DFSC 1317 Introduction to Digital Forensics and Information Assurance

1. DFSC 1317 Introduction to Digital Forensics and Information Assurance 02 NumberING Systems: bInary, decImal, and hexadecImal

2. Basic operation performed by a computer • Arithmetic Operations: Addition, subtraction, multiplication and division • Logical operations: the sign or the comparative magnitude of two numbers • Data transfer: Moving data from one location to another in the memory. • Input-output operations: Controlling the reading/writing of information into or out of the computer

3. On Digital Computers Digital computers store numbers in an entity (or unit) called a word which consists of a string of binary digits, or bits.Various number systems are used to represent mathematical numbers. Some commonly used number systems are: hexadecimal (base 16), decimal (base 10), octal (base 8), and binary (base 2). For example, in a decimal system the number 8,410 is represented in powers of ten as: 8103 + 4102 +1101 + 0100 = 8000 + 400 + 10 + 0 = 8,410

4. On Digital Computers (Cont’d) A method known as the doubling procedure is as follows. Given a decimal number N, it can be decomposed as: N = 2Q1 + R1 (Q1 = N/2 - remainder) Q1 = 2Q2+ R2 (Q2 = Q1/2 - remainder) . . . . . . Qk = 0 + Rk+1 etc. The corresponding binary number is obtained by writing the remainders Rk+1, Rk, ... , R1 in the reverse order as: B = Rk+1RkRk-1... R1

5. Example Convert the decimal number N = 8,410, to a binary number. Solution: Perform sequential division by 2 as follows: 8,410 = (24,205) + 0 65 = (232) + 1 4,205 = (22,102) + 1 32 = (216) + 0 2,102 = (21,051) + 0 16 = (28) + 0 1,051 = (2525) + 1 8 = (24) + 0 525 = (2262) + 1 4 = (22) + 0 262 = (2131) + 0 2 = (21) + 0 131 = (265) + 1 1 = (20) + 1 The binary equivalent of 8,410 is then given by collecting the remainder digits from the last to the first: 10000011011010 = 11213 + 0212 + 0211 + 0210 + 029 + 028 + 127 + 126+ 025 +124 + 123 + 022 + 121 + 020

6. Representations of Numbers Numbers are usually represented using the normal form notation. That is, x = m.10E for 10-1 < m < 1 where for x 0, m is called the mantissa and E is the exponent. By convention, the number zero has the normal notation, 0.100.

7. Significant Digits If a number is written in standard decimal, floating-point form, or in normal form such that: x = 0.d1 d2 d3... dk10n with d1 0 anddk 0, we say that this number has k significant digits (or significant figures) which indicates those digits that can be used with a confidence relative to the true value of the number.

8. Significant Digits (Cont’d) Note that the zeros which are used only to shift the decimal point are not counted as significant figures. The leading zeros may or may not be significant. For example, x = 0.0002815 has 4 significant figures! x = 1,200 may have 4 significant figures! Some examples are: 46.45072800 = 0.46450728102 (with 8significant digits) -335.12 = -0.33512103 (with 5 significant digits) 0.00517 = 0.51710-3 (with 3 significant digits) 0.74 = 0.74100 (with 2 significant digits)

9. Computer Representation of Numbers The decimal equivalent of the binary number represented in Figure 1 is given by: -(026 + 025 + 024 + 123 + 022 + 121 + 120) = -(0 + 0 + 0 + 8 + 0 + 2 + 1) = -11 • Figure 1 Binary representation of an integer using an • 8 bit word (or Byte)

10. Example Determine the largest integer that can be represented by an 8 bit machine. Solution: Imax = +(126 + 125 + 124 + 123 + 122 + 121 + 120) = +( 64 + 32 + 16 + 8 + 4 + 2 + 1 ) = +(127) = +(27 - 1) In general: Imax = +[2(n -1) - 1]; Imin = -[2(n -1) - 1] For a binary computer utilizing 32 bit words, Imax= 2,147,483,647

11. Floating-Point Representation A floating-point number is written as: x = (sign)m.b(sign)E where m is the mantissa, b is the base (b = 2 for a binary system), and E is the exponent.

12. Example Determine the smallest, positive, nonzero, floating point number that can be represented by an eight bit machine using binary system with one bit spared for the sign of the mantissa, one bit for the sign of the exponent, and two bits for the digits of the exponent: Solution: m = +(023 + 022 + 021 + 120) m = +( 0 + 0 + 0 + 1 ) = 1 E = -[(121) + (120)] = -(2+1) = - 3 Number = 12-3 (which is equal to 0.1250 in decimal system)

13. Decimal Numeral System • Base 10 positional notation • 1010 = 1*103 + 0*102 + 1*101 + 0*100 = 1*1000+ 0*100 + 1*10 + 0*1 = 1010 • Number set : (0, 1, 2, …, 8, 9) • Operations: addition, multiplication, etc. • Numbers of same value are longer than Hex but shorter than Bin • Not so easy to convert into Bin or Hex

14. Binary Numeral System • Base 2 positional notation • 1010 (B) = 1*23 + 0*22 + 1*21 + 0*20 = 1*8 + 0*4 + 1*2 + 0*1 = 10 (in decimal) • Number set : (0 & 1) • Operations: addition, multiplication, etc. • Addition: 1001 + 1100 = 10101 • Good for computer systems – logical gates with only two different values or states; Can be easily converted into Hex (4 Bin bits 1 Hex bit)

15. Hexadecimal Numeral System • Base 16 positional notation • 1010 (H) = 1*163 + 0*162 + 1*161 + 0*160 = 1 *4096 + 0* 256 + 1* 16 + 0 * 1 = 4112 (in decimal) • Number set : (0 ~ 9, A, B, C, D, E, F) • Operations: addition, multiplication, etc. • Numbers of same value are shorter (vs. binary and decimal) • Can be easily converted into Binary (1 Hex bit  4 Bin bits)

16. Binary  HexadecimalDecimal