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Introduction to Computing CS 101 Autumn 2004

Introduction to Computing CS 101 Autumn 2004. Chapter 1 Introduction to Computing. Decimal Number System. Example 3 N = 4790 P = 10 4 – 4790 = 5210 Conclusions Leave least significant 0’s as 0’s Subtract least significant non-zero digit from 10

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Introduction to Computing CS 101 Autumn 2004

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  1. Introduction to ComputingCS 101Autumn 2004

  2. Chapter 1Introduction to Computing

  3. Decimal Number System • Example 3 • N = 4790 • P = 104 – 4790 = 5210 • Conclusions • Leave least significant 0’s as 0’s • Subtract least significant non-zero digit from 10 • Subtract each of the remaining digit from 9 • Compliment of a compliment returns the original number

  4. Decimal Number System • 10’s compliment signed numbers • No signs in 10’s compliment representation • 00  99 represents -50  49 • Smallest number is -50 • Biggest number is 49 • A number is negative if its MSD ≥ 10/2 = 5 • To find magnitude of a negative number, take its 10’s compliment MSD = Most Significant Digit

  5. Decimal Number System • Subtraction using 10’s compliment representation • Compute 35 – 42 • 35 – 42 = (+35) + (-42)  35 + 58 in 10’s compliment • Now 35 + 58 = 93 • 93 is a negative number • 10’s compliment of 93 = 07, so 93 represents -07 • Finally 35 – 42 = -7

  6. Decimal Number System • Subtraction using 10’s compliment representation • Compute 48 – 32 • 48 – 32 = (+48) + (-32)  48 + 68 in 10’s compliment • Now 48 + 68 = 116 • Ignore the 3rd digit, since we are doing two digit arithmetic • 16 is a positive number • Finally 48 - 32 = 16

  7. Decimal Number System • Subtraction using 10’s compliment representation • Compute 35 – 8 • 35 – 8 = (+35) + (-08)  35 + 92 in 10’s compliment • 8 must be represented in 2 digits as 08 (10’s compliment of 8 is 2 in one digit, while 10’s compliment of 08 is 92) • Now 35 + 92 = 127 • Ignore the 3rd digit, since we are doing two digit arithmetic • 27 is a positive number • Finally 35 – 8 = 27

  8. Binary Number System • Base 2 system (2 numbers: 0, 1) • Counting 000001010011100101110111…

  9. Binary Number System Compare with decimal system n zeros

  10. Binary Number System • Expanded form (110010)2 = 1x25 + 1x24 + 1x21 = 32 + 16 + 2 = (50)10 • Addition

  11. Binary Number System • Subtraction (M-N) • Case I: M ≥ N • Case II: M < N • Compute N – M • Multiply answer by -1

  12. Binary Number System • Compliment of a binary number (2’s compliment) • Let N be an n bit number (1010 is 4 bit number) • 2’s compliment of N is given as P = 2n – N • 10000 – 1010 = 0110 • 1s  0s and 0s  1s, except the least significant 1 and lagging 0s’ • Fixed number of bits in a computer • Compute 2’s compliment of 1010 in 8-bit arith. • 0000 1010  1111 0110

  13. Binary Number System • Subtraction using 2’s complement • To evaluate M – N • Write M & N in the same number of bits • Let P = 2’s complement of N • Calculate S = M + P • If there is an “end carry”, discard it. Subtraction is complete • If there is no “end carry”, take 2’s complement of S and associate a negative sign with the result.

  14. Binary Number System • Example • Evaluate the following using 2’s complement. Verify in decimal system. • 1010100 – 1000011  84 – 67 = 17 • M=1010100, N=1000011, P=0111101, S=M+P

  15. Binary Number System • Example • Evaluate the following using 2’s complement. Verify in decimal system. • 1000011 - 1010100  67 - 84 = -17 • M=1000011, N=1010100, P=0101100, S=M+P 2’s compliment

  16. Decimal & Binary conversion • Convert (289)10 into binary • Convert (10111010)2 into decimal • Convert (0.289)10 into binary • Convert (0.10111010)2 into decimal

  17. 2 289 144 1 72 0 36 0 18 0 9 0 4 1 2 0 1 0 Decimal & Binary conversion (289)10 = (1 0010 0001)2

  18. Decimal & Binary conversion • (1011 1010)2 = (?)10 • 1x27 + 1x25 + 1x24 + 1x23 + 1x21 • 128 + 32 + 16 +8 + 2 = 186 • (1011 1010)2 = (186)10

  19. 0.289 X 2 0 0.578 X 2 1 0.156 X 2 0 0.312 X 2 0 0.624 x2 1 0.248 Decimal & Binary conversion • (0.289)10 = (?)2 (0.289)10 = (0.01001…)2 Convert 0.01001 to decimal. (0.01001) 2= 1x2-2 + 1x2-5 = 0.250 + 0.03125 = 0.28125 (0.01001)2 = (0.28125)10

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