Calculation with general numbers.
A computation is just counting, unit by unit, forwards or backwards. It becomes tiring when the numbers are large. Hence people grouped objects as they counted and treated groups in the same way as they treated units. Creation of such super-units simplified computations. You remember results of certain basic tasks. Instead of counting in the task 3 + 4 one by one four times from 3 to 7, you remember this result (addition table). In place of repeated addition as in 4 + 4 + 4, you memorize the result as a product (multiplication table). In the position system, you only remember quite a small range of numbers. You apply them also to larger numbers. You have the rules:
| Definition | Computing stage 1 |
| Addition - abbreviated advance counting | term+term=sum |
| a+b=c | |
| Subtraction - inverse addition - reverse counting | term-term=difference |
| a-b=c |
Given the task 8 - 3, you can ask by how many units you must count forward from 3 to 8. A merchant who must give change for 17 Baht out of 20 Baht counts forwards from 17 to 20 or backwards from 20 to 17.
| Definition | Computing stage 2 |
| Multiplication - sum of equal terms | factor×factor = product |
| a · b = c | |
| Inverse multiplication - subtraction of divisor | dividend:divisor=quotient |
| a : b = c |
Algebra replaces these definitions by rules. For example:
a/b is the number which multiplied by b yields a: (a/b)·b = a.
Since general numbers have definite values, you apply to them these definitions and rules. For example, by the definition of multiplication,
| 2 · 3= 2 + 2 + 2 | 8 · 5 = 5 · 5 · 5 · 5 · 5· 5 · 5 · 5 |
Hence,
| a · 3 = a + a + a | b · 5 = b + b + b + b + b. |
and
| a·2+a·3=a+a+a+a+a=a·5 | b·4+b·2=b·6 | c·4+c·5=c·9 | x·5 - x·2=x·3. |
Since the sequence of factors in a product is arbitrary, you place numbers first, followed by the general numbers in alphabetical sequence:
| 3·a not a·3 | a·b·c not b·a·c or c·a·b |
You can omit the multiplication signs (·) between numbers and general numbers and write 3a instead of 3·a. You have done so already with fractions when you wrote 4¾ instead of 4+¾.
The number 4 in 4a is called a coefficient; it is omitted, if a = 1! As follows from the above, the letter a in 4a is a number and not a dimension like cm! The expression 4 cm is not the same as the product 4a.
Tasks:
Invent always your own, similar examples! Make sure that there do not occur subtractions such as 3a - 4a!
| 1. | 4b + 6b = 10b | 2. | ½ c + ¾ c = 1¼ c | 3. | 6a - 5 a = a |
4. 2a + 3b + 5c. In a given problem, a, b, c denote different numbers!
5. Compute 3a + 7b + 5a + 3b! The sequence of the terms is arbitrary. Hence
3a + 5a + 7b + 3b = 8a + 10b.
6. 3a + 8b - 2a - 7b = 3a - 2a + 8b - 7b = a + b.
| 7. | 2·3·5=6·5=30 | 2·3·5=2·15=30 | 2·3·5=3·10=30 | 2·a·3=6a |
In a product
of several factors,
you can replace two or more
factors by a product without changing the final result.
| 8. | 4a·6b=24ab | 9. | 4a·3b·2c=24abc | 10. | 14·15/7=2·15 | 11. | 14·15/5=14·3 | 12. | 4a/2=2a |
Instead of dividing a product by a number, you can also divide one of its factors by that number and multiply the result by the other factors.
| 13. | 25a/5=5a | 14. | ab/b = a | 15. | 90/3·5 = 30/5 = 6 or 90/3·5 = 18/3 = 6 |
If a product of several factors is to be divided by numbers, you can at first divide by one number, then another, etc.; the sequence of operations is indifferent.
| 16. | abc/bc = a | 17. | 4bc/2c = 2b |