Operations with Decimals: Free Response

Give both numbers in a calculation the same number of decimal places first, so that no column is left with an empty slot to misjudge. Here $23.4$ becomes $23.40$, and the addition is worked from the right.

The hundredths give $0 + 6 = 6$. The tenths give $4 + 8 = 12$, so write $2$ and carry $1$ into the ones. The ones give $3 + 9 + 1 = 13$, so write $3$ and carry $1$ into the tens. The tens give $2 + 1 = 3$, and the point drops straight down.

$23.40 + 9.86 = 33.26$

The subtraction needs the same padding: $15.2$ becomes $15.20$. Without that hundredths zero there is nothing in the top row for the $7$ to be taken from.

The hundredths need $0 - 7$, which cannot be done, so borrow one tenth, which is ten hundredths: $10 - 7 = 3$, and the tenths drop from $2$ to $1$. The tenths now read $1 - 4$, so borrow one whole, which is ten tenths: $11 - 4 = 7$, and the ones drop from $5$ to $4$. The ones read $4 - 8$, so borrow one ten: $14 - 8 = 6$, and the tens are left with $1 - 1 = 0$.

$15.20 - 8.47 = 6.73$

Check the subtraction by adding back, which is the cheapest check there is: $6.73 + 8.47 = 15.20$, so the difference undoes correctly.

A whole number has a decimal point too. It sits at the right-hand end with nothing after it, so $6$ is $6.00$ and the two zeros fill the tenths and hundredths columns. They are not decoration: without them the bottom row has digits standing under nothing at all.

Subtract from the right. The hundredths need $0 - 5$, and the tenths column above is also $0$, so the borrow has to travel from the ones. One whole is ten tenths, so $6.00$ becomes $5$ ones, $9$ tenths and $10$ hundredths. Now the columns work: $10 - 5 = 5$ in the hundredths, $9 - 8 = 1$ in the tenths, and $5 - 1 = 4$ in the ones.

$6 - 1.85 = 4.15$

Check by adding back: $4.15 + 1.85 = 6.00$.

Each zero was holding a column open so that the borrow had somewhere to land. The tenths zero received the whole that was broken up, and the hundredths zero received the tenth that was broken up after it.

Read each number by its places. $23.4$ is $2$ tens, $3$ ones and $4$ tenths, while $9.86$ is $9$ ones, $8$ tenths and $6$ hundredths. Adding them means gathering the pieces of the same size, because a single column can only ever total pieces of one size:

$(20 + 3 + 0.4) + (9 + 0.8 + 0.06)$

Addition lets those six terms be reordered and rebracketed, so the ones gather with the ones and the tenths with the tenths, while the lone tens term and the lone hundredths term stay as they are. Each column of the written calculation is exactly one of those groups. That is the whole content of the alignment rule, and it is word for word the reason ones go under ones and tens under tens in whole-number arithmetic. Nothing has been added for decimals; the point is only a marker that makes the same grouping visible further down the place-value line.

Now push the two numbers to the right so their last digits share a column. Reading the columns from the right, the first would combine $4$ tenths with $6$ hundredths, the second $3$ ones with $8$ tenths, and the third $2$ tens with $9$ ones.

$0.4 + 0.06, \qquad 3 + 0.8, \qquad 20 + 9$

Not one of those three is a sum of like pieces: each of them pairs a size with the size one step smaller. Take the first. Four tenths and six hundredths are quantities of different sizes, so writing a single digit under them claims a total of something that was never counted. They can certainly be combined by regrouping, since four tenths is forty hundredths and the two together are forty-six hundredths, but that is a calculation and not a column, and no single digit records it. The arrangement has stopped tracking what each digit is worth, which is the one job the columns exist to do.

Ignore both points and multiply the digits as whole numbers: $35 \times 6 = 210$. Now count the decimal places in the factors, since that count is the only question left. The factor $0.35$ has two places and $0.6$ has one, so the product needs $2 + 1 = 3$ places.

Three places from the right of $210$ puts the point in front of the $2$, which needs a zero written before the point:

$0.35 \times 0.6 = 0.210 = 0.21$

The zero on the end may be written or left off without changing the value.

The second product runs the same way. Multiply as whole numbers, $47 \times 23$, which is $47 \times 20 = 940$ plus $47 \times 3 = 141$:

$940 + 141 = 1081$

The factor $4.7$ has one decimal place and $0.23$ has two, so the product needs $1 + 2 = 3$ places. Three from the right of $1081$ lands the point between the $1$ and the $0$:

$4.7 \times 0.23 = 1.081$

This time the zero cannot be left out, so it is not the same kind of zero as the one on the end of the first answer. Which zeros come off and which do not is the subject of the last question in this set.

The multiplier $100$ has two zeros, so the point slides two places to the right:

$0.407 \times 100 = 40.7$

The multiplier $1000$ has three zeros, so the point slides three places. The digits of $3.6$ run out after one place, so zeros fill the columns behind it: $3.6$ goes to $36$, then $360$, then $3600$.

$3.6 \times 1000 = 3600$

Now the check. A whole number has no decimal places at all, so the counting rule says the product of $0.407$ and $100$ needs $3 + 0 = 3$ places, and the whole-number multiplication is $407 \times 100 = 40700$. Three places from the right of $40700$ gives

$0.407 \times 100 = 40.700 = 40.7$

which is the same answer the slide gave. So the slide is not a separate rule; it is the counting rule with a multiplier that adds nothing to the count.

The reason the digits move at all is place value. Multiplying by ten makes every digit worth ten times as much, so the tenths digit becomes a ones digit and the ones digit becomes a tens digit. Each zero in the multiplier is one more step up that line, which is why two zeros move the point two places and three move it three.

Read each factor by its place value. The number $0.35$ is $35$ hundredths and $0.6$ is $6$ tenths, so each factor is a whole number sitting over ten or over a hundred:

$0.35 = \frac{35}{100}, \qquad 0.6 = \frac{6}{10}$

Multiplying fractions multiplies the tops and multiplies the bottoms:

$\frac{35}{100} \times \frac{6}{10} = \frac{35 \times 6}{100 \times 10} = \frac{210}{1000}$

Two things happened on that line, and they are the two halves of the rule. The tops multiplied to give exactly the whole-number product the algorithm asks for, $35 \times 6 = 210$. The bottoms multiplied to give a thousand, and a thousand underneath is three decimal places.

That is where the addition comes from. A factor's decimal places are the zeros in its bottom: two zeros for hundredths, one for tenths. Multiplying the two bottoms writes all of those zeros together in one number, and putting two zeros with one zero gives three. Nothing else could happen to them, which is why the counts add rather than, say, taking whichever is larger.

The same run through the other pair confirms it. Here $4.7$ is $47$ tenths and $0.23$ is $23$ hundredths, so one zero meets two:

$\frac{47}{10} \times \frac{23}{100} = \frac{1081}{1000}$

three zeros underneath, three decimal places.

Now the cost of a miscount. Suppose the point in the first product were placed two from the right of $210$ instead of three. That reports $2.10$ where the number is $0.210$, which is dividing the whole-number product by a hundred instead of by a thousand:

$\frac{210}{100} = 2.10, \qquad \frac{210}{1000} = 0.210$

A hundred is a tenth of a thousand, so the report is ten times too big. Every place left out of the count multiplies the answer by ten again, which is why the count is worth doing twice.

One metre costs $3.60$ dollars, so $4.5$ metres cost $3.6 \times 4.5$ dollars. The change is the note less everything paid for, which is the wire and one connector, so it is $25 - (3.6 \times 4.5 + 2.85)$. The brackets are what hold that expression together, since the two purchases have to be added before anything is taken from the note. Taking them away one at a time needs no brackets at all, and $25 - 3.6 \times 4.5 - 2.85$ is just as good an expression for the same change.

Multiply the wire as whole numbers: $36 \times 45$ is $36 \times 40 = 1440$ plus $36 \times 5 = 180$, so $36 \times 45 = 1620$. One decimal place in $3.6$ and one in $4.5$ make two, so the point goes two from the right:

$3.6 \times 4.5 = 16.20$

The wire costs $16.20$ dollars. Reading the price as $3.60$ instead, with two places, gives $360 \times 45 = 16200$ and a count of three places, which is $16.200$: the same amount, since the extra zero in the factor multiplies the whole-number product by ten and adds one to the count at the same time.

Now add the connector with the points aligned. The hundredths give $0 + 5 = 5$; the tenths give $2 + 8 = 10$, so write $0$ and carry $1$; the ones give $6 + 2 + 1 = 9$; the tens give $1$.

$16.20 + 2.85 = 19.05$

The order comes to $19.05$ dollars. Write the note as $25.00$ so every column is filled, and subtract. The hundredths need $0 - 5$ and the tenths above are $0$, so the borrow travels from the ones: $25.00$ becomes $24$ ones, $9$ tenths and $10$ hundredths. Then $10 - 5 = 5$ in the hundredths and $9 - 0 = 9$ in the tenths; the ones read $4 - 9$, so borrow a ten for $14 - 9 = 5$, and the tens are left with $1 - 1 = 0$.

$25.00 - 19.05 = 5.95$

The change is $5.95$ dollars. Check by adding back: $19.05 + 5.95 = 25.00$.

Asking how many metres $27$ dollars buys is asking how many times the price of one metre fits inside $27$ dollars, and that is a division: $27 \div 3.6$.

The divisor is $3.60$, which is $3.6$, so it has one decimal place. Move its point one place right to make it whole, and move the dividend's point one place right as well:

$27 \div 3.6 = 270 \div 36$

Now divide. Since $36 \times 7 = 252$ and $270 - 252 = 18$, the whole part of the quotient is $7$ with $18$ left over. The dividend has run out of digits, so append a zero after its point and keep going, which turns the $18$ left over into $180$ tenths. Since $36 \times 5 = 180$, the tenths digit is $5$ and nothing is left:

$270 \div 36 = 7.5$

Check by multiplying back: $36 \times 75 = 2700$, and two decimal places give $3.6 \times 7.5 = 27.00$, which is the money spent.

So $27$ dollars buys $7.5$ metres of the wire.

The division asked how many times the price of one metre fits inside the money spent, and each of those fits buys one metre. So the quotient is counted in metres, and its decimal part is a part of a metre of wire. It is not a part of a dollar, and it is not a remainder in the whole-number sense either.

Back it up by pricing that part directly. Half a metre costs

$0.5 \times 3.6 = 1.80$

dollars, since $5 \times 36 = 180$ and two places put the point two from the right. Seven whole metres cost $7 \times 3.6 = 25.20$ dollars, and

$25.20 + 1.80 = 27.00$

which is the whole order, so the decimal part is genuinely the extra half metre.

Now why the scaled division still answers the money question. Write the division as a fraction and multiply the top and the bottom by ten:

$\frac{27}{3.6} = \frac{27 \times 10}{3.6 \times 10} = \frac{270}{36}$

Multiplying the top and the bottom of a fraction by the same nonzero number leaves its value alone, because it is multiplying by ten over ten, which is one. Read in the words of the situation, the two new numbers are both amounts of money counted in tenths of a dollar instead of in dollars: $270$ tenths of a dollar spent, and $36$ tenths of a dollar for one metre. Changing the unit both amounts are measured in cannot change how many times one of them fits inside the other, and that count is the answer the question wanted. This is why the numbers being divided are allowed to stop looking like prices: only their ratio is being asked about, and the ratio never moved.

The divisor $0.6$ has one decimal place, so one move to the right makes it the whole number $6$. Move the dividend's point one place as well, which turns $8.4$ into $84$:

$8.4 \div 0.6 = 84 \div 6$

That is a whole-number division you already know:

$84 \div 6 = 14$

Check it by multiplying back, which uses the counting rule: $6 \times 14 = 84$, and $0.6$ has one decimal place while $14$ has none, so the product needs one place, giving $0.6 \times 14 = 8.4$. That is the dividend, so the quotient is right.

It is worth noticing that the answer is larger than the number being divided. Dividing by something smaller than one asks how many small pieces fit inside a larger quantity, and six tenths is small, so it fits many times over.

The first step is sound. Making the divisor whole is exactly what the rule asks for, and $0.6$ does become $6$ by one move of the point. The step that fails is the second one, where the dividend has to travel the same distance and did not.

Work out what the student's division actually gives. Dividing $8.4$ by the whole number $6$ needs no sliding at all: the point in the quotient goes straight above the point in the dividend. Six goes into $8$ once with $2$ over, so the ones digit is $1$; the $2$ left over is $20$ tenths, together with the $4$ tenths making $24$ tenths, and six goes into $24$ four times exactly:

$8.4 \div 6 = 1.4$

Check: $6 \times 1.4 = 8.4$, so that is an honest answer to some question. It is just not this one.

Which question is it an answer to? It says how many sixes fit inside $8.4$, when the question asked how many six tenths fit inside $8.4$. Those are different questions with different sizes of answer, and the relationship between them is the diagnosis. Six is ten times as big as six tenths, so it fits into $8.4$ one tenth as often. Dividing by a number ten times too big returns a quotient ten times too small.

So the report misses by a factor of ten, and that factor was readable from the student's own first step alone: the divisor was multiplied by ten and nothing was done to compensate. It agrees with part A, where the correct quotient is ten times the student's, but the factor did not have to be found that way.

Write the division as a fraction, since that is what a division is: $a \div b$ is $\tfrac{a}{b}$. From the equivalent-fractions rule, multiplying the top and the bottom by the same nonzero number gives an equal fraction, because it is multiplying by a disguised one. On this division,

$\frac{8.4}{0.6} = \frac{8.4 \times 10}{0.6 \times 10} = \frac{84}{6}$

and since $\tfrac{10}{10} = 1$, the value has not moved. Moving both points one place right is precisely multiplying both numbers by ten, so it is one of these harmless moves, and moving both points two places multiplies both by a hundred, which is equally harmless.

Moving only one point is a different act entirely. Multiplying only the top by ten multiplies the whole fraction by ten; multiplying only the bottom by ten divides the whole fraction by ten. Nothing cancels, because there is nothing on the other side of the bar to cancel with. The factor is ten for each place the lone point travels, so a single stray place is a factor of ten, two stray places a factor of a hundred, and the direction is decided by which of the two numbers moved: the quotient grows when the dividend moves alone and shrinks when the divisor does. That is the general reason the two steps travel together: they are one act of multiplying by one, split across two lines of writing.

Now the second division. The divisor $0.06$ has two decimal places, so both points move two places to the right. The divisor becomes $6$. The dividend $8.4$ has only one digit after its point, so after one move it is $84$ and its digits have run out; a zero must be appended for the second move, giving $840$. Appending a trailing zero to a whole number here is not a free choice of notation, it is the second multiplication by ten:

$8.4 \div 0.06 = 840 \div 6 = 140$

Check by multiplying back: $6 \times 140 = 840$, and $0.06$ has two decimal places while $140$ has none, so the product needs two places, giving $0.06 \times 140 = 8.40$. That is the dividend, so $140$ is right. Comparing the two divisions is instructive on its own: the same dividend divided by a divisor ten times smaller gives a quotient ten times larger.

For the first pair, multiply as whole numbers: $5 \times 4 = 20$. Each factor has one decimal place, so the product needs $1 + 1 = 2$ places, and two places from the right of $20$ puts the point in front of the $2$:

$0.5 \times 0.4 = 0.20$

That is the form the rule hands you. The zero on the end can come off, so the simplified form is $0.2$.

For the second pair, $25 \times 12$ is $25 \times 10 = 250$ plus $25 \times 2 = 50$, so $25 \times 12 = 300$. Each factor has two decimal places, so the product needs $2 + 2 = 4$ places. Four places from the right of $300$ needs a zero written between the point and the $3$:

$0.25 \times 0.12 = 0.0300$

The two zeros on the end can come off, so the simplified form is $0.03$. The zero standing between the point and the $3$ cannot, which is the subject of part C.

Take the promise first. The factors of the first product have one decimal place each, so the classmate's version promises two places; the factors of the second have two each, so it promises four.

Now count what the simplified forms show. The first, $0.2$, shows one decimal place. The second, $0.03$, shows two:

$0.20 = 0.2, \qquad 0.0300 = 0.03$

So the promise fails in both cases, and it fails by more in the second than the first. That looks damning until you see where the failure is not. The point was in the right column both times: the rule placed it correctly in $0.20$ and in $0.0300$, and neither of those is wrong. What changed afterwards was only the writing.

So the classmate's version is a true statement about one thing and a false statement about another. It is exactly right about where the point goes in the whole-number product: that position is settled by the total of the factors' places, and this is the rule the whole algorithm rests on. It is not a statement about the simplified answer, because when the whole-number product ends in zeros, those zeros land at the right-hand end of the decimal and may be taken off, which lowers the count of places that appear on the page.

A wording that is safe in both places names what is being counted: the total of the factors' decimal places tells you how many places from the right of the whole-number product the point goes. The simplified answer shows that many places unless the whole-number product ends in one or more zeros, and each of those zeros that lands after the point may be dropped without changing the value.

That qualification is worth keeping narrow. It is only the zeros at the right-hand END of the decimal that come off. In the second answer, $0.0300$ ends in two zeros that go, while the zero between the point and the $3$ stays, and the count of places drops from four to two rather than to nothing.

A digit's worth comes from its column, and a column is fixed by how far that digit stands from the point. A zero counts none of its own column wherever it sits, so the only question a zero really raises is what happens to the OTHER digits when it is taken out.

Take the zero on the end of $0.20$. No digit lies beyond it, so it is holding nothing away from the point: the $2$ keeps the tenths column, the point stays where it is, and the total is untouched.

$0.20 = 0.2, \qquad 8.5 = 8.50$

Those are two writings of one number, and the line may be read in either direction. That single permission is what lets a product's trailing zero be dropped and what lets a short number be padded before an addition or subtraction: the two moves are one move going opposite ways.

The zero in $0.07$ is doing a different job. It contributes nothing itself, but the $7$ stands beyond it, held out in the hundredths column. Take the zero away and the $7$ slides one column in towards the point, into a column worth ten times as much:

$0.07 \to 0.7$

and $0.7$ is ten times $0.07$. A zero inside a number works the same way: in a product written $1.081$, the zero is what keeps the $8$ in the hundredths column and the $1$ in the thousandths.

The zero in $840$ is the case that catches a careless test, because nothing whatever is written to its right. But a whole number's point waits at its right-hand END, just past that zero, so the digits held away from the point are the $8$ and the $4$. Take the zero out and both slide one column in:

$840 \to 84$

and $84$ is a tenth of $840$. This is not a curiosity: it is exactly why appending a zero to $84$ counts as a second multiplication by ten when a divisor of $0.06$ has to be cleared, and why multiplying $3.6$ by a thousand gives $3600$ and not $36$.

So the test needs two conditions, one for each of the two ways the cases above went wrong. A zero may be dropped when it sits AFTER the point and no nonzero digit stands anywhere to its right. Then nothing is held away from the point by it and no digit moves. Fail either condition and some digit is depending on it.

One zero is deliberately left out of that test: the one written in front of the point in $0.20$ or $0.07$. Nothing stands on the far side of it from the point, so leaving it off would not change the value, and $.07$ names the same number as $0.07$. It is written anyway, always, because a point with no digit before it is easy to miss altogether. That zero is a convention of writing, not a column being held.

The test sorts a whole run of zeros just as cleanly. In $0.0300$ the last two zeros each sit after the point with no nonzero digit to the right of them, so both come off, while the first has a $3$ beyond it and stays:

$0.0300 = 0.030 = 0.03$

That settles the padding zero in $8.50$. It sits after the point, and no nonzero digit stands to its right, so it passes both conditions: it holds nothing away from the point, and writing it changes how the number looks rather than what it is worth. That is why padding is safe, and why the whole subtraction may be worked on the padded form with the answer still an answer about $8.5$.