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# 1.0 number bases form 5

Number Bases Form 5

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### 1.0 number bases form 5

1. 1. Chapter 1 : Number Bases • 1.1 a : Stating Numbers in Base Two, Eight and Five • 1.1 b : Value of a Digit of a Number in Base 2, 8 and 5 • 1.1 c : Writing Numbers in Base 2, 8 and 5 in Expanded Notation • 1.1 d : Converting Numbers in Base 2, 8 and 5 to Base 10 and Vice Versa • 1.I e : Converting from One Base to Another • 1.1 f : Addition and Subtraction in Base Two
2. 2. Chapter 1 Number Bases Number in Base Two, Eight and Five1 1 a 1.1 Stating Numbers in Base Two, Eight and Five • The numbers we use daily are in base 10. The place value of numbers in base 10 are as shown below. 103 =1000 102 =100 101=10 100 =1 Place value Number in base 10 9 7 0 39 7 0 3
3. 3. 103 =1000 102 =100 101=10 100 =1 Place value Number in base 109 7 0 3 The place value of the digit 7 in the number 9703 is 100 • The place value of any digit of a number is a fixed value and does not change with the value of the digit. • There is no place value equal to zero. • The smallest place value of all number bases is “ones”. • The place value of 3 in the number 9703 is 1.
4. 4. 103 =1000 102 =100 101=10 100 =1 Place value Number in base 109 7 0 3 • There are 10 digits that can be written in any place value column for numbers in base 10. • The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 • The digit value or value of digit varies with the place value and the digit
5. 5. 103 =1000 102 =100 101=10 100 =1 Place value Number in base 109 7 0 3 The value of the digit 9 is 9 x 1000 = 9000 Digit Place value of 9 Value of the digit 9
6. 6. 103 =1000 102 =100 101=10 100 =1 Place value Number in base 109 7 0 3 The value of the digit 0 is 0 x 10 = 0 Digit Place value of 0 Value of the digit 0
7. 7. • Numbers in base 2 have their respective place values as shown below 22 = 21 = 20 = 2 x 2 = 4 2 1 Place Value • There are only 2 digits in base 2 : 0 and 1
8. 8. • Numbers in base 8 have their respective place values as shown below 82 = 81 = 80 = 8 x 8 = 64 8 1 Place Value • There are only 8 digits in base 8 : 0, 1, 2, 3, 4, 5, 6 and 7
9. 9. • Numbers in base 5 have their respective place values as shown below 52 = 51 = 50 = 5 x 5 = 25 5 1 Place Value • There are only 5 digits in base 5 : 0, 1, 2, 3 and 4
10. 10. 24= 16 23= 8 22= 4 21= 2 20= 1 BASE 2 Place Value of Numbers in Base 2 Number in Base 10 0 0 2 = 2 + 0 1 1 01 11 1 0 0 0 0 000 00 1 11 111 11 3 = 2 + 1 4 = 4 + 0 + 0 5 = 4 + 0 + 1 6 = 4 + 2 + 0 7 = 4 + 2 + 1 8 = 8 + 0 + 0 + 0 9 = 8 + 0 + 0 + 1 1 1
11. 11. 83= 512 82= 64 81= 8 80= 1 BASE 8 Place Value of Numbers in Base 8 Number in Base 10 0 0 2 1 1 2 3 4 6 01 2 5 7 3 3 4 5 6 7 8 = 8 + 0 19 = 2 x 8 + 3
12. 12. 54= 625 53= 125 52= 25 51= 5 50= 1 BASE 5 Place Value of Numbers in Base 5 Number in Base 10 0 0 2 1 1 2 3 4 1 1 02 3 0 1 21 2 3 4 5 = 5 + 0 6 = 5 + 1 7 = 5 + 2 10 = 2 x 5 + 0 17 = 3 x 5 + 2
13. 13. 910 = 8 + 0 + 0 + 1 =10012 Read as “one zero zero one base 2” 910 = 8 + 1 = 118 Read as “one one base 8” 910 = 5 + 4 =145 Read as “one four base 5” • Numbers in base 2 are also known as binary numbers • Numbers in base 8 are also known as octal numbers • Numbers in base ten are also known as denary numbers
14. 14. State two numbers in base two after 11102EXAMPLE 24=16 23=8 22=4 21=2 20=1 Base 10 1 1 1 0 8+4+2+ 0 = 14 SOLUTION 8+4+2+ 1=15 1 1 1 1 16+0+0+ 0=16 0 0 0 01 11112 and 100002
15. 15. State a number before and after 218 in base 8EXAMPLE SOLUTION 81=8 80=1 Base 10 2 1 2 x 8 + 1 = 17 Before 2 x 8 + 0 = 162 0 After 2 x 8 + 2 = 182 2 208 and 228
16. 16. State two numbers after 435 in base 5EXAMPLE SOLUTION 52=25 51=5 50=1 Base 10 4 3 4 x 5 + 3 = 23 4 x 5 + 4 = 244 4 1 x 25 + 0 + 0 = 251 0 0 445 and 1005
17. 17. 1.1 b Value of A Digit of A Number in Base Two, Eight and Five Value of a digit = The digit x Place value of a digit State the value of the underlined digit in each of the following numbers (a)11012 (b) 40328 (c)12435 EXAMPLE
18. 18. SOLUTION Place Value 23=8 22=4 21=2 20=1 Number 1 1 0 1 Value of Digit 1 x 4 = 4 The value of the digit “1” in 11012 is 4
19. 19. SOLUTION Place Value 83=512 82=64 81=8 80=1 Number 4 0 3 2 Value of Digit 0 x 64 = 0 The value of the digit “0” in 40328 is 0
20. 20. SOLUTION Place Value 53=125 52=25 51=5 50=1 Number 1 2 4 3 Value of Digit 4 x 5 = 20 The value of the digit “4” in 12435 is 20
21. 21. 1.1 c Writing Numbers in Base Two, Eight and Five in Expanded Notation • A number written in expanded notation refers to the sum of the value of the digits that make up the number . • Let us write 42510 in expanded notation Place Value 102 101 100 Number 425 Therefore, 42510 written in expanded notation is as follows 42510 = 4 x 102 + 2 x 101 + 5 x 100
22. 22. 1.1 c Writing Numbers in Base Two, Eight and Five in Expanded Notation • Let us write 3748 in expanded notation Place Value 82 81 80 Number 3 7 4 Therefore, 3748 written in expanded notation is as follows 3748 = 3 x 82 + 7 x 81 + 4 x 80
23. 23. 1.1 c Writing Numbers in Base Two, Eight and Five in Expanded Notation • Let us write 110012 in expanded notation Place Value 24 23 22 21 20 Number Therefore, 110012 written in expanded notation is as follows 110012 = 1 x 24 + 1 x 23 + 0 x 22 + 0 x 21 + 1 x 20 1 1 00 1
24. 24. 1.1 c Writing Numbers in Base Two, Eight and Five in Expanded Notation • Let us write 41035 in expanded notation Place Value 53 52 51 50 Number Therefore, 41035 written in expanded notation is as follows 41035 = 4 x 53 + 1 x 52 + 0 x 51 +3 x 50 4 1 0 3
25. 25. 1.1 d Converting Numbers in Base Two, Eight and Five to Base 10 and Vice Versa Steps to convert numbers in base 2, 8 and 5 to base 10 are as follows 1. Write the number in expanded notation 2. Simplify the expanded notation into a single number EXAMPLE Convert each of the following numbers to a number in base 10 a. 110012 b. 3748 c. 41035
26. 26. EXAMPLE Convert each of the following numbers to a number in base 10 a. 110012 b. 3748 c. 41035 SOLUTION a. 110012 = 1 x 24 + 1 x 23 + 0 x 22 + 0 x 21 + 1 x 20 = 2510 b. 3748 = 3 x 82 + 7 x 81 + 4 x 80 = 25210 c. 41035 = 4 x 53 + 1 x 52 + 0 x 51 +3 x 50 = 52810
27. 27. 1.1 d Converting Numbers in Base Two, Eight and Five to Base 10 and Vice Versa Steps to convert numbers in base 10 to base 2, 8 or 5 are as follows 1. Perform repeated division until the quotient is 0 2. Write the number in the new base by referring to the remainders from bottom to the top EXAMPLE Convert 1810 to numbers in base two, eight and five
28. 28. SOLUTION 1810 to Base 2 18 9 4 2 1 0 2 2 2 2 2 R1 R0 R0 R1 R0 1810 = 100102
29. 29. SOLUTION 1810 to Base 5 18 3 0 5 5 R3 R3 1810 = 335
30. 30. SOLUTION 1810 to Base 8 18 2 0 8 8 R2 R2 1810 = 228
31. 31. 1.1 e Converting Numbers from One Base to Another The following steps are used to convert from one base to another 1. Convert the number to a number in base 10 2. Use repeated division to convert the number in base 10 to the respective bases EXAMPLE Convert a. 1102 to number in base 5 b. 325 to number in base 2 c. 1278 to number in base 5 d. 2035 to number in base 8
32. 32. SOLUTION a.1102 to number in base 5 1102 = 1 x 22 + 1 x 21 + 0 x 20 = 610 6 1 0 5 5 R1 R1 1102 = 115
33. 33. SOLUTION b. 325 to number in base 2 325 = 3 x 51 + 2 x 50 = 1710 17 8 4 2 1 0 2 2 2 2 2 R1 R0 R0 R0 R1 325 = 100012
34. 34. SOLUTION c. 1278 to number in base 5 1278 = 1 x 82 + 2 x 81 + 7 x 80 = 8710 87 17 3 5 5 R2 R2 1278 = 3225 5 0 R3
35. 35. SOLUTION d. 2035 to number in base 8 2035 = 2 x 52 + 0 x 51 + 3 x 50 = 5310 53 6 0 8 8 R6 R5 2035 = 658
36. 36. Binary Octal Decimal 000 0 0 001 1 1 010 2 2 011 3 3 100 4 4 101 5 5 110 6 6 111 7 7
37. 37. Converting binary to octal: Counting from right to left, draw a line between every group of 3-bits. The most significant group may not have exactly three bits, so you can just pretend the others are zeros. Now convert each group of three to a single, octal digit. The conversion of a 3-bit number to an octal number is easy. You can memorize the patterns easily and, even if you forget, they are not hard to figure out. The resulting octal digits, when written together in the same order, are the equivalent binary number. Here's an example which converts the binary number '11111010' to its equivalent octal number.
38. 38. Converting octal to binary: This is simply the reverse of the above process. For every octal digit, just write down the 3-bit pattern that represents it. Here is an example which converts octal number 6252 to binary.
39. 39. Convert 1000111012 to number in base 8 solution 101011100 101011100 534 1000111012 = 4358
40. 40. Convert 5318 to number in base 2 solution 5318 = 1010110012 531135 001011101
41. 41. 1.1f Addition and Subtraction in Base Two • To add two numbers in base two, the following addition rules are important
42. 42. 1 0 1 1 + 1 1 1 0 _______10 1 0 1 11 12 + 12 = 102 12 + 12 = 102 12 + 12 + 12 = 112
43. 43. 1.1f Addition and Subtraction in Base Two • To subtract two numbers in base two, the following subtraction rules are important 02 - 02 = 12 - 02 = 12 - 12 = 102 - 12 = 02 12 02 12
44. 44. 1 0 1 1 - 1 1 0 _______10 10 1 0 102 -12 = 12 02 - 02 = 02
45. 45. BASE
46. 46. BASE Binary BIN (b) Octal OCT (o) Denary DEC (d)
47. 47. Example 1 MODE BASE 3 3 4 6 2 = ln OCT 8181 x2 DEC 2x o d Convert 14628 to a number in base 10
48. 48.  To clear the Base-n specification MODE COMP 1 1Press 1X  To continue the Base-n specification Press ON
49. 49. Convert 11012 to a number in base 8 Example 2 MODE BASE 3 3 1 0 1 = log BIN 151 ln OCT 2x b o
50. 50.  To clear the Base-n specification MODE COMP 1 1Press 1X  To continue the Base-n specification Press ON
51. 51. Example 3 Convert 146210 to a number in base 8 MODE BASE 3 3 4 6 2 = x2 DEC 26661 ln OCT 2x d o
52. 52. Example 4 Calculate 10012 + 1112, stating your answer as a number in base 2 MODE BASE 3 3 0 0 1 + log BIN 1 2x b 1 1 = 10000 1 b
53. 53.  To clear the Base-n specification MODE COMP 1 1Press 1X  To continue the Base-n specification Press ON