1. DigitValue Denotation NumericValue Denotation 2thousands1,000 0hundreds100 4tens10 8ones1 Computer Organization Basic Computer Concepts Number Systems Concepts  “Deka” – Greek word denoting a value equal to ten Table 1: Decimal Example 2,048 TABLE 2: The value denotation of a number “place” or position  Given a base-n number system, the value denotation of a position m is: m Value denotation 1 2 1 x n 3 1 x n x n 4 1 x n x n x n M 1 x n x n... x n (n multiplied M-1 times)0

2. Basic Computer Concepts Number Systems Concepts Example A:  Finding the value of a number in any number system Given a base-n system and a number PQRST, the value is given by: T x 1 + S x n + R x n x n + Q x n x n x n + P x n x n x n x n Example B:  Given a base-8 system and a number 56472, the value is given by: Value 2x1= 2 7 x 8 = 56 4 x 64 = 256 6 x 512 = 3,072 5 x 4,096 = 20,480 Step 2x1 +7x8 +4x8x8 +6x8x8x8 +5x8x8x8x8 Total: 23,866 Computer Organization

3. Basic Computer Concepts Number Systems Concepts TABLE 3: General method for obtaining the value denotation of a base-n numeral Value denotation V m The value denotation of a position m in a base-n number system is: value denotation V m = n (m-1) Decimal value N of a base-n numeral The decimal value N of a number p k p k-1 p k-2,..,p 2 p 1 in a base-n number system is given by: numerical value N = i=1 ? i=k p i V i = i=1 ? i=k p i n (i-1)

4. Computer Organization Basic Computer Concepts Number Systems Concept Examples: 1. Derive the decimal value of 640 9 Solution: From Table 3: 640 9 = 0 x 9 0 + 4 x 9 1 + 6 x 9 2 = 0 x1+ 4 x9+ 6 x9x9 = 0 + 36 + 486 = 522 10 2. Derive the decimal value of 640 12 Solution: From Table 3: 640 12 = 0 x 12 0 + 4 x 12 1 + 6 x 12 2 = 0 x 1 + 4 x 12 + 6 x 12 x 12 = 0 + 48 + 864 = 912 10

5. Computer Organization Basic Computer Concepts Number Systems Concept Using the notation of Table 3: 3. Derive the decimal value of 640 16 Solution: From Table 3: 640 16 = i=1 ? i=3 pi 16 (i-1) = 0 x 16 0 + 4 X 16 1 + 6 X 16 2 Value = 0 + 64 + 1536 = 1600

6. Computer Organization Basic Computer Concepts Hexadecimal, Octal, & Binary Systems 1. Derive the decimal value of 110111012 Solution: 110111012 = 1 x 2 0 + 0 x 2 1 + 1 x 2 2 + 1 x 2 3 + 1 x 2 4 + 0 x 2 5 + 1 x 2 6 + 1 x 2 7 Value = 1 + 0 + 4 + 8 + 16 + 0 + 64 + 128 = 221 10 2. Derive the decimal value of 54176023 8 Solution: 54176023 8 = 3 x 8 0 + 2 x 8 1 + 0 x 8 2 + 6 x 8 3 + 7 x 8 4 + 1 x 8 5 + 4 x 8 6 + 5 x 8 7 Value = 3 + 16 + 0 + 3072 + 28672 + 32768 + 1048576 + 10485760 = 11598867 10

7. Computer Organization Basic Computer Concepts Hexadecimal, Octal, & Binary Systems 3. Derive the decimal value of 24CF 16 Solution: 24CF 16 = F x 16 0 + C x 16 1 + 4 x 16 2 + 2 x 16 3 Value = 15 + 192 + 1024 + 8192 = 942310 10

8. 4848 3838 2828 1818 0808 4,096512 64 81 05 1 -- 4848 3838 2828 1818 0808 4,096 512 6481 0 5 --- Computer Organization Basic Computer Concepts Converting: Decimal to Another Number System  Given a decimal number M, to convert to a number system with base - n: As an example, let M = 2,654 and n = 8

9. 4848 3838 2828 1818 0808 40965126481 05136 4848 3838 2828 1818 0808 40965126481 0513 Computer Organization Basic Computer Concepts Converting: Decimal to Another Number System 2,65410 = 51368

10. Computer Organization Basic Computer Concepts Converting: Decimal to Another Number System Derive the octal value of 10112 Solution: a. Convert 1011 2 to decimal: 1011 2 = 1 x 2 0 + 1 x 2 1 + 0 x 2 2 + 1 x 2 3 decimal value = 1 + 2 + 0 + 8 = 11 10 b. Convert 11 10 to octal: : 11/8 = 1 remainder 3 : 3/1 = 3 remainder 0 octal value = 138

11. Computer Organization Basic Computer Concepts Converting: Decimal to Another Number System Shortcut Method Derive the octal value of 10111011 2 Solution: Group the binary digits into three : 10111011 2 = 10 111 011 2 = 010 111 011 2 421 Now010= 2 421 111= 7 421 011 = 3 Therefore 010 111 011 2 = 273 8

12. Computer Organization Basic Computer Concepts Converting: Decimal to Another Number System Shortcut Method Derive the hexadecimal value of 10111011 2 Solution: Group the binary digits into four : 10111011 2 = 1011 1011 2 8 421 Now 1 0 1 1 = 11 8421 1 0 1 1= 11 But 11 16 = B; therefore 10111011 2 = BB 16