Decimal Place Value: Free Response

A digit's contribution is decided by the column it sits in, so name the columns before touching the digits. Immediately left of the point is the ones column; to the right the columns run tenths, hundredths, thousandths, so the three digits after the point take those three names in order.

That fixes all four places at once: the $6$ is in the ones, the $0$ is in the tenths, the $7$ is in the hundredths, and the $3$ is in the thousandths. Multiplying each digit by the value of its column gives the four contributions, in the same order:

$6, \qquad 0, \qquad \frac{7}{100}, \qquad \frac{3}{1000}$

The digit that contributes nothing is the $0$ in the tenths column, since zero tenths is zero however large a tenth is. That is a statement about what it adds to the sum, and nothing more; whether the digit can be spared is a different question, which part C asks.

Expanded form is one term per column. The zero term may be written or dropped, since it adds nothing either way:

$6.073 = 6 + \frac{0}{10} + \frac{7}{100} + \frac{3}{1000} = 6 + \frac{7}{100} + \frac{3}{1000}$

Now read the second sum backwards into columns. Each denominator names a column: $\tfrac{9}{10}$ is nine tenths, so the $9$ belongs in the first column after the point, and $\tfrac{4}{1000}$ is four thousandths, so the $4$ belongs in the third.

The hundredths column is not mentioned in the sum at all, and an unmentioned column is an empty one. It still has to be written, because the $4$ can only be read as thousandths if some digit stands between it and the tenths:

$2 + \frac{9}{10} + \frac{4}{1000} = 2.904$

Writing $2.94$ instead would put the $4$ in the hundredths column, where it would name four hundredths, and that is a different number from the one the sum describes.

Deleting a digit is not a quiet act. The point stays where it is, so every digit to the right of the deleted one slides one column to the left, and one column to the left is worth ten times as much. Track the two digits that move.

In $6.073$ the $7$ is in the hundredths column and the $3$ is in the thousandths. In $6.73$ the $7$ has landed in the tenths and the $3$ in the hundredths, so each is worth ten times what it was. Gathering each part after the point over a single denominator makes the comparison plain:

$6.073 = 6 + \frac{73}{1000}, \qquad 6.73 = 6 + \frac{73}{100}$

Those two fractional parts are not equal, since $\tfrac{73}{100} = \tfrac{730}{1000}$, so the claim is false. The whole parts agree and the string of digits after the point is the same in both, and neither of those things decides a value. The columns do.

So what the zero is for is now visible. It contributes nothing to the sum, and that was never its job. Its job is to occupy the tenths column, which forces the $7$ out into the hundredths and the $3$ out into the thousandths, where this number needs them. A zero that holds a column open like that is called a placeholder, and a placeholder cannot be removed.

Name the columns first. The digit before the point is in the ones column, and the two after it are in the tenths and the hundredths, so the three values are

$8, \qquad \frac{8}{10}, \qquad \frac{8}{100}$

Now take the two steps one at a time. Going from $8$ to $\tfrac{8}{10}$ multiplies the value by $\tfrac{1}{10}$, which is the same as dividing it by ten. Going from $\tfrac{8}{10}$ to $\tfrac{8}{100}$ multiplies by $\tfrac{1}{10}$ again:

$\frac{1}{10} \times \frac{8}{10} = \frac{8}{100}$

So one multiplier serves both steps. The first of them, from the ones column to the tenths, is the step that crosses the point, and it is the step a reader might expect to behave differently. It does not.

Look at what the bundling rule actually says: ten of one column make one of the column to its left. Ten ones make one ten, ten tens make one hundred. It is a statement about a pair of neighbouring columns, and it names no particular pair, so it applies to every pair there is.

Read in the direction it is stated, one step to the left multiplies a place value by ten. A step to the right is the undoing of a step to the left, and undoing a multiplication by ten is a division by ten. So one step right divides the place value by ten, wherever the step is taken:

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

Those are the tenths and the hundredths, produced by the rule the classmate already accepts rather than by a new one.

The boundary deserves a check in the rule's own words, since the objection is really about the boundary. Do ten tenths make one whole, the way ten ones make one ten?

$10 \times \frac{1}{10} = 1$

They do, exactly. So the ones column and the tenths column are a bundling pair like any other, and the tenths place is not a new invention but the next thing the old rule produces.

That leaves the point itself. It is not a column and it holds no digit; it is a mark showing a reader where the ones column ends, so that the columns can be told apart on the page at all. A mark cannot interrupt a relation between two columns, because the relation was never about the mark.

Take the two contributions first. A $7$ in the hundreds column is worth $700$, because one unit of that column is a hundred wholes. A $7$ in the hundredths column is worth $\tfrac{7}{100}$, because one unit of that column is one part of a whole cut into a hundred equal parts:

$700 \qquad \text{against} \qquad \frac{7}{100}$

Now count the steps from the ones column. Going left, ones to tens to hundreds is two steps, each multiplying by ten. Going right, ones to tenths to hundredths is two steps, each dividing by ten. The two columns therefore sit the same distance from the ones column in opposite directions, which is why their names are mirror images of one another.

The arithmetic of that mirror is a single line: the two place values multiply to one.

$100 \times \frac{1}{100} = 1$

And that is what the ending records. Hundreds counts a hundred wholes; hundredths cuts one whole into a hundred equal parts and counts one of them. The ending -ths turns a count of wholes into the name of a part, which is exactly the relation the line above shows: each of the pair undoes the other, and they meet at the ones column between them.

There are three digits after the point, so the last of them sits in the thousandths column, and the name of that last column is what the whole string after the point is counted in. The rule for reading a decimal aloud takes the whole-number part first, then the word and for the point, then the digits after the point as one whole number, then the name of that last column. Here the whole-number part is zero, so the reading is zero and one hundred sixty-four thousandths of a gram.

The same fact about the last column gives the fraction, because the number of digits after the point is the number of zeros in the denominator:

$0.164 = \frac{164}{1000}$

Now reduce, and the prime factorizations show what to divide by:

$164 = 2 \times 2 \times 41, \qquad 1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$

The factors common to both are two twos, so the greatest common factor is $4$:

$\frac{164}{1000} = \frac{164 \div 4}{1000 \div 4} = \frac{41}{250}$

The piece weighs $\tfrac{41}{250}$ of a gram, and since $41$ is prime and is not a factor of $250$, that fraction is in lowest terms.

Each spoken form names its last column, and that name fixes where the final digit has to land. Fill every column between the point and that digit, writing a zero wherever nothing was counted.

Four tenths puts the $4$ in the first column after the point, so nothing else is needed. Four hundredths puts the $4$ in the second column, so the tenths column has to be held open with a zero. Forty thousandths puts the last digit of $40$ in the third column, so the $4$ lands in the hundredths and the $0$ in the thousandths:

$9.4, \qquad 9.04, \qquad 9.040$

Now compare the three masses. The second and third are the same, which the fractions confirm:

$9.04 = 9 + \frac{4}{100}, \qquad 9.040 = 9 + \frac{40}{1000} = 9 + \frac{4}{100}$

The first is a different mass, since $\tfrac{4}{10}$ is ten times $\tfrac{4}{100}$. Note that the third could have been written $9.04$ just as well, since its trailing zero changes nothing; what the words fix is the column the final digit is named in, not how many zeros get written after it.

Two tickets can therefore be worded quite differently and record the same mass, while two that sound almost alike can name parts of a gram standing in a ten to one ratio. Both whole-gram parts here are $9$, so it is the parts after the point, not the masses themselves, that stand in that ratio.

Start from what the ticket committed the writer to. The word thousandths names the third column after the point, so the $5$ had to land there, with the tenths and hundredths columns held open by zeros:

$8.005 = 8 + \frac{5}{1000}$

The clerk wrote the $5$ in the first column after the point, the tenths, which is two columns to the left of where the ticket sent it:

$8.5 = 8 + \frac{5}{10}$

The whole-gram part is right in both figures, so the whole of the error sits in the part after the point. Ask how many times larger the clerk's part is by dividing one by the other:

$\frac{5}{10} \div \frac{5}{1000} = \frac{5}{10} \times \frac{1000}{5} = 100$

So the clerk's part after the point is one hundred times the part the ticket asked for, which is what two columns cost in a base-ten system: ten for each step.

The correct entry is $8.005$ grams. The general repair is to write the columns down before the digits: name the column the words end in, put the last digit there, and fill everything between it and the point with zeros.

The number of digits after the point is the number of zeros in the denominator, so start there and reduce each fraction with the greatest common factor.

For $0.940$ there are three digits after the point, so the denominator is a thousand, and the greatest common factor of $940$ and $1000$ is $20$:

$0.940 = \frac{940}{1000} = \frac{940 \div 20}{1000 \div 20} = \frac{47}{50}$

For $0.094$ there are also three digits after the point, but the greatest common factor of $94$ and $1000$ is only $2$:

$0.094 = \frac{94}{1000} = \frac{47}{500}$

Both deletions produce the same string of digits, $0.94$, which shows two digits after the point and so sits over a hundred:

$0.94 = \frac{94}{100} = \frac{47}{50}$

Now compare each result with the number it came from. Deleting the final zero of $0.940$ produced $\tfrac{47}{50}$, which is exactly what $0.940$ was already worth, so that deletion left the value alone. Deleting the zero inside $0.094$ turned $\tfrac{47}{500}$ into $\tfrac{47}{50}$, and since $\tfrac{47}{50} = \tfrac{470}{500}$, that deletion multiplied the value by ten.

Two deletions of the same digit, from two decimals built out of the same three figures, therefore do entirely different things.

First see exactly what a deletion does. The columns after the point are counted from the point, and the point does not move, so the digits to the LEFT of a deleted digit stay in the columns they were in. Every digit to its RIGHT slides one column to the left, and a column to the left is worth ten times as much, so whatever those digits contributed together is multiplied by ten.

That one observation settles both directions, and the test it gives is this: a zero written after the decimal point may be deleted without changing the value exactly when every digit to its right is also a zero, the case of no digits at all to its right included.

Take the direction where the deletion is harmless. If every digit to the right of the zero is a zero, those columns contribute nothing before the deletion and nothing after it, because ten times nothing is nothing. The digits to the left have not moved, and the zero itself contributed nothing either way, so no term in the sum has changed:

$0.4700 = 0.470 = \frac{47}{100}$

Now the other direction. Suppose some digit to the right of the zero is not a zero. Then the digits to its right contribute some amount above nothing, and after the deletion they contribute ten times that amount instead, since each of them has moved one column left. An amount above nothing is never equal to ten times itself, so the total has to change, which is the case part A ran:

$0.094 = \frac{94}{1000} \quad \text{becomes} \quad 0.94 = \frac{940}{1000}$

So the condition is not merely enough to make the deletion safe, it is required: as soon as one nonzero digit stands to the right of the zero, the deletion moves it and the value moves with it.

The difference is what the columns are counted from.

After the decimal point, the point is written on the page and the columns are counted outward from it. A zero at the far right end has no digit to its right, and the point does not move when the zero goes, so no digit changes column at all.

A whole number has no point written. Its ones column is its right-hand end, and every other column is counted from there, so rubbing out the last digit brings a different digit to the right-hand end and every remaining digit drops one column:

$3{,}700 = 3{,}000 + 700, \qquad 370 = 300 + 70$

The $3$ has gone from the thousands column to the hundreds and the $7$ from the hundreds to the tens, so each contributes one tenth of what it contributed before, and the number itself is one tenth of what it was.

So the two kinds of trailing zero are doing different jobs. The trailing zeros of a whole number are placeholders: they hold the other digits away from the ones column, and they are the only record of which column each digit is in. A trailing zero after the point holds nothing away from anything, because the written point already fixes where the columns begin, so that zero is free to go.

Comparing column against column is easiest when every number shows the same columns, so pad each one with trailing zeros until all four have three digits after the point. That step is safe because a trailing zero after the point changes no value, as the fractions confirm: $\tfrac{9}{10} = \tfrac{900}{1000}$.

$0.190, \qquad 0.900, \qquad 0.109, \qquad 0.910$

Now read from the left. The tenths is the leftmost column where these numbers differ: $0.190$ and $0.109$ have $1$ tenth each, while $0.900$ and $0.910$ have $9$ tenths each, so both of the first pair are below both of the second pair, whatever comes later.

Settle each pair by moving one column right. In the lower pair the hundredths are $9$ and $0$, so $0.109$ is the smaller. In the upper pair the hundredths are $0$ and $1$, so $0.900$ is the smaller. Chaining the four:

$0.109 < 0.19 < 0.9 < 0.91$

Notice which number came out smallest. It is the one showing the most digits after the point, and the one showing the fewest is not the smallest either, so the length of a decimal settles nothing at all.

Pad the shorter decimal so the columns line up. $0.7$ becomes $0.7000$, and over a denominator of ten thousand the two numbers are

$0.7000 = \frac{7000}{10000}, \qquad 0.6999 = \frac{6999}{10000}$

With one denominator the numerators decide, and $7000 > 6999$, so $0.7 > 0.6999$. The tenths is the first column where the two differ, with $7$ against $6$.

Now show that the three nines never had a chance. Everything after the tenths column in $0.6999$ is nine hundredths, nine thousandths and nine ten-thousandths, which come to

$\frac{9}{100} + \frac{9}{1000} + \frac{9}{10000} = \frac{900 + 90 + 9}{10000} = \frac{999}{10000}$

To overturn a shortfall of one in the tenths column, that tail would have to be worth at least one tenth. Over ten thousand, one tenth is $\tfrac{1000}{10000}$:

$\frac{999}{10000} < \frac{1000}{10000} = \frac{1}{10}$

The tail falls one ten-thousandth short. And this is the largest tail those three columns could hold, since $9$ is the largest digit there is, so no filling of them could have closed the gap.

That is the general fact behind comparing from the left: in a decimal that stops after finitely many columns, one unit of any column is worth more than every column to its right can supply together. Adding further columns to the right makes the shortfall smaller without ever removing it, which is why the first differing column always decides.

Take $0.19$ and $0.9$, two of the numbers from part A. The rule reads their digit strings as the whole numbers $19$ and $9$, and since $19 > 9$ it claims that $0.19$ is the larger. Writing both over a hundred says otherwise:

$0.19 = \frac{19}{100}, \qquad 0.9 = \frac{90}{100}$

So $0.9$ is the larger and the rule has the comparison backwards. One case is enough to sink a rule as stated.

What went wrong is that the rule reads the two digit strings without asking which column each string ends in. The string $19$ ends in the hundredths column and the string $9$ ends in the tenths, so the two strings are counting pieces of different sizes, and comparing them as whole numbers compares counts of pieces that are not comparable.

The rule is safe when the pieces are the same size, and two things secure that. The whole-number parts have to be equal, or the comparison is not decided after the point at all: on $1.2$ against $0.9$ the rule compares $2$ with $9$ and gets that backwards too. And the two decimals have to show the same number of digits after the point, so that both strings end in the same column. Given those two, each number is that same whole-number part plus a string over one shared denominator, and comparing the strings is exactly comparing numerators over that denominator. (A rule can of course come out right by luck on a particular pair without either condition, which is precisely why luck is not a rule.)

The step that secures the second condition is padding with trailing zeros, which costs nothing because it changes no value. Padded to two places the strings above become $19$ and $90$, and $19 < 90$, which now agrees with the truth. Padding is not a patch on a broken rule; it is the reason a column-by-column comparison works at all.