Decimals: Chapter Test

Read outwards from the point. Immediately to its left is the ones column and beyond that the tens; immediately to its right the columns continue as tenths, hundredths and thousandths, each one a tenth of the column before it.

$27.406 \longrightarrow 2 \text{ tens}, \; 7 \text{ ones}, \; 4 \text{ tenths}, \; 0 \text{ hundredths}, \; 6 \text{ thousandths}$

A digit's value is the digit multiplied by what its column is worth, so the $4$ contributes four tenths and the $6$ contributes six thousandths. The zero column still has to be counted, because it is what puts the $6$ in the third place rather than the second.

Expanded form writes one term per column, each digit over the denominator its column names, and the zero term may be dropped because it adds nothing.

$27.406 = 20 + 7 + \frac{4}{10} + \frac{0}{100} + \frac{6}{1000} = 20 + 7 + \frac{4}{10} + \frac{6}{1000}$

Run the second sum the other way. Its terms name two tens, seven ones, four hundredths and six thousandths, so the tenths column has no term and needs a zero to hold it open.

$20 + 7 + \frac{4}{100} + \frac{6}{1000} = 27.046$

Read the whole-number part, say "and" for the point, then read the digits after the point as though they were a whole number and finish with the name of the column the last digit sits in.

$27.406 = \text{twenty-seven and four hundred six thousandths}$

The reason the last column supplies the name is that it is the smallest one in use, so every other column can be rewritten in terms of it and the digits then collect into a single fraction.

$\frac{4}{10} + \frac{0}{100} + \frac{6}{1000} = \frac{400}{1000} + \frac{0}{1000} + \frac{6}{1000} = \frac{406}{1000}$

The first column cannot do that job, because a thousandth is not a whole number of tenths: counting the whole string in tenths would leave a part of a tenth over, so the numerator could not come out whole.

Watch out. Naming it from the first column instead would read this time as "four hundred six tenths", which is more than forty seconds.

Pad every amount to two places so the cents column is never empty, then work down the column with the points aligned.

$40.00 - 12.45 = 27.55$

The second payment has one decimal place, so it is padded before anything is taken away.

$27.55 - 7.80 = 19.75$

The receipt is added the same way.

$19.75 + 3.06 = 22.81$

So the column closes at $22.81$ dollars. Combining the three entries first gives the same figure, since $12.45 + 7.80 - 3.06 = 17.19$ and $40.00 - 17.19 = 22.81$.

Pad the new amount to two places and take it from the running balance.

$22.81 - 22.90$

The amount being taken away is the larger of the two, so the result is below zero, and the size of the gap is the difference the other way round.

$22.90 - 22.81 = 0.09$

So the column closes at $-0.09$ dollars: nine cents short. The two figures agree in the dollars and part company in the cents, which is why the shortfall is so small.

The reason column arithmetic works is not the layout, it is what the layout enforces. Every column holds digits of a single place value, so each column sum is a sum of like quantities. Take the second and third of part A's entries and write them out by column.

$\left(7 + \frac{8}{10}\right) + \left(3 + \frac{0}{10} + \frac{6}{100}\right) = 10 + \frac{8}{10} + \frac{6}{100}$

Tenths gather with tenths and hundredths with hundredths, and nothing else can happen. A row of digit strings has no columns, so there is nothing to stop a tenth being added to a hundredth, and the point at the end is placed by guess rather than by dropping straight down.

The amount at risk is the one whose length differs from its neighbours. Two of part A's three entries carry two decimal places and $7.8$ carries one, so it is the one whose digits would slide into the wrong columns.

Watch out. Padding is what removes the risk, not care. Once $7.8$ is written as $7.80$ every amount has the same shape and there is nothing left to misjudge.

The bar means divide, so work out $9 \div 16$. Sixteen does not fit into $9$, so the quotient starts with a zero and a point and the division runs into $9.0000$.

$90 \div 16 = 5 \text{ r } 10, \quad 100 \div 16 = 6 \text{ r } 4, \quad 40 \div 16 = 2 \text{ r } 8, \quad 80 \div 16 = 5 \text{ r } 0$

The remainder has reached zero, so the division stops and the digits collected are the decimal.

$\frac{9}{16} = 0.5625$

A size check agrees: $\tfrac{9}{16}$ is a little over a half, and so is $0.5625$.

Three digits follow the point, so the last column is thousandths and the digits sit straight over a thousand with no arithmetic at all.

$0.096 = \frac{96}{1000}$

Now reduce. The greatest common factor of $96$ and $1000$ is $8$.

$\frac{96}{1000} = \frac{96 \div 8}{1000 \div 8} = \frac{12}{125}$

Check that nothing is left: $125$ is three fives and $12$ is two twos and a three, so they share no factor above one and the fraction is finished.

Call the width the fraction names $w$. The meaning of the bar is that $w$ is the number which, multiplied by the denominator, rebuilds the numerator, and that is exactly the definition of the quotient.

$16 \times w = 9 \quad \text{and} \quad 16 \times (9 \div 16) = 9$

So the fraction and the quotient are the same number, and a conversion that carries the division out to its end cannot land anywhere else. Long division writes that one number down a digit at a time, each digit the correct count for its own column, so the string it produces is the number itself.

Approximation enters when the division is stopped early. If the remainder has not reached zero and the digits are cut off, or the last digit kept is rounded, then what is written down differs from the true value by whatever the tail was worth. Cutting off always lands below the true value; rounding may land either side of it.

Watch out. "The digits ran out" and "I stopped writing digits" are different events. The first ends a conversion exactly; the second ends it approximately.

Ignore both points, multiply as whole numbers, then count the places the two factors carry between them.

$65 \times 184 = 11{,}960$

The rate carries two decimal places and the time carries one, so the product carries three.

$0.65 \times 18.4 = 11.960 = 11.96$

The trailing zero the rule produced is at the right-hand end, so it may be dropped. A size check agrees: two thirds of eighteen is about twelve.

The divisor has two decimal places, so move both points two places right. That multiplies each number by a hundred, and a hundred over a hundred is one, so the quotient does not move.

$46.8 \div 0.65 = 4680 \div 65$

The dividend needed a trailing zero for its second move, which never changes a value. Now it is whole-number division.

$4680 \div 65 = 72$

Check it back by multiplying: $0.65 \times 72 = 46.8$, which is the tank.

In the multiplication nothing was moved. Stripping the points and multiplying as whole numbers changes each factor by a known power of ten, and reading each factor as a whole number over ten or a hundred shows what happens to the answer.

$\frac{65}{100} \times \frac{184}{10} = \frac{65 \times 184}{1000}$

The denominators multiply, so their zeros add, and the count of places in the product is fixed in advance by the counts in the factors. That is why the point can be counted in at the end.

In the division the points had to move, and they had to move together, because a quotient is a ratio.

$\frac{46.8}{0.65} = \frac{46.8 \times 100}{0.65 \times 100} = \frac{4680}{65}$

Multiplying the top and the bottom by the same number leaves a ratio alone, which is what makes the move legal, and it means the quotient the long division produces is already the answer.

Moving only one point is not forbidden, it is just more work. Moving the divisor's point alone multiplies the quotient by a known power of ten, so the division still gives a usable number, but its point then has to be moved back the same number of places before it can be reported. Moving both is what makes the answer right immediately, with nothing left to undo, and that is the whole reason the rule pairs the two moves.

Watch out. The two rules are not opposites of each other. Both come from the same fact, that a decimal is a whole number over a power of ten; they differ because multiplying denominators and dividing them do different things.

The hundredths place holds the $9$, so the deciding digit is the thousandths digit immediately to its right, which is $6$.

$6 \ge 5 \;\Rightarrow\; \text{round up}$

Rounding up means raising the hundredths digit by one, but $9$ hundredths plus one hundredth is ten hundredths, which does not fit in a single column. It carries, exactly as in addition: the hundredths column goes to $0$ and one tenth is added next door.

$6.7961 \approx 6.80$

The final $1$ never mattered: everything past the thousandths adds less than one thousandth, so it cannot move the number back below the midpoint $6.795$. The trailing zero is kept, because "to the nearest hundredth" means two decimal places.

To the nearest hundredth, the deciding digit is the thousandths $3$, which is below five, so the hundredths digit stays as it is and everything after it goes.

$0.4938 \approx 0.49$

To the nearest tenth, the deciding digit is the hundredths $9$, which is five or more, so the tenths digit rises from $4$ to $5$.

$0.4938 \approx 0.5$

The two answers differ in what they retain. The first keeps two columns and so still records that the rate is a little under a half; the second keeps one column and rounds that detail away, and it is the very column the first answer kept that pushed it up.

Watch out. Rounding the first answer again would be a second rounding, not the same one. Rounding to a NAMED place always starts from the measured figure. Rounding for an estimate is a different job, and there any convenient nearby number will do.

Round both figures to something you can multiply in your head, then compare the size of the result with the figure on the sheet.

$0.5 \times 7 = 3.5 \text{ litres}$

The written total is in the threes, and so is the estimate, so they are the same size and the total survives the check. Had the sheet read a tenth of that, or ten times it, the estimate would have caught it at once, because a misplaced point moves an answer by a whole factor of ten.

What the check cannot do is confirm the figure. The estimate is a range, not a value, and many wrong totals of the right size pass it.

$3.36 \text{ and } 3.30 \text{ and } 3.42 \text{ all pass the same check}$

Only the exact product settles the digits, and it comes to just under $3.36$, so the sheet is in fact close.

Watch out. "Passed the size check" is not "is correct". The check has one job, catching an answer of the wrong magnitude, and it does that job whether or not the digits are right.

Pad all three to the same number of places, which changes no value, and then read from the left.

$4.110, \qquad 4.108, \qquad 4.100$

All three agree in the ones and the tenths, so those columns settle nothing. Comparing the first two, the hundredths differ, $1$ against $0$, and that is enough.

$4.110 > 4.108$

Comparing the second and third, the hundredths tie at $0$, so the comparison moves one column right to the thousandths, where $8$ meets $0$.

$4.108 > 4.100$

So the ranking is $4.11$, then $4.108$, then $4.1$, and the two comparisons were decided in different columns.

Once two decimals part company in some column, the winner of that column has a lead of at least one unit of it. The question is whether the rest of the digits can make that lead up, and they cannot.

These three measurements all stop, so each tail is a finite string of digits. Every column is one tenth of the one to its left, so the largest such a tail can be is nine of the next column, plus nine of the one after that, down to its last digit, which never reaches one full unit of the column that was decided.

$\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} < 1$

So the first differing column settles the comparison outright and the digits after it are along for the ride. Longer strings arise from measuring more finely, and a finely measured short jump is still short.

Watch out. Padding the other way makes this obvious: writing $4.1$ as $4.100$ gives it the same length without changing it at all, so length cannot have been carrying information about size.

Pad the fourth jump to three places so it can be lined up with the others. Its thousandths column is empty, so it reads as a zero.

$4.1\square 0 \qquad \text{against} \qquad 4.110, \; 4.108, \; 4.100$

The smudge sits in the hundredths, and part A found that the hundredths column is where the top of the ranking is decided, so the missing digit is the deciding one. Work through what it could be.

$\square = 0 \Rightarrow 4.100, \qquad \square = 1 \Rightarrow 4.110, \qquad \square \ge 2 \Rightarrow 4.120 \text{ or more}$

A smudged $0$ makes the jump equal to the shortest of the three. A smudged $1$ makes it equal to the longest. Anything from $2$ upwards puts it clear of all three. Those are three different positions in the ranking, so no placement can be made without reading the digit.

Watch out. It is tempting to argue that the smudged jump is at least as long as $4.1$ and so cannot be last, but a tie for last is still last, so even that much is not settled.

Reduce first, because that is the form the rule is about. The two terms share $3$ and $7$.

$\frac{63}{84} = \frac{63 \div 21}{84 \div 21} = \frac{3}{4}$

Now prime factorize the reduced denominator.

$4 = 2 \times 2$

The only prime present is $2$, so the rule predicts a decimal that ends. Note what reducing did: as written, $84 = 2 \times 2 \times 3 \times 7$ carries two forbidden primes, and both of them cancelled against the numerator, so testing the written form would have given the wrong answer.

Watch out. The rule is a statement about a fraction in lowest terms. Applied to a fraction that has not been reduced it is not a weaker rule, it is a false one.

Reduce first: both terms are divisible by $3$.

$\frac{42}{99} = \frac{42 \div 3}{99 \div 3} = \frac{14}{33}, \qquad 33 = 3 \times 11$

Neither prime is $2$ or $5$, so the rule predicts a decimal that repeats. Now divide and watch the remainders.

$140 \div 33 = 4 \text{ r } 8, \qquad 80 \div 33 = 2 \text{ r } 14$

The remainder $14$ is the numerator the division started from, so from here every step repeats what has already happened and the digits cycle. The block that recurs is the pair $42$.

$\frac{42}{99} = 0.424242\ldots = 0.\overline{42}$

The prediction and the division agree, as they must, and the division also shows why: a remainder can never be zero here, so it has to revisit an earlier value instead.

The claim joins two statements with "exactly when", so both have to be tested.

Direction one: only twos and fives in the written denominator, therefore the decimal ends. This one survives. Reducing a fraction divides both terms by a common factor, which can only remove primes from the denominator, never introduce a new one. So a denominator built from twos and fives still has only twos and fives after reducing, and the lowest-terms rule then applies directly. The second fraction is consistent with it too: $99$ carries a $3$ and an $11$, the decimal repeats, and nothing is contradicted.

Direction two: the decimal ends, therefore the written denominator has only twos and fives. This one fails, and one case is enough to sink it.

$\frac{63}{84} = \frac{3}{4} = 0.75, \qquad 84 = 2 \times 2 \times 3 \times 7$

The decimal ends while the written denominator carries two forbidden primes, so the claim is false in that direction. The repair is a single phrase.

$\text{a fraction IN LOWEST TERMS ends exactly when its denominator's only primes are } 2 \text{ and } 5$

Watch out. A rule that is right in one direction and wrong in the other is not half right. Used to predict, it will still send a student who has not reduced to the wrong answer, which is exactly what part A would have done.

Pad the shorter number at its right-hand end so that both numbers show three decimal places, then work column by column with the points aligned.

$14.900 + 6.375 = 21.275$

The thousandths give $5$, the hundredths $7$, the tenths $9 + 3 = 12$, so a $2$ is written and one is carried into the ones.

For the subtraction the whole number is padded the same way, with a point and three zeros.

$20.000 - 6.375 = 13.625$

Every column of the top number is a zero, so the borrow has to travel: it takes one from the tens, which leaves nine ones, then nine tenths, then nine hundredths, and ten thousandths to take the $5$ from.

Nothing is wrong with the writer's column work, so the fault is earlier. Look at what number their columns must have held.

$20.384 - 6.375 = 14.009$

So the top row of their column was $14.009$, not $14.900$: they gave the shorter number its two extra zeros immediately after the point instead of at the far end. That is a different number, one whose $9$ has been pushed from the tenths column down into the thousandths.

$14.009 = 14 + \frac{9}{1000}, \qquad 14.900 = 14 + \frac{9}{10}$

Padding is only allowed because a zero at the right-hand end opens a column and leaves it empty. A zero anywhere else displaces the digits after it, which changes the value, so the correct sum is the one part A produced.

A borrow is a re-bundling, not a new rule. One unit of any column is exactly ten units of the column to its right, because that ratio of ten is what builds the whole system on both sides of the point.

$1 = \frac{10}{10}, \qquad \frac{1}{10} = \frac{10}{100}, \qquad \frac{1}{100} = \frac{10}{1000}$

So when a column has too little to subtract from, one unit is fetched from its left neighbour and arrives as ten of its own units. If that neighbour is a zero it has nothing to give, so it borrows in turn, and the request travels left until it reaches a digit that is not zero.

Padding is what makes the journey possible. Written as $20$, the number has no tenths, hundredths or thousandths columns on the page at all, and there is nowhere for the borrow to pass through. Written as $20.000$ every column the chain needs exists, so the request travels from the thousandths up to the tens, and the tens digit is not zero, so the chain terminates there.

Watch out. The zeros do not create anything. They record columns that were always there and always empty, which is exactly why writing them changes no value.

The divisor has two decimal places, so move both points two places right. Each number is multiplied by a hundred, and a hundred over a hundred is one, so the quotient is untouched.

$32.76 \div 0.78 = 3276 \div 78$

Now it is whole-number division.

$3276 \div 78 = 42$

Check it by multiplying back, counting the places as the product rule requires.

$0.78 \times 42 = 32.76$

That is the whole roll, so nothing is left over, as the setting promised.

Move both points one place right so the divisor is whole, then divide.

$32.76 \div 1.2 = 327.6 \div 12 = 27.3$

The quotient is not a whole number, and only whole strips can be cut, so the machine produces $27$ of them. To find what is left, work out how much those strips used and take it from the roll.

$1.2 \times 27 = 32.4, \qquad 32.76 - 32.4 = 0.36$

So $0.36$ metres remain, which is less than one strip's length, as it has to be, since another whole strip would otherwise have been cut.

A division of this shape answers one question: how many of the smaller amount fit inside the larger one.

$\frac{\text{length of the roll}}{\text{length of one strip}} = \text{number of strips}$

Both divisions asked that of the same roll, so what changed between them was only the size of the piece being counted.

Now reason from the meaning rather than from arithmetic. The roll is fixed, so the total length to be handed out is fixed. Making each strip longer means each one takes a bigger share of that total, so the roll cannot be shared among MORE strips than before. The count cannot rise.

It does not have to fall, though, because only whole strips are cut. A small rise in the setting can leave the whole-strip count exactly where it was and simply eat into the leftover.

$1.21 \times 27 = 32.67, \qquad 32.67 < 32.76$

At a setting of $1.21$ metres, twenty-seven strips still fit inside the roll, so the count has not moved at all. A larger rise does bring it down, and the two settings already worked show that much, since the longer of them gave the smaller count.

Watch out. The count falling is not the same as the leftover falling. A longer setting can easily leave more behind, because what remains is whatever is too short to make one more strip, and "one more strip" is now a longer thing.