16} = 0.5$ and less than one whole, and $0.6875$ is more than $0.5$ and less than $1$ as well.

The test mistakes a convenient list for a complete one. Ten, a hundred and a thousand are the first three numbers that are a $1$ followed by zeros, and they are the three worth trying by hand, but the list runs on for ever. Ten thousand is the next one, and $16$ divides it:

$10000 = 16 \times 625.$

So scale by $625$:

$\frac{11}{16} = \frac{11 \times 625}{16 \times 625} = \frac{6875}{10000} = 0.6875,$

which agrees with the long division in part B, four zeros in the denominator matching four decimal places. The fraction can be written over a $1$ followed by zeros after all. It just needs a bigger one than the three the shortcut tries first.

So what does the shortcut's failure tell you? Only that the denominator divides none of those first three, which is a fact about how far you happened to look. The question of whether the decimal ends is settled somewhere else, by the rule that a fraction in lowest terms gives a decimal that ends exactly when the only prime factors of its denominator are $2$ and $5$. Reduce first, then factorize. Here $11$ is prime and does not divide $16$, so $\tfrac{11}{16}$ is already in lowest terms, and

$16 = 2 \times 2 \times 2 \times 2.$

Every prime factor is a $2$, so the decimal ends, and the four $2$s are exactly why it took four places to get there: a thousand carries only three $2$s, while ten thousand carries four. Trying ten, a hundred and a thousand by hand is a good habit. Reading a verdict out of their failure is not.

Sharing $35$ metres equally between $28$ panels is a division, and the fraction bar is how a division is written, so one panel is $\tfrac{35}{28}$ of a metre.

Reduce it before doing anything else. Both numbers are multiples of $7$, since $35 = 5 \times 7$ and $28 = 2 \times 2 \times 7$, so the greatest common factor is $7$:

$\frac{35}{28} = \frac{35 \div 7}{28 \div 7} = \frac{5}{4}.$

The reduced denominator $4$ divides a hundred, since $100 = 4 \times 25$, so the shortcut applies and no long division is needed:

$\frac{5}{4} = \frac{5 \times 25}{4 \times 25} = \frac{125}{100} = 1.25.$

One panel is $1.25$ metres wide. The size is believable, since $28$ panels of a little over a metre should come to a little over $28$ metres, and multiplying back confirms it exactly:

$28 \times 1.25 = 35.$

That multiplication is also a check on the reducing, since a slip there would not have rebuilt the bolt.

First reduce, because both the test and the division want lowest terms. Factorizing gives $35 = 5 \times 7$ and $18 = 2 \times 3 \times 3$, which share no prime, so $\tfrac{35}{18}$ is already in lowest terms.

Now divide $35$ by $18$. Eighteen fits into $35$ once, using $18$ and leaving $17$, so the whole part is $1$ and the division carries the remainder $17$ past the point into $35.000$.

Eighteen into $170$ tenths goes $9$ times, since $9 \times 18 = 162$, leaving a remainder of $8$. Eighteen into $80$ hundredths goes $4$ times ($72$), leaving a remainder of $8$ again. That second $8$ is the signal to stop: the same remainder must produce the same next digit and the same next remainder, so from here the division hands back a $4$ for ever:

$\frac{35}{18} = 35 \div 18 = 1.9444\ldots = 1.9\overline{4}.$

The bar covers the $4$ alone. The $9$ in the tenths place came from the remainder $17$, which never returns, so the $9$ happens exactly once and stands outside the repeating block. The prediction agrees with the division, since $18 = 2 \times 3 \times 3$ carries two $3$s and this fraction is in lowest terms.

Compare the two widths against what the table will take. The $28$-panel width came out as $1.25$ metres, a decimal that ends after two places, so it can be typed in exactly. The $18$-panel width is $1.9\overline{4}$ metres, whose $4$s never stop, so there is no exact decimal to type at all. Only the cut into $28$ panels can be set on this table.

So the assistant is wrong about one of the two, and the reason is worth being precise about. Here are the two starting denominators factorized:

$28 = 2 \times 2 \times 7, \qquad 18 = 2 \times 3 \times 3.$

Both do carry a prime other than $2$ and $5$, exactly as the assistant says, and yet the two widths behave differently. The observation is true and the conclusion drawn from it is not, so the test must be missing a step.

The missing step is the reduction. The rule is that a fraction in lowest terms gives a decimal that ends exactly when the only prime factors of its denominator are $2$ and $5$, and until the fraction is in lowest terms its denominator is not the one the rule is talking about. For the first cut the numerator brought a $7$ of its own:

$\frac{35}{28} = \frac{5}{4}, \qquad 4 = 2 \times 2.$

The $7$ in $28$ cancelled against the $7$ in $35$, leaving a denominator of nothing but $2$s, so the decimal ends. For the second cut nothing cancels, since $35 = 5 \times 7$ and $18 = 2 \times 3 \times 3$ share no prime, so $18$ is the denominator the rule speaks about, its two $3$s are still there because there was no reduction to remove them, and the decimal repeats.

The working order is therefore fixed: reduce, then factorize, then decide. Reading a verdict off a denominator that has not been reduced is the one way to get this test wrong, and it goes wrong in only one direction, because a prime that was going to cancel makes a decimal that ends look as though it repeats, never the other way about.

The rule to apply is that a fraction in lowest terms gives a decimal that ends exactly when the only prime factors of its denominator are $2$ and $5$. The words in lowest terms say what has to be done first.

So reduce. Factorize both numbers to find the greatest common factor:

$45 = 3 \times 3 \times 5, \qquad 72 = 2 \times 2 \times 2 \times 3 \times 3.$

They share $3 \times 3 = 9$, so

$\frac{45}{72} = \frac{45 \div 9}{72 \div 9} = \frac{5}{8}.$

Now the rule has a fraction it applies to. Factorize the new denominator:

$8 = 2 \times 2 \times 2.$

Every prime factor is a $2$, so the prediction is that the decimal ends. Notice what became of the two $3$s in $72$: they were cancelled by the two $3$s the numerator brought, so they were never a problem in the first place.

Now convert and see. The denominator $8$ divides a thousand, since $1000 = 8 \times 125$, so scaling is quicker than dividing:

$\frac{5}{8} = \frac{5 \times 125}{8 \times 125} = \frac{625}{1000} = 0.625.$

The decimal ends after three places, as predicted, and three places is no accident: three $2$s in the denominator needed three $5$s brought alongside them to complete a thousand.

Reduce first again. Factorize both numbers:

$33 = 3 \times 11, \qquad 45 = 3 \times 3 \times 5.$

The greatest common factor is $3$, so

$\frac{33}{45} = \frac{33 \div 3}{45 \div 3} = \frac{11}{15}.$

Factorize the new denominator:

$15 = 3 \times 5.$

The $5$ is allowed and the $3$ is not, so this fraction in lowest terms carries a prime other than $2$ and $5$ and the prediction is that its decimal repeats.

Watch what the reduction managed. The denominator began with two $3$s. Cancelling removes copies of a prime one for one against copies the numerator brings, and $33 = 3 \times 11$ brought a single $3$, so one $3$ went and one stayed. The numerator's $11$ could cancel nothing, since $45$ has no factor of $11$. A reduction can never remove more copies of a prime than the numerator holds, which is why the outcome here differs from part A even though both denominators started with $3$s in them.

Now convert. There is no shortcut, since $15$ divides none of ten, a hundred or a thousand, so divide $11$ by $15$. Fifteen does not fit into $11$, so the quotient starts $0$ and a point and the remainder $11$ carries past it. Fifteen into $110$ tenths goes $7$ times, since $7 \times 15 = 105$, leaving a remainder of $5$. Fifteen into $50$ hundredths goes $3$ times ($45$), leaving a remainder of $5$ again, so the same digit comes round for ever:

$\frac{11}{15} = 11 \div 15 = 0.7333\ldots = 0.7\overline{3}.$

The prediction holds, and the bar sits over the $3$ alone, because the $7$ came from the opening remainder $11$, which never returns.

An "exactly when" is two statements, and they can meet different fates, so take them one at a time.

The direction that holds. Suppose the denominator's only prime factors are $2$ and $5$. Then it divides some $1$ followed by zeros, because ten is $2 \times 5$, a hundred is $2 \times 2 \times 5 \times 5$, a thousand is three $2$s times three $5$s, and the list goes on, so one of them always has at least as many $2$s and at least as many $5$s as the denominator does. Scaling to it turns the fraction into a decimal that ends, so it does not repeat. Nothing in that argument asked the fraction to be reduced, and it is not fussy about it. Take $\tfrac{12}{50}$, which is not in lowest terms, and whose denominator is $2 \times 5 \times 5$:

$\frac{12}{50} = \frac{12 \times 2}{50 \times 2} = \frac{24}{100} = 0.24.$

That ends. So read in the direction "if it repeats, there must be a prime other than $2$ and $5$ down there", the classmate's rule is sound.

The direction that fails. Read the other way round, the rule says a prime other than $2$ and $5$ in the denominator forces the decimal to repeat, and a single case brings that down. Part A's fraction has $72 = 2 \times 2 \times 2 \times 3 \times 3$, two $3$s in plain sight, and yet

$\frac{45}{72} = \frac{5}{8} = 0.625,$

a decimal that ends after three places. The $3$s were never in the denominator of the number; they were in that particular way of writing it, and the numerator cancelled them.

The repair. Attach the condition the rule was always about: a fraction in lowest terms repeats exactly when its denominator has a prime factor other than $2$ and $5$. With those words in place both directions hold, because a reduced denominator has no cancellable primes left in it to be misread. In practice the condition is an instruction about order: reduce, then factorize, then decide.

It is worth noticing that the repair is needed on one side only. The direction that held did not need it, because a denominator of nothing but $2$s and $5$s still has nothing but $2$s and $5$s after reducing, since reducing only removes factors. All the harm an unreduced fraction can do is to make a decimal that ends look as though it repeats, and never the reverse.

Eleven does not fit into $4$, so the quotient starts with $0$ and a point, and the division carries the remainder $4$ past the point into the tenths.

Eleven into $40$ tenths goes $3$ times, since $3 \times 11 = 33$, leaving a remainder of $7$. Eleven into $70$ hundredths goes $6$ times ($66$), leaving a remainder of $4$. That $4$ is the remainder the division started with, so everything from here has to happen again in the same order: the next digit must be $3$, then $6$, then $3$, without end.

$\frac{4}{11} = 4 \div 11 = 0.363636\ldots = 0.\overline{36}.$

The bar covers both digits, because the block that repeats is the pair $36$ and it begins in the very first decimal place. No digit stands outside it.

That the division could never end was predictable before it began: $4$ and $11$ share no factor, so $\tfrac{4}{11}$ is already in lowest terms, and $11$ is a prime other than $2$ and $5$.

Run the division and keep every remainder. Twelve does not fit into $7$, so the quotient begins $0$ and a point and the remainder $7$ carries past it.

Twelve into $70$ tenths goes $5$ times ($60$), leaving a remainder of $10$. Twelve into $100$ hundredths goes $8$ times ($96$), leaving a remainder of $4$. Twelve into $40$ thousandths goes $3$ times ($36$), leaving a remainder of $4$ again.

So the remainders are $7$, $10$, $4$, $4$. The first repeat is the $4$, and it repeats after one step rather than after three, so the cycle is one digit long and the digit it produces is the $3$:

$\frac{7}{12} = 7 \div 12 = 0.58333\ldots = 0.58\overline{3}.$

The $5$ and the $8$ came from the remainders $7$ and $10$, and neither of those ever comes back, so those two digits happen once each and belong outside the bar. The classmate's bar swept them into the block.

A comparison of digits settles it with no theory at all. The reported $0.\overline{583}$ claims the whole block $583$ repeats, so its digits run

$0.583583583\ldots$

while the true value runs $0.583333\ldots$. They agree for three places and part company at the fourth, where the reported value has a $5$ and the division produces a $3$. Two decimals that differ in any place are different numbers, so $0.\overline{583}$ is not $\tfrac{7}{12}$.

Why a repeat is forced. Every step of a long division leaves a remainder, and a remainder is always smaller than the number you are dividing by. So the remainders a given division can produce form a short list fixed before the division starts: for a divisor of $12$ they are the whole numbers from $0$ to $11$, and for any divisor they run from $0$ up to one less than the divisor. If a remainder of $0$ ever turns up, the division stops and the decimal ends. If $0$ never turns up, the division takes step after step for ever while drawing from that same short list, so some remainder has to be used a second time.

And a remainder decides everything that follows it. The next digit is how many times the divisor fits into ten times the remainder, and what is left over is the next remainder. Part B shows both halves of that on one number:

$100 = 12 \times 8 + 4, \qquad 40 = 12 \times 3 + 4,$

where the remainder $4$ arrives and then reproduces itself, which is exactly why its digit $3$ goes on for ever. So once a remainder returns, the digits after it return in the same order, and there is no room left for a tail without a pattern.

How long the block can be. For a divisor of $12$ the possible remainders are

$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.$

A remainder of $0$ would end the division rather than continue it, so a division that repeats has only the eleven values $1$ through $11$ to work with, and it must revisit one of them within twelve steps. The repeating block therefore cannot be longer than $11$ digits. Counting the whole list of twelve instead, with the $0$ left in, gives a bound of $12$: also correct, just less sharp. Either is an honest answer to the counting question, and neither is a claim that any block actually gets that long. Dividing by $12$ in fact never produces a block longer than one digit, as part B found. A ceiling reached by counting is still worth having, because it is known before any dividing is done.

Why the bar need not start at the point. The block begins over the digit produced by the first occurrence of whichever remainder later comes back, not over the digit produced when it returns, and there is no reason for that first occurrence to be the opening remainder. In part A it was: the division began with a remainder of $4$ and came back to $4$, so the block started immediately and the bar covered every decimal digit. In part B it was not: the remainders $7$ and $10$ each appeared once and never again, and only the $4$ came back, so the two digits they produced stood outside the block. A remainder that never returns produces its digit exactly once, and a digit that happens once cannot go under a bar whose whole claim is that it happens for ever.