4.4: Multiplication and Division in Base Systems
- Multiply and divide in bases other than 10.
- Identify errors in multiplying and dividing in bases other than 10.
Just as in Addition and Subtraction in Base Systems, once we decide on a system for counting, we need to establish rules for combining the numbers we’re using. This includes the rules for multiplication and division. We are familiar with those operations in base 10. How do they change if we instead use a different base? A larger base? A smaller one?
In this section, we use multiplication and division in bases other than 10 by referencing the processes of base 10, but applied to a new base system.
Multiplication in Bases Other Than 10
Multiplication is a way of representing repeated additions, regardless of what base is being used. However, different bases have different addition rules. In order to create the multiplication tables for a base other than 10, we need to rely on addition and the addition table for the base. So let’s look at multiplication in base 6.
Multiplication still has the same meaning as it does in base 10, in that is 4 added to itself six times, .
So, let’s apply that to base 6. It should be clear that 0 multiplied by anything, regardless of base, will give 0, and that 1 multiplied by anything, regardless of base, will be the value of “anything.”
Step 1: So, we start with the table below:
| * | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | |||
| 3 | 0 | 3 | ||||
| 4 | 0 | 4 | ||||
| 5 | 0 | 5 |
Step 2: Notice
| * | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 10 | ||
| 3 | 0 | 3 | 10 | |||
| 4 | 0 | 4 | ||||
| 5 | 0 | 5 |
Step 3:
Next, we need to fill in
| * | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 10 | 12 | |
| 3 | 0 | 3 | 10 | |||
| 4 | 0 | 4 | 12 | |||
| 5 | 0 | 5 |
Step 4:
Finally,
| * | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 10 | 12 | 14 |
| 3 | 0 | 3 | 10 | |||
| 4 | 0 | 4 | 12 | |||
| 5 | 0 | 5 | 14 |
Step 5:
A similar analysis will give us the remainder of the entries. Here is
This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that
| * | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 10 | 12 | 14 |
| 3 | 0 | 3 | 10 | 13 | 20 | 23 |
| 4 | 0 | 4 | 12 | 20 | 24 | 32 |
| 5 | 0 | 5 | 14 | 23 | 32 | 41 |
Notice anything about that bottom line? Is that similar to what happens in base 10?
To summarize the creation of a multiplication in a base other than base 10, you need the addition table of the base with which you are working. Create the table, and calculate the entries of the multiplication table by performing repeated addition in that base. The table needs to be drawn only the one time.
Create the multiplication table for base 7.
- Answer
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Step 1: Let’s apply the process demonstrated and outlined above to find the base 7 multiplication table. It should be clear that 0 multiplied by anything, regardless of base, will give 0, and that 1 multiplied by anything, regardless of base, will be the value of “anything.” So, we start with the table below:
* 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 3 0 3 6 4 0 4 5 0 5 6 0 6 Step 2: Notice
2 × 2 = Example 4.29,4 . So, we add that to our table:6 + 2 = 11 6 + 2 = 11 * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 3 0 3 6 4 0 4 11 5 0 5 6 0 6 Step 3: Next, we need to fill in
. Using repeated addition,2 × 5 2 × 5 if we use our base 7 addition rules. So, we add that to our table:2 × 5 = 2 + 2 + 2 + 2 + 2 = 11 + 2 = 13 2 × 5 = 2 + 2 + 2 + 2 + 2 = 11 + 2 = 13 * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 13 3 0 3 6 4 0 4 11 5 0 5 13 6 0 6 Step 4: Finally,
. And so we add that to our table:2 × 6 = 2 + 2 + 2 + 2 + 2 + 2 + 2 = 13 + 2 = 15 2 × 6 = 2 + 2 + 2 + 2 + 2 + 2 + 2 = 13 + 2 = 15 * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 13 15 3 0 3 6 4 0 4 11 5 0 5 13 6 0 6 15 Step 5: A similar analysis will give us the remainder of the entries. Here is
demonstrated:4 7 × 5 7 4 7 × 5 7 4 7 × 5 7 = 4 7 + 4 7 + 4 7 + 4 7 + 4 7 = 11 7 + 11 7 + 4 7 = 22 7 + 4 7 = 26 7 4 7 × 5 7 = 4 7 + 4 7 + 4 7 + 4 7 + 4 7 = 11 7 + 11 7 + 4 7 = 22 7 + 4 7 = 26 7 This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that
4 7 + 4 7 = Example 4.29. Using those addition rules, the rest of the table is given below:11 7 * 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 11 13 15 3 0 3 6 12 15 21 24 4 0 4 11 15 22 26 33 5 0 5 13 21 26 34 42 6 0 6 15 24 33 42 51
Create the multiplication table for base 4.
Create the multiplication table for base 12.
- Answer
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Let’s apply the repeated addition to base 12. Here is
demonstrated:7 12 × 9 12 7 12 × 9 12 7 12 × 9 12 = 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 = 12 12 + 12 12 + 12 12 + 12 12 + 7 12 = 48 12 + 7 12 = 53 12 7 12 × 9 12 = 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 + 7 12 = 12 12 + 12 12 + 12 12 + 12 12 + 7 12 = 48 12 + 7 12 = 53 12 This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that
7 12 + 7 12 = Example 4.31. Using those addition rules, the rest of the table is given below:12 12 * 0 1 2 3 4 5 6 7 8 9 A B 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 9 A B 2 0 2 4 6 8 A 10 12 14 16 18 1A 3 0 3 6 9 10 13 16 19 20 23 26 29 4 0 4 8 10 14 18 20 24 28 30 34 38 5 0 5 A 13 18 21 26 2B 34 39 42 47 6 0 6 10 16 20 26 30 36 40 46 50 56 7 0 7 12 19 24 2B 36 41 48 53 5A 65 8 0 8 14 20 28 34 40 48 54 60 68 74 9 0 9 16 23 30 39 46 53 60 69 76 83 A 0 A 18 26 34 42 50 5A 68 76 84 92 B 0 B 1A 29 38 47 56 65 74 83 92 A1 Table 4.9
Create the multiplication table for base 14.
The multiplication table in base 2 below is as minimal as the addition table in the solution for Table 4.6. Since the product of 1 with anything is itself, the following multiplication table is found.
| * | 0 | 1 |
| 0 | 0 | 0 |
| 1 | 0 | 1 |
As with the addition table, we can use the multiplication tables and the addition tables to perform multiplication of two numbers in bases other than base 10. The process is the same, with the same carry rules and placeholder rules.
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Calculate
.45 6 × 24 6 45 6 × 24 6 -
Calculate
.101 2 × 110 2 101 2 × 110 2
- Answer
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Step 1:
Use the base 6 multiplication table (Table 4.8) and, when necessary, the base 6 addition table (Table 4.4).
Set up this calculation using columns:
4 5 x 2 4 Step 2: Multiply the 1s digits, 5 and 4, using the base 6 multiplication table (Table 4.8). There we see the result is 32 6 . So, we enter the 2 and carry the 3.
3 4 5 x 2 4 2 Table 4.8]), then . So, we enter the 31.24 6 + 3 6 = 31 6 24 6 + 3 6 = 31 6 3 4 5 x 2 4 3 1 2 Step 4: Now we move on to the 2 in the “tens” place in the bottom value. We multiply the 2 6 and the 5 6 , and we get 14 6 . So, we enter the 4 and carry the 1.
Step 5: Next up, we multiply the 2 and the 4, and then add 1. This gives us
. We enter those on that second line.12 6 + 1 6 = 13 6 12 6 + 1 6 = 13 6 1 4 5 x 2 4 3 1 2 1 3 4 0 Step 6: Now we add down the columns.
1 4 5 x 2 4 1 3 1 2 1 3 4 0 2 0 5 2 Step 7: The 3 and the 3 add to 10 in base 6, so we enter the 0 and carry the 1. We now have the result:
45 6 × 24 6 = 2052 6 45 6 × 24 6 = 2052 62.
Step 1: Use the base 2 multiplication table (Table 4.10) and, when necessary, the base 2 addition table in the solution for Example 4.29. Set up this calculation using columns:
1 0 1 x 1 1 0 Step 2: Using the pattern established above, and the processes from multiplication from base 10, we find the following:
1 0 1 x 1 1 0 0 0 0 1 0 1 1 0 1 Step 3: Adding down the columns results in the following:
1 0 1 x 1 1 0 0 0 0 1 0 1 1 0 1 1 1 1 1 0 So,
.101 2 × 110 2 = 11110 2 101 2 × 110 2 = 11110 2
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Step 1:
Use the base 6 multiplication table (Table 4.8) and, when necessary, the base 6 addition table (Table 4.4).
1. Calculate \(43_6 \times 52_6\).
2. Calculate \(11101_2 \times 11_2\).
Summarizing the process of multiplying two numbers in different bases, the multiplication table is referenced. Using that table, the multiplication is carried out in the same manner as it is in base 10. The addition rules for the base will also be referenced when carrying a 1 or when adding the results for each digit’s multiplication line.
Calculate
- Answer
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Step 1: Use the base 12 multiplication table in the solution for Example 4.39 and, when necessary, the base 12 addition table in the solution for Table 4.7. Set up this calculation using columns:
3 A × 7 4 Step 2: First, the 4 is multiplied by 3A, resulting in the first line.
3 A × 7 4 1 3 4 Step 3: Now we move on to the 7 in the “tens” place in the bottom value.
5 3 A x 7 4 1 3 4 2 2 A 0 Step 4: Now we add down the columns.
3 A x 7 4 1 3 4 2 2 A 0 2 4 1 4 Step 5: The 3 and the A add to 11 in base 12, so we enter the 1 and carry the 1.
We now have the result:
3 A 12 × 74 12 = 2414 12 3 A 12 × 74 12 \(3 \mathrm{~A}_{12} \times 74_{12}=2414_{12}\)= 2414 12
Calculate \(\mathrm{B} 3_{12} \times 47_{12}\).
Division in Bases Other Than 10
Just as with the other operations,
division in a base
other than 10, the process of division in a base other than 10 is the same as the process when working in base 10. For instance,
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Calculate
.14 6 ÷ 5 6 14 6 ÷ 5 6 -
Calculate
5 A 12 ÷ 7 12 5 A 12 ÷ 7 12
- Answer
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Looking at the multiplication table for base 6 (Table 4.8), we see that
. Using that, we know that5 6 × 2 6 = 12 6 5 6 × 2 6 = 12 6 .14 6 ÷ 5 6 = 2 6 14 6 ÷ 5 6 = 2 6 -
Looking at the multiplication table for base 12 in the solution for Example 4.39, we see that
. Using that, we know that7 12 × A 12 = 5 A 12 7 12 × A 12 = 5 A 12 .5 A 12 ÷ 7 12 = A 12 5 A 12 ÷ 7 12 = A 12
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Looking at the multiplication table for base 6 (Table 4.8), we see that
Calculate \(10_6 \div 3_6\)
Calculate \(50_{12} \div \mathrm{A}_{12}\).
Errors in Multiplying and Dividing in Bases Other Than Base 10
The types of errors encountered when multiplying and dividing in bases other than base 10 are the same as when adding and subtracting. They often involve applying base 10 rules or symbols to an arithmetic problem in a base other than base 10. The first type of error is using a symbol that is not in the symbol set for the base.
Explain the error in the following calculation, and determine the correct answer:
- Answer
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Since the problem is in base 6, the symbol set available is 0, 1, 2, 3, 4, and 5. The 8 in the answer is clearly not a legal symbol for base 6. Looking back to the base 6 multiplication table (Table 4.8), we see that
.4 6 × 2 6 = 12 6 4 6 × 2 6 = 12 6
Explain the error in the following calculation and determine the correct answer: \(13_4 \times 21_4=54_4\)
The second type of error is using a base 10 rule when the numbers are not in base 10. For instance, in base 17,
Explain the error in the following calculation. Determine the correct answer:
- Answer
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If this problem was a base 10 problem, this would be the correct answer. However, in base 12, \(8_{12} \times 7_{12}\) is not 56 , but is instead 48 . To correct this error, carefully use the multiplication table for base 12. If properly used, the correct answer would be \(18_{12} \times 7_{12}=\mathrm{B} 8_{12}\).
Explain the error in the following calculation. Determine the correct answer: \(49_{14} \times 9_{14}=441_{14}\)
Check Your Understanding
- To create the multiplication table for a given base, what should be used?
- What are the differences between multiplying in base 10 and multiplying in a different base?
- When dividing in a base other than base 10 , what table is referenced?
- Compute \(24_6 \times 53_6\).
- Compute \(32_{14} \div 4_{14}\).
- How do you know an error has occurred in a base 5 multiplication question if the answer obtained was 285 ?
- What are two common ways to determine an error is committed when computing in s base other than base \(10 ?\)