4.1: Trading and Place Value

Let's exploring place value in this problem set. We'll start off by working with cups, pints, quarts, half-gallons, and gallons. We are using these because they can represent the relationship between place values in binary, the place-value number system with only two digits: 1 and 0. It will help if you create your own manipulatives for this. Cut out many small strips of paper of the same length, and write "C" on them. Those are your cups. Cut out two strips of paper that are each twice the size of the previous strips. Write "P" on them to represent pints. Continue this process until you have two strips representing gallons. If you don't have paper long enough, you can double the width instead of the length. See the examples below. You may also download this manipulatives file and print it. Then just cut out the shapes! If you do not have the ability to do this, you can use drawings. Size is important, so if you use drawings, make sure you keep the sizes consistent.

Figure \(\PageIndex{1}\): Strips (or shapes) representing cups, pints, quarts, half-gallons, and gallons

Problem 1

Fill in the blanks with the correct numeral to represent the relationship for each pair.

1. There are _____ cups in a pint.
2. There are _____ pints in a quart.
3. There are _____ quarts in a half-gallon.
4. There are _____ half-gallons in a gallon.

Problem 2

Suppose you were sick for nine days and a friend came over three times a day, each time leaving you a cup of soup. Suppose you never ate any of the soup, but instead kept putting each cup in the refrigerator. After those nine days were up, your refrigerator was stocked with all these individual cups of soup. You decide to consolidate these into as few containers as possible. You have five different kinds of containers: those holding cups, pints, quarts, half-gallons and gallons. Fill in each blank with the correct numeral.

1. Using your manipulatives or drawings, represent aside three cups a day for nine days. There are _____ cups represented.
2. Trade in the cups for as many pints as you can. There are now _____ pint(s) and _____ cup(s) of soup. This is a total of _____ containers of soup.
3. Trade in the pints for as many quarts as you can. There are now _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.
4. Trade in the quarts for as many half-gallons as you can. There are now _____ half-gallon(s), _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.
5. Trade in the half-gallons for as many gallons as you can. There are now _____ gallon(s), _____ half-gallon(s), _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup
6. Write down the information from part e in a consolidated fashion by filling in each space in the chart with the correct numeral.

   G	H	Q	P	C

7. Originally, from part (a), there were _____ cups of soup. By consolidating (see part e and f), the number of containers was reduced to only _____ containers of soup!

What is interesting about this problem is you have just done a conversion from a number written in base ten to a number written in base two! The 27 cups, is 27 ones, which is 2 tens and 7 ones. That's why in base ten, the leftmost digit is 2 and the rightmost digit is 7. Within 27, there is 1 sixteen, 1 eight, 0 fours, 1 two, and 1 one. In binary, the first five places (right to left) are ones, twos, fours, eights, and sixteens. So, 27 in base ten is 11011 in binary (or base two). Notice that the number of containers of soup is the sum of digits in the binary number and the binary number is precisely the numbers in the chart you filled in above!

Problem 3

Repeat Problem 2, but change the circumstances so that the friend only brings over two cups of soup a day for nine days.

a. Using your manipulatives or drawings, represent two cups a day for nine days. There are _____ cups.

b. Trade in the cups for as many pints as you can. There are now _____ pint(s) and _____ cup(s) of soup. This is a total of _____ containers of soup.

c. Trade in the pints for as many quarts as you can. There are now _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.

d. Trade in the quarts for as many half-gallons as you can. There are now _____ half-gallon(s), _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.

e. Trade in the half-gallons for as many gallons as you can. There are now _____ gallon(s), _____ half-gallon(s), _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.

f. Write down the information from part e by filling in each space in the chart with the correct number. G H Q P C

See if you can notice a pattern with what happens while doing these trades (or conversions). Which base ten number did you just convert into a binary number?

Now try the next problem. 

Problem 4

Repeat Problem 2 one more time, but change the circumstances so that the friend brings over five cups of soup a day for nine days. Also, let's assume you also own a 2 gallon container, which we'll refer to as a double-gallon. See if you can do this without manipulatives or drawings. It is okay if you cannot.

a. There are _____ cups.

b. Trade in the cups for as many pints as you can. There are now _____ pint(s) and _____ cup(s) of soup. This is a total of _____ containers of soup.

c. Trade in the pints for as many quarts as you can. There are now _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.

d. Trade in the quarts for as many half-gallons as you can. There are now _____ half-gallon(s), _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.

e. Trade in the half-gallons for as many gallons as you can. There are now _____ gallon(s), _____ half-gallon(s), _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.

f. Trade in the gallons for as many double-gallons as you can. There are now _____ double-gallon(s), _____ gallons, _____ half-gallon(s), _____ quart(s), _____ pint(s), and _____ cup(s) of soup. This is a total of _____ containers of soup.

g. Write down the information from part f by filling in each space in the chart with the correct number. G H Q P C

In the previous problems, exchanges or trades were being made any time you had two of one container. You were basically employing the grouping technique we see when counting objects and writing them with base ten numerals. When employing the grouping technique, we are working in a particular base, depending on how many it takes to form a group. In the exercises you just completed, you were working in base two.

To write a numeral in a given base, the base is written out in words to the right and a little below the numeral, as in \(11011_{\text{two}}\). The only time you do not have to write the base is when it is a base ten numeral because we are so used to that. However, it is okay and often good to write the "ten" anyway when doing conversions.

• In Problem 3, the base ten numeral \(18_{\text{ten}}\) (two cups for nine days) is written as \(10010_{\text{two}}\) in base two (or binary).
• In exercise 4, the base ten numeral \(45_{\text{ten}}\) (five cups for nine days) is written as \(101101_{\text{two}}\) in base two.

It is imperative that you read and say these numerals in different bases correctly. It's much easier to make mistakes if you say the numeral wrong and much easier to avoid mistakes if you say the numeral correctly. The following paragraph explains the proper way to read or say the numerals.

In base ten, we have abbreviated ways of saying numerals out loud. For instance, "13" is read "thirteen". But, "\(13_{\text{five}}\)" is read "One, three, base five" and "\(246201_{\text{eight}}\)" is read "two, four, six, two, zero, one, base eight." It's extremely important that you learn to read and say "\(13_{\text{five}}\)" as "one, three, base five" and not as "thirteen, base five"! There is no "thirteen" in base five. Why? When you say thirteen, it refers to the base ten numeral meaning one group of ten and three ones ("thirteen" is essentially an abbreviation of "three and ten"). Even though when we read it, we say the word "base", do not write the word "base".

Problem 5

Write out in words how to say each of the following numerals.

1. \(302_{\text{six}}\) is said out loud as _____
2. \(1011_{\text{two}}\) is said out loud as _____
3. \(435_{\text{seven}}\) is said out loud as _____

Problem 6

Write the given numerals

1. "Five, zero, one, six, base eight" is written as _____
2. "one, zero, one, zero, zero, one, base two" is written as _____

When you see a base two (or binary) numeral like \(11011_{\text{two}}\), you need to think about what each place value stands for. In all bases, the rightmost place value is the units or ones place. Next, since the "two" to the right of the numeral tells you this is a base two numeral, each place value as you move to the left increases by a multiple of 2. To the right is a chart showing the first eight place values in base two. They are written in base ten.

128

64

32

16

8

4

2

1

To check that the base two numeral \(11011_{\text{two}}\) really is \(27_{\text{ten}}\), put it into a chart with just five place values showing (since it's a five-digit numeral) and check the total value of the numeral. Look at the numeral written to the right. The place value for each digit is written below each digit of the numeral. The numeral is written in bold so it does not get confused with the place values.

\[\begin{align*} \text{(1 group of 16)} + \text{(1 group of 8)} + \text{(0 groups of 4)} + \text{(1 group of 2)} + \text{(1 group of 1)} &= (\mathbf{1} \times 16) + (\mathbf{1} \times 8) + (\mathbf{0} \times 4) + (\mathbf{1} \times 2) + (\mathbf{1} \times 1) \\[4pt] &= 16 + 8 + 2 + 1 \\[4pt] &= \mathbf{27}.\end{align*} \nonumber \]

All right! It really works!

Let's convert \(\mathbf{10 \ 011 \ 010}_{\mathbf{two}}\) to base ten. Try it on your own, then check your answer below.

This numeral converts to 128 + 16 + 8 + 2 = 154.

Problem 7

Fill in the chart to show what values are missing in this base two place value chart.

128

64

32

16

8

4

2

1

Problem 8

Convert each base two numeral to base ten.

a. \(10 \ 011_{\text{two}}\)

b. \(1 \ 000 \ 001_{\text{two}}\)

c. \(111 \ 111_{\text{two}}\)

Let's convert some numerals in other bases. Consider the base five numeral, \(24_{\text{five}}\). First, we need to establish the place values in base five just like we did for base two.

Problem 9

Fill in the chart to show what values are missing in this base five place value chart.

25

1

In base five, \(\mathbf{24}_{\mathbf{five}}\) is a two digit numeral. Note there are 2 groups of 5 and 4 units or \(2 \times 5 + 4 \times 1 = 10 + 4\) = 14.

Problem 10

Convert \(\mathbf{13}_{\mathbf{five}}\) to a base ten numeral. Remember not to read this as "thirteen"!!!

Let's look at a tougher problem (only tougher in that it is bigger and has more room for error). Here is how to convert \(\mathbf{31204}_{\mathbf{five}}\) to base ten. Try it on your own then continue reading. It is okay if you get stuck; it is important that you try first because it prepares your brain before reading the correct answer.

\[\begin{align*} \mathbf{3} \times 625 + \mathbf{1} \times 125 + \mathbf{2} \times 25 + \mathbf{0} \times 5 + \mathbf{4} \times 1 &= 1875 + 125 + 50 + 4 \\[4pt] &= \mathbf{2054}. \end{align*} \nonumber \]

Let's take the base three numerals \(\mathbf{1002021}_{\mathbf{three}}\) and \(\mathbf{22122}_{\mathbf{three}}\) and convert to base ten. First, we should make a place value chart for base three.

Problem 11

Fill in the missing place values for base three.

243

9

Try to convert \(\mathbf{1002021}_{\mathbf{three}}\) on your own, then continue reading. Note that in the solution below, the 0 terms are skipped. If this makes it more confusing, rewrite this solution but with the 0 terms included.

\[\mathbf{1} \times 729 + \mathbf{2} \times 27 + \mathbf{2} \times 3 + \mathbf{1} = 729 + 54 + 6 + 1 = \mathbf{790}. \nonumber \]

Try too convert \(\mathbf{22122}_{\mathbf{three}}\) on your own, then continue reading.

\[\mathbf{2} \times 81 + \mathbf{2} \times 27 + \mathbf{1} \times 9 + \mathbf{2} \times 3 + \mathbf{2} = 162 + 54 + 9 + 6 + 2 = \mathbf{233}. \nonumber \]

Problem 12

Convert the following base three numerals to base ten.

1. \(200 \ 112_{\text{three}}\) = ____
2. \(12 \ 002 \ 110_{\text{three}}\) = ____
3. \(1 \ 111 \ 111_{\text{three}}\) = ____

 Knowing what the value of any place other than the ones place in a numeral can tell you which base you're working in. See if you can determine the base from the given values of the places in the following problem.

Problem 13

Figure out the base in each place value chart and fill in the missing place values for each

a. Base ____       36

b. Base ____         7

c. Base ____       81

d. Base ____       64

e. Base ____     64

f. Base ____   10000

g. Base ____         12

h. Base ____         11

Another way to write out the values in a place value chart for a given base is by using exponents. Each place value is a power of the base. When writing it out this way, the actual base ten values are not explicitly written out. The following is another way to write a place value chart for base seven:

\(7^{11}\)

\(7^{10}\)

\(7^{9}\)

\(7^{8}\)

\(7^{7}\)

\(7^{6}\)

\(7^{5}\)

\(7^{4}\)

\(7^{3}\)

\(7^{2}\)

\(7^{1}\)

\(7^{0}\)

Did you remember that \(7^0 = 1\)? Any nonzero number raised to the zero power equals 1. For example, \(3^{0} = 1, 15^{0} =1, 1^{0} = 1, 10^{0} = 1, 546^{0} = 1\) and so on.

Problem 14

Consider base \(n\), where \(n\) is any integer greater than 1. Look at the place value chart below and fill in the missing place values for base \(n\).

\(n^{11}\)

\(n^{7}\)

\(n^{3}\)

\(n^{0}\)

If you are converting very large numbers, it might be useful to write out a place value chart using the base written to a given exponent to save time and space.

For instance, let's say you were asked to convert the numeral \(100 \ 200 \ 020 \ 000 \ 100_{\text{three}}\) to base ten (some spaces have been added to make it easier to read this number). There are a lot of zeros for most of the place values. In fact, this fifteen digit base three numeral only has four nonzero digits. We know each place value is a power of three, since this is a base three numeral. In base ten, this numeral is:

\[\begin{align*} 1 \times 3^{14} + 2 \times 3^{11} + 2 \times 3^{7} + 1 \times 3^{2} &= 1 \times 4782969 + 2 \times 177147 + 2 \times 2187 + 1 \times 9 \\[4pt] &= 4782969+ 354294 + 4374 + 9 \\[4pt] &= 5,141,646 \end{align*} \nonumber \]

Problem 15

\(100 \ 200 \ 020 \ 000 \ 100_{\text{three}}\) can be written as \(1 \times 3^{14} + 2 \times 3^{11} + 2 \times 3^{7} + 1 \times 3^{2}\). This is called expanded notation. Pay close attention to the fact that the 1 in the third place from the right is \(3^2\) (not \(3^3\)). Write the following numerals in expanded notation:

1. \(3 \ 000 \ 600 \ 020 \ 000 \ 000_{\text{eight}}\) = ____
2. \(3 \ 000 \ 040 \ 020 \ 000 \ 000_{\text{five}}\) = ____
3. \(400 \ 030 \ 000 \ 000 \ 002_{\text{eleven}}\) = ____
4. \(100 \ 100 \ 000 \ 010 \ 000_{\text{two}}\) = ____

Problem 16

Refer to the numeral, \(3 \ 040 \ 001 \ 000 \ 002_{\text{nine}}\). For each nonzero digit in the numeral, write the place value of that digit as a power of nine. Do not write out the actual base ten numeral. The first one is done for you.

1. 2 is in the \(9^0\) place.
2. 1 is in the ____ place.
3. 4 is in the ____ place.
4. 3 is in the ____ place.

If a number is written in expanded notation, you can reverse the process and write it as a number in the base used. For instance, \(4 \times 7^{13} + 5 \times 7^{10} + 2 \times 7^{8} + 3 \times 7^{3} + 6 \times 7^{0}\) can be written as the base seven numeral \(40 \ 050 \ 200 \ 003 \ 006_{\text{seven}}\). Notice that a number in the ninth place from the right is really the eighth power of the base, as shown by the placement of 2 in the numeral \(40 \ 050 \ 200 \ 003 \ 006_{\text{seven}}\). This is because the first place represents the base to the zero power. In other words, the counting does not start at one, but at zero.

Problem 17

Rewrite the following numbers that are in expanded notation to a numeral in the base indicated:

1. \(2 \times 5^{9} + 4 \times 5^{8} + 1 \times 5^{6} + 3 \times 5^{2}\) is written in base five as ____.
2. \(3 \times 8^{10} + 4 \times 8^{7} + 1 \times 8^{3} + 6 \times 8^{2}\) is written in base eight as ____.
3. \(7 \times 12^{9} + 8 \times 12^{5} + 4 \times 12^{3}\) is written in base twelve as ____.
4. \(2 \times 3^{8} + 1 \times 3^{6} + 1 \times 3^{4} + 2 \times 3^{2}\) is written in base three as ____.

Problem 18

a. If a base six numeral has eighteen digits, what is the place value of the first (leftmost) digit written as a power of six? ____

b. If a base nine numeral has twenty digits, what is the place value of the first (leftmost) digit written as a power of nine? ____

We now consider the number of distinct digits that exist in each base system.

Take another the following numerals:

\(1 \ 100 \ 101_{\text{two}}\)

\(212 \ 201_{\text{three}}\)

\(42101_{\text{seven}}\)

\(80034_{\text{nine}}\)

Did you notice that there are only 0's and 1's in base two numerals? That's because if there was a 2 or higher for one of the placeholders, it could be traded in for the next place value to the left. Using physical models at the beginning of this problem set, two cups could be exchanged for a pint, two pints could be exchanged for a quart, etc. Although we only went up to the double-gallon which contained 32 cups, there is really no limit to the number of place values a numeral has in a place value system. Also, one doesn't have to come up with a new name for each place value (unlike how new symbols need to be created to represent bigger and bigger numbers in, say, Roman numerals, without writing a lot of symbols.

Think about any base ten numeral. There are ten possible digits (or symbols) for each placeholder in the numeral 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. There is no separate single digit for the number ten (10); it is made up of a 1 and 0, which means one group of ten and zero units! In base two, there are only two possible digits for each place value in the numeral: 0 and 1. There is no need or use for the symbol (or digit) "2" when writing a numeral in base two because it is represented as \(10_{\text{two}}\).

Problem 19

How many possible digits can each place value in base three have? _____

What are those possible digits? ____

Problem 20

How many possible digits can each place value in base four have? _____

What are those possible digits? ____

Problem 21

How many possible digits can each place value in base five have? _____

What are those possible digits? ____

Problem 22

How many possible digits can each place value in base six have? _____

What are those possible digits? ____

Problem 23

How many possible digits can each place value in base eight have? _____

What are those possible digits? ____

Problem 24

Can the digit (symbol) "7" be used in a base seven numeral? _____

Explain your answer.

Problem 25

a. What is the lowest base in which the digit "6" may appear? _____

b. What is the highest base in which the digit "6" may appear? _____

Do you see a pattern developing? Just wait until you see what happens in base eleven, twelve, and thirteen!

Problem 26

How many possible digits for each place value do you need for base eleven? _____

Problem 27

How many possible digits for each place value do you need for base twelve? _____

Problem 28

How many possible digits for each place value do you need for base thirteen? _____

In base eleven, you need eleven different symbols (or digits) for possible placeholders. The problem we encounter is that we only have these ten recognizable digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Remember that 10 is made up of two separate digits; it isn't a single symbol! We need to introduce a new symbol to represent ten in base eleven and higher. The convention is to use the letter "T" or "A". Similarly, we use the letter "E" or "B" to represent the number eleven in base twelve and higher and the letter "W" or "C" to represent the number twelve in base thirteen and higher. Since the second place value from the right represents the base you are working in, you never need a separate symbol to represent the value of the base or any number higher than the base. That's why there are only 0s and 1s and no 2s in base two and why there are only 0s, 1s, 2s, 3s, and 4s and no 5s in base five, etc.

The next few problems are worked exactly like those you have done so far, except the bases are higher than ten. We will use the new "digits" T, E, and W for ten, eleven, and twelve.

Below is a value chart for base twelve.

144

12

1

Let's convert the numeral \(TE5_{\text{twelve}}\) to base ten.

\(TE5_{\text{twelve}} = 10 \times 144 + 11 \times 12 + 5 \times 1 = 1440 + 132 + 5 = \mathbf{1577}\).

Let's try a pair of examples: \(2T9_{\text{eleven}}\) and \(2109_{\text{eleven}}\).

To convert \(2T9_{\text{eleven}}\) to base ten, first consider the base eleven place value chart shown.

1331

121

11

1

\(2T9_{\text{eleven}} = 2 \times 121 + 10 \times 11 + 9 \times 1 = 242 + 110 + 9 = \underline{361}\).

Now, let's convert \(2109_{\text{eleven}}\) to base ten. Looking at the base eleven place value chart, \(2109_{\text{eleven}}\) represents \(2 \times 1331 + 1 \times 121 + 0 11 + 9 \times 1 = 2662 + 121 + 0 + 9 = 2792\).

Let's try another one. We will convert \(TEW_{\text{thirteen}}\) to base ten. First, we should fill in a place value chart for base thirteen.

2197

169

13

1

\(TEW_{\text{thirteen}} = 10 \times 169 + 11 \times 13 + 12 \times 1 = 1690 + 143 + 12 = \underline{1845}\)

Problem 29

Convert each of the following to a base ten numeral.

a. \(47E_{\text{twelve}}\) =	b. \(T74_{\text{eleven}}\) =	c. \(TTT_{\text{thirteen}}\) =
d. \(2034_{\text{twelve}}\) =	e. \(1025_{\text{eleven}}\) =	f. \(1028_{\text{thirteen}}\) =

Problem 30

Convert each of the following to a base ten numeral.

a. \(110_{\text{two}}\) =	e. \(110_{\text{six}}\) =	i. \(110_{\text{eleven}}\) =
b. \(110_{\text{three}}\) =	f. \(110_{\text{seven}}\) =	j. \(110_{\text{twelve}}\) =
c. \(110_{\text{four}}\) =	g. \(110_{\text{eight}}\) =	k. \(110_{\text{thirteen}}\) =
d. \(110_{\text{five}}\) =	h. \(110_{\text{nine}}\) =	l. \(110_{\text{twenty}}\) =

When using base systems for numerals, there is an easy way to compare number sizes that did not work for Roman numerals. Two positive integers written in the same base can be compared by first looking at the length of the numbers and, if they are the same length, looking at the digits starting from left to right. But what if they are written in different bases. Sometimes we have to convert them to the same base to compare them. But not always. Let's finish up this problem set by thinking about this.

Problem 31

Without doing any computations, circle which numeral has the larger value. Explain your reasoning. Try to be mathematically clear in your explanation. Notice the sequence of digits is the same for each numeral. Only the base is different.

\[13 \ 201 \ 154 \ 320 \ 050 \ 146_{\text{eight}} \ \mathbf{OR} \ 13 \ 201 \ 154 \ 320 \ 050 \ 146_{\text{eleven}} \nonumber \]