Base 3
Write numbers in base 3
Do arithmetic in base 3
Explain the "links" between base 3 and decimal numbers
Devise and use problem solving strategies to explore situations mathematically (be systematic).
This problem aims to give a better idea of our decimal system by looking at a comparable system – the base 3 system. By seeing how base 3 numbers work students should have a better appreciation of base 10 numbers, the decimal numbers.
Students should also see that numbers may be represented in more than one way. Sometimes one system has an advantage over another. For instance base 2 initially looks worthless. Why would you want to write a number in descending powers of 2? Well it turns out that base 2 is very useful in telecommunications. Electricity can be made to pulse or not to pulse through circuits. By using the pulse to represent 1 and the non-pulse to represent 0, numbers can be transmitted along a line. And once that has been done, letters can be transmitted too. This is because we can code letters by numbers. Hence the base 2 system allows us to send messages by phone lines or as radio waves.
There are other uses of the general idea of bases. For instance we represent vectors in three dimensions in terms of the unit vectors i, j and k. In seven dimensions we can get away with seven unit vectors and so on for higher dimensions.
If we think for a moment about functions, we might try to represent them in terms of other simpler functions. Powers of x are quite simple. Can we represent any function as a polynomial in x? The answer is yes, provided the original function is not too complicated or bizarre and if we are prepared to accept a close approximation. Being able to represent functions as polynomials has some obvious advantages. The students will meet some of these at university level.
The Problem
We say that the number 1202 is a number in base 3 when we write it as 1 x 33 + 2 x 32 + 0 x 3 + 2 x 30. This is equal to 27 + 18 + 2 = 47 in our normal base, base 10.
- What is the value of the base 3 number 21021?
- What is the sum of the two base 3 numbers in base 3?
2 2 0 2 1 +2 1 2 1 0
- Write the base 10 number 582 in base 3.
Teaching sequence
- Introduce the problem by first talking about decimal numbers.
What is 102? How about 107?
What do the digits 1, 2 and 0 actually stand for in the number 120?
Can you expand 2376 in terms of powers of 10? - Then talk about base 3 numbers.
What is 34? How about 38?
What do the digits 1, 2, 0 actually stand for in the base 3 number 120?
Can you expand the base 3 number 21102 in terms of powers of 3?
What does the base 3 number 220011 equal in base 10? - Pose the problem for the children to tackle in their groups.
- Check that the groups are making progress. You may need to give them simpler problems to help them to work out parts b) and c).
- Get a few of the groups to report to the class on their answers and how they got them.
- Allow students time to write up what they did and to explain why they did what they did.
Extension to the problem
Find a way of changing any base 10 number into a base 3 number.
Solution
- 21021 = 2 x 34 + 1 x 33 + 0 x 32 + 2 x 3 + 1 = 162 + 27 + 6 + 1 = 196.
- Note that 1 + 2 = 10 in base 3 and 2 + 2 = 11. So
2 2 0 2 1 +2 1 2 1 0 12 1 0 0 1
- Key numbers in base 3 are 1, 3, 9, 27, 81, 243, 729, … In other words, the powers of 3.
Now the biggest power of 3 less than 582 is 243 and 2 x 243 = 486 is less than 582. So the base 3 form of the number will start 2… = 2 x 35 + …
582 – 486 = 96.
81 is the biggest power of 3 less than 96.
So 582 starts 21… = 2 x 35 + 1 x 34 + …
Since 96 – 81 = 15 which equals 9 + 6,
then 582 = 2101… = 2 x 35 + 1 x 34 + 0 x 33 + 1 x 32 + 2 x 3 + 0 x 30.
Solution to the extension
We show a particular example of a general method
|
3 |
582 |
|
|
3 |
194 |
0 |
|
3 |
64 |
2 |
|
3 |
21 |
1 |
|
3 |
7 |
0 |
|
3 |
2 |
1 |
|
0 |
2 |
The threes on the left divide the corresponding numbers in the centre. The quotient is written in the line below and the remainder goes on the right.
So let’s start off with 299.
|
3 |
299 |
3 |
299 |
3 |
299 |
|||||
|
99 |
2 |
3 |
99 |
2 |
3 |
99 |
2 |
|||
|
33 |
0 |
3 |
33 |
0 |
||||||
|
11 |
0 |
|
3 |
299 |
3 |
299 |
|||
|
3 |
99 |
2 |
3 |
99 |
2 |
|
|
3 |
33 |
0 |
3 |
33 |
0 |
|
|
3 |
11 |
0 |
3 |
11 |
0 |
|
|
3 |
3 |
2 |
3 |
3 |
2 |
|
|
3 |
1 |
0 |
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|
3 |
0 |
1 |
In the first table we divide 299 by 3 get 99 (in the table) and remainder of 2 (to the right). This 2 turns out to be the last digit on the right of the base 3 form of 299. Then 3 into 99 goes 33 times (in the table) and the remainder is 0 (to the right). This remainder of 0 becomes the second digit from the right in the final base 3 form of the number. We keep going down.
This algorithm changes numbers in base 10 to numbers in base 3. Just check, 582 in base 3 is 210120 (we’ve seen that already). But 299 in base 3 is 102002.
Why does this algorithm work?
If you are on top of this base 3 work try working in base 4 or any other base you can think of.
| Attachment | Size |
|---|---|
| Base.pdf | 39.34 KB |
| BaseMaori.pdf | 48.92 KB |
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