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Published byXavier Moser Modified over 3 years ago

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Numbering Systems i = R n -1 * r n-1 + R n -2 * r n-2 + ….. + R 2 * r 2 + R 1 * r 1 + R 0 * r 0, integer f = R -1 * r -1 + R -2 * r -2 + ….. + R -m * r -m, fraction 0 <= r i < R. R = 1 unary (Roman numerals), 12 10 = 111111111111 1 R = 2 binary (computers), 12 10 = 1100 2 = 1100 R = 8 octal 12 10 = 14 8 R = 10 decimal R = 16 hexadecimal 12 10 = C 16 (hex digits are: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F) Number does not change a value by padding it with 0s to the left for integer or to the right for fraction, e.g.: 29 10 = 0029 10 = 11101 = 011101, or.3515625 10 =.0101101 =.010110100 If R i = R k then going from R to R i is by grouping k R digits, e.g. from binary to octal group bits by 3 (from the right for integer or from the left for fraction), or to hexadecimal group bits by 4: 29 10 = 011101 = 35 8 = 00011101 = 1D 16 or.3515625 10 =.010110100 =.264 8 =.01011010 =.5A 16. Since integer and fraction part are independent in any system, mixed number can be converted by converting integer and fraction separately and adding them back together, e.g.: 29.3515625 10 = 11101.0101101 = 35.264 8 = 1D.5A 16

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i = 2 n -1 * r n-1 + 2 n -2 * r n-2 + ….. + 2 2 * r 2 + 2 1 * r 1 + 2 0 * r 0, integer f = 2 -1 * r -1 + 2 -2 * r -2 + ….. + 2 -m * r -m, fraction 0 <= r i < 2. Binary to decimal use the above formulas. Decimal to binary use the following: for integer successive division by 2 and taking the remainder until quotient = 0, e.g.: 29 :2 and write bits bottom up: 11101. For fractions successive multiply by 2 and 14 1 LSB 07 0 03 1 01 1 00 1 MSB Binary to Decimal and vice versa take integer, e.g.: x2 3515625 and read bits from the top,.3515625 10 = MSB 0 703125.0101101 1 40625 0 8125 1 625 1 25 0 5 LSB 1 0

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Negative Numbers (twos complement arithmetic) k = - 2 n * s + 2 n -1 * b n-1 + 2 n -2 * b n-2 + ….. + 2 2 * b 2 + 2 1 * b 1 + 2 0 * b 0 Sign extension (padding by s) to the left does not change the value: e.g.: -4 = 1100 = 11100 = 11111100, +4 = 0100 = 00100 = 00000100. k c = (k complement) > 0 if s=1 the number is negative: if s=0 the number is positive: k > 0 Addition is regular if both s = 0 or s = 1. However, subtraction reduces to addition: i - k = i + (- 2 n + k c ) = - 2 n + (i + k c ) Examples: i = 01010 = 10 i = 01010 = 10 + j = 10010 = -14 + j = 11100 = - 4 11100 = - 4 00110 = 6 Overflow: if carry into sign bit is different than carry out of sign bit. Examples: i = 10110 = -10 i = 01010 = 10 + j = 10010 = -14 + j = 01110 = 14 01000 11000 carry10 carry01

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ADDING INTEGERS 1. POS. + POS. = POS. 2. NEG. + NEG. = NEG. 3. POS. + NEG. OR NEG. + POS. SUBTRACT TAKE SIGN OF BIGGER ABSOLUTE VALUE.

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