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Number Systems - Part II

Number Systems - Part II. CS 215 Lecture # 6. Conversion from Decimal. Let N be the number to be converted. We can write N as N = q*b + r where q = quotient b = the base into which N is to be converted r = remainder. Example 6.1. Convert 80 10 to binary (base 2) 80 = 40*2 + 0

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Number Systems - Part II

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  1. Number Systems - Part II CS 215 Lecture # 6

  2. Conversion from Decimal Let N be the number to be converted. We can write N as N = q*b + r where q =quotient b =the base into which N is to be converted r =remainder

  3. Example 6.1 Convert 8010 to binary (base 2) 80 = 40*2 + 0 40 = 20*2 + 0 20 = 10*2 + 0 10 = 5*2 + 0 5 = 2*2 + 1 2 = 1*2 + 0 1 = 0*2 + 1 Answer: 10100002

  4. Example 6.2 Convert 8010 to octal(base 8) 80 = 10*8 + 0 10 = 1*8 + 2 1 = 0*8 + 1 Answer: 8010 = 1208 Also, from binary: 8010 = 10100002 = 1 010 000 = 120

  5. Example 6.3 Convert 8010 to hexadecimal(base 16) 80 = 5*16 + 0 5 = 0*16 + 5 Answer: 8010 = 5016 Also, from binary: 8010 = 10100002 = 101 0000 = 50

  6. Conversion from Decimal Decimal  Binary or Hexadecimal Repeated division by 2 or 16 Example: Convert 292 to hexadecimal. 292/16 = 18 R 4 18/16 = 1 R 2 1/16 = 0 R 1 292 = 12416 When the quotient is less than 16, the process ends

  7. Conversion from Decimal • Convert 42 to hexadecimal • 42/16 = 2 R 10 (but 10 = a16) • 2/16 = 0 R 2 42 = 2a16 • Convert 109 to binary • 109/2 = 54 R 1 • 54/2 = 27 R 0 • 27/2 = 13 R 1 • 13/2 = 6 R 1 • 6/2 = 3 R 0 • 3/2 = 1 R 1 • 1/2 = 0 R 1 109 = 11011012

  8. Shortcut: convert to hexadecimal • It is sometimes easier to convert from binary to hexadecimal, then to decimal, instead of converting directly to decimal • 11101010110110 • = 0011 1010 1011 0110 • = 3 a b 6 • = 3*16^3 + 10*16^2 + 11*16 + 6 • = 3*4096 + 10*256 + 176 + 6 • = 12198 + 2560 + 182 • = 12198 + 2742 = 14940

  9. Shortcut: convert to hexadecimal • The converse is also true: converting from decimal to hexadecimal, then to binary generally requires far fewer steps than converting directly to binary • 278/16 = 17 R 6 • 17/16 = 1 R 1 • 1/16 = 0 R 1 • 278 = 11616 = 0001 0001 01102 = 1000101102

  10. Example 6.4 Convert the following: 5010 = ____16 5010 = ____8 5010 = ____2 5010 = ____5

  11. Conversion of Fractions • A decimal improper fraction a/b is converted into a different base b by • converting it into a mixed fraction a/b = d e/f • e.g. 13/8 = 1 5/8 • convert d into the required base using the techniques we have already discussed • convert e/f (now a proper fraction) into the required base by following the steps below • the answer is in the format ...f1f0.f-1f-2...

  12. Example 6.5 Convert 2 5/8 to binary 2*(5/8) = 1 + 1/4 2*(1/4) = 0 + 1/2 2*(1/2) = 1 + 0 Answer: 10.101

  13. Answer: 11.10011001... Example 6.6 Convert 3.6 to binary 2*.6 = 1 + .2 2*.2 = 0 + .4 2*.4 = 0 + .8 2*.8 = 1 + .6 2*.6 = 1 + .2 2*.2 = 0 + .4 2*.4 = 0 + .8 2*.8 = 1 + .6

  14. Converting back into decimal 10.101 fraction part = 1*(1/2)1 + 0*(1/2)2 + 1*(1/2)3 = .625 or 5/8 integer part = 1*(21) + 0*(20) = 2 Answer: 2 5/8

  15. Scientific Notation Given a number N, we can express N as N = s*rx where s = mantissa or significand r = radix x = exponent

  16. The mantissa can be made an integer by multiplying it with a fixed power of the radix and subtracting the appropriate integer constant from the exponent • For example, 2.358 * 101 = 2358 * 10-2

  17. Normalization • Standard scientific notation defines the normal form to meet the following rule. 1 £ s < r where s = significand r = radix • A representation is in normal form if the radix point occurs just to the right of the first significant symbol

  18. Example 6.7 2358 * 10-2 = 2.358 * 101 1525 * 10-1 = 1.525 * 102

  19. Representing Numbers • Choosing an appropriate representation is a critical decision a computer designer has to make • The chosen representation must allow for efficient execution of primitive operations • For general-purpose computers, the representation must allow efficient algorithms for • addition of two integers • determination of additive inverse

  20. Representing Numbers • With a sequence of N bits, there are 2N unique representations • Each memory cell can hold N bits • The size of the memory cell determines the number of unique values that can be represented • The cost of performing operations also increases as the size of the memory cell increases • It is reasonable to select a memory cell size such that numbers that are frequently used are represented

  21. Binary Representation • The binary, weighted positional notation is the natural way to represent non-negative numbers. • MAL numbers the bits from right to left, beginning with 0 as the least significant digit. 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

  22. 0 7 1 6 2 5 3 4 Modulo Arithmetic • Consider the set of number {0, … ,7} • Suppose all arithmetic operations were finished by taking the result modulo 8 • 3 + 6 = 9, 9 mod 8 = 1 • 3 + 6 = 1 • 3*5 = 15, 15 mod 8 = 7 • 3 * 5 = 7

  23. 0 7 1 6 2 5 3 4 Modulo Arithmetic: Additive Inverse • What is the additive inverse of 7? • 7 + x = 0 • 7 + 1 = 0 • 0 and 4 are their own additive inverses

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