maths.free › Number Theory › 4. Number Representation and Calculation › Multiplication and Division in Base Systems
Multiplication and Division in Base Systems
Multiply and divide in bases other than 10.
Learning Objectives
After completing this section, you should be able to:
- Multiply and divide in bases other than 10.
- Identify errors in multiplying and dividing in bases other than 10.
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 \(4\times 6\) is 4 added to itself six times, \(4\times 6=4+4+4+4+4+4\).
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 \(2\times 2=4\) is there. But we didn’t hit a problematic number there (4 works fine in both base 10 and base 6). But what is \(2\times 3\)? If we use the repeated addition concept, \(2\times 3=2+2+2=4+2\). According to the base 6 addition table (), \(4+2=10\). So, we add that to our table:
| * | 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 \(2\times 4\). Using repeated addition, \(2\times 4=2+2+2+2=10+2=12\) (if we use our base 6 addition rules). So, we add that to our table:
| * | 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, \(2\times 5=2+2+2+2+2=12+2=14\). And so we add that to our table:
| * | 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 |
| * | 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 |
Creating a Multiplication Table for a Base Higher Than 10
Try it.
Create the multiplication table for base 12.
Solution
Let’s apply the repeated addition to base 12. Here is \({7}_{12}\times {9}_{12}\) demonstrated: \[{7}_{12}\times {9}_{12}={7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}={12}_{12}+{12}_{12}+{12}_{12}+{12}_{12}+{7}_{12}={48}_{12}+{7}_{12}={53}_{12}\]
This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that \({7}_{12}+{7}_{12}={12}_{12}\) and then applying the addition processes we’ve always known, but with the base 12 table in the solution for . Using those addition rules, the rest of the table is given below:
| * | 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 |
| * | 0 | 1 |
| 0 | 0 | 0 |
| 1 | 0 | 1 |
Condensed — the full section is in OpenStax Contemporary Mathematics.
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, \(72\div 9=8\) because, we know that \(9\times 8=72\). So, for many division problems, we are simply looking to the multiplication table to identify the appropriate multiplication rule.
Dividing with a Base Other Than 10
Try it.
- Calculate \({14}_{6}\div {5}_{6}\).
- Calculate \(5{A}_{12}\div {7}_{12}\)
Solution
- Looking at the multiplication table for base 6 (), we see that \({5}_{6}\times {2}_{6}={14}_{6}\). Using that, we know that \({14}_{6}\div {5}_{6}={2}_{6}\).
- Looking at the multiplication table for base 12 in the solution for , we see that \({7}_{12}\times {A}_{12}=5{A}_{12}\). Using that, we know that \(5{A}_{12}\div {7}_{12}={A}_{12}\).
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.
Identifying an Illegal Symbol in a Base Other Than Base 10
Try it.
Explain the error in the following calculation, and determine the correct answer:
\[{4}_{6}\times {2}_{6}={8}_{6}\]Solution
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 (), we see that \({4}_{6}\times {2}_{6}={12}_{6}\).
The second type of error is using a base 10 rule when the numbers are not in base 10. For instance, in base 17, \({6}_{17}\times {9}_{17}={54}_{17}\) would be incorrect, even though in base 10, \(6\times 9=54\). That rule doesn’t apply in base 17.
Identifying an Error in Arithmetic in a Base Other Than Base 10
Try it.
Explain the error in the following calculation. Determine the correct answer:
\[{18}_{12}\times {7}_{12}={126}_{12}\]Solution
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 \({18}_{12}\times {7}_{12}=B{8}_{12}\).
Key Concepts
- Multiplication tables for bases other than 10 can be built using the same processes that are used in base 10, including using repeated addition and the addition table for the base.
- Multiplication in bases other than base 10 use the same processes as multiplication in base 10, but use the multiplication table for that base.
- Basic division in bases other than base 10 use the same processes as basic division in base 10, where the missing factor process is used.
Projects
Go online. Google “additive number systems.”What system comes up?
- Describe the additive system you found.
Using Google, identify three more additive systems of numbers.
- Compare and contrast the systems you found. For instance, how many times can a symbol be used before a new symbol is used.
- Identify three situations where additive systems are still used.
Use Google to determine what base computers use.
Were other bases attempted for use in computers?
Determine why the base used in computers is appropriate.
Determine how the base used in computers is related to the circuitry in computers.
Determine how Boolean logic and the base used in computers are related, and might be identical.
There is research into using quibits in computers. Find out what quibits are and how can they improve computing speed.
Using Google, find three cultures, other than Babylonian or Mayan, that use base systems other than 10.
- Tell what base is used for each system.
- If possible, determine why the culture used that base system.
- Choose one of those systems. Explain that base system. Be sure to address whether the system is additive, place-value based, a blend of the two, and if it employs a zero.
Using any resources available to you, determine the history of 0 in at least three different numbering systems. Address at least when and why such a development occurred and why a 0 is vital to the use of a positional system.
Using any resources available to you, find at least three numbering systems from sub-Saharan Africa, Australia, China, or the Pacific Islands. Explore if they are positional or additive systems (or combinations!), the terminology of the system, if they used a 0, and what base they employed (if positional).
Practice (7)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Create the multiplication table for base 7.
උත්තරය හෙළි කරන්න
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 \(2\times 2=4\) is there. But we didn’t hit a problematic number there (4 works fine in both base 10 and base 6). The same is true for \(2\times 3\) and \(3\times 2\), which equal 6. But what is \(2\times 4\)? If we use the repeated addition concept, \(2\times 4=2+2+2+2=6+2\). According to the base 7 addition table in the solution for , \(6+2=11\). So, we add that to our table:
* 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 \(2\times 5\). Using repeated addition, \(2\times 5=2+2+2+2+2=11+2=13\) if we use our base 7 addition rules. So, we add that to our table:
* 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, \(2\times 6=2+2+2+2+2+2+2=13+2=15\). And so we add that to our table:
* 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 \({4}_{7}\times {5}_{7}\) demonstrated: \[{4}_{7}\times {5}_{7}={4}_{7}+{4}_{7}+{4}_{7}+{4}_{7}+{4}_{7}={11}_{7}+{11}_{7}+{4}_{7}={22}_{7}+{4}_{7}={26}_{7}\]
This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that \({4}_{7}+{4}_{7}={11}_{7}\) and then applying the addition processes we’ve always known, but with the base 7 table in the solution for . Using those addition rules, the rest of the table is given 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 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 12.
උත්තරය හෙළි කරන්න
Let’s apply the repeated addition to base 12. Here is \({7}_{12}\times {9}_{12}\) demonstrated: \[{7}_{12}\times {9}_{12}={7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}+{7}_{12}={12}_{12}+{12}_{12}+{12}_{12}+{12}_{12}+{7}_{12}={48}_{12}+{7}_{12}={53}_{12}\]
This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that \({7}_{12}+{7}_{12}={12}_{12}\) and then applying the addition processes we’ve always known, but with the base 12 table in the solution for . Using those addition rules, the rest of the table is given below:
* 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 -
- Calculate \({45}_{6}\times {24}_{6}\).
- Calculate \({101}_{2}\times {110}_{2}\).
උත්තරය හෙළි කරන්න
Step 1: Use the base 6 multiplication table () and, when necessary, the base 6 addition table ().
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 (). There we see the result is 326. So, we enter the 2 and carry the 3.
3 4 5 x 2 4 2 Step 3: So, now we multiply the 4 and the 4, then add the 3 (just as you would do if multiplying two base 10 numbers!). \({4}_{6}\times {4}_{6}={24}_{6}\) (from the base 6 table []), then \({24}_{6}+{3}_{6}={31}_{6}\). So, we enter the 31.
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 26 and the 56, and we get 146. 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 \({12}_{6}+{1}_{6}={13}_{6}\). We enter those on that second line.
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:
\({45}_{6}\times {24}_{6}={2052}_{6}\).- Step 1: Use the base 2 multiplication table () and, when necessary, the base 2 addition table in the solution for . 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, \({101}_{2}\times {110}_{2}={11110}_{2}\).
-
Calculate \(3{A}_{12}\times {74}_{12}\).
උත්තරය හෙළි කරන්න
Step 1: Use the base 12 multiplication table in the solution for and, when necessary, the base 12 addition table in the solution for . 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: \(3{A}_{12}\times {74}_{12}={2414}_{12}\).
-
- Calculate \({14}_{6}\div {5}_{6}\).
- Calculate \(5{A}_{12}\div {7}_{12}\)
උත්තරය හෙළි කරන්න
- Looking at the multiplication table for base 6 (), we see that \({5}_{6}\times {2}_{6}={14}_{6}\). Using that, we know that \({14}_{6}\div {5}_{6}={2}_{6}\).
- Looking at the multiplication table for base 12 in the solution for , we see that \({7}_{12}\times {A}_{12}=5{A}_{12}\). Using that, we know that \(5{A}_{12}\div {7}_{12}={A}_{12}\).
-
Explain the error in the following calculation, and determine the correct answer:
\[{4}_{6}\times {2}_{6}={8}_{6}\]උත්තරය හෙළි කරන්න
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 (), we see that \({4}_{6}\times {2}_{6}={12}_{6}\).
-
Explain the error in the following calculation. Determine the correct answer:
\[{18}_{12}\times {7}_{12}={126}_{12}\]උත්තරය හෙළි කරන්න
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 \({18}_{12}\times {7}_{12}=B{8}_{12}\).
Symbols used here
The exponent b must be raised to for x; ln uses base e.
Add a_k for k = 1 up to n.
Multiply a_k for k = 1 up to n.
Naturals, integers, rationals, reals, complex numbers.
n divides a − b; a and b have the same remainder.
b is a multiple of a; the largest number dividing both.
Count of 1..n coprime to n; number of primes up to x.
The remainders 0…n−1 with clock arithmetic.
What is left after dividing a by n.
How to: 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.
- Calculate
- Calculate
- Calculate
- Calculate
- Looking at the multiplication table for base 6 (
- Looking at the multiplication table for base 12 in the solution for
Questions people ask
Why are primes so important?
Every integer factors into primes in exactly one way, so primes are the atoms of multiplication. Cryptography relies on that factoring being easy to state and hard to do.
How do I tell whether a big number is prime?
Trial division up to the square root works for small numbers. For large ones, probabilistic tests (Miller–Rabin) give an answer that is wrong with negligible probability, and deterministic tests (AKS) exist but are slower.
ඔයාගේම උත්සහ කරන්න
Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
තවත් Number Theory
Prime factorisationPrime numbersGCD and LCMModular arithmeticDivisorsSequencesNumber basesDiophantine equationsFermat's little theorem and Euler's theoremRSA: cryptography from number theory