This problem solving activity has a number focus.
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?
Write the base 10 number 582 in base 3.
This problem enables students to better understand our decimal system by looking at a comparable system – the base 3 system.
Students should see that numbers can be represented in more than one way. Sometimes one system has an advantage over another.
For instance 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. Once that has been done, letters can be transmitted too. This is because we can code letters by numbers. In this way the base 2 system allows us to send messages by phone lines or as radio waves.
There are other uses 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 use seven unit vectors and so on for higher dimensions.
The base 3 system (ternary) has 3 digits: 0, 1, and 2. Your students might find it interesting to investigate other number base systems.
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?
Write the base 10 number 582 in base 3.
Find a way of changing any base 10 number into a base 3 number.
2 2 0 2 1 |
+2 1 2 1 0 |
12 1 0 0 1 |
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.
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.
Begin 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 | ||||
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 is 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 base 3 is well understood, try working in base 4 or any other base you can think of.
Printed from https://nzmaths.co.nz/resource/base-3 at 3:05pm on the 20th April 2024