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Base conversions

Base conversions. (129) 10 = ( ? ) 6. 129 = Q 1 6 + R 1. Q 1 = 21 R 1 = 3 LSB. 21 = Q 2 6 + R 2. Q 2 = 3 R 2 = 3 LSB+1. 3 = Q 3 6 + R 3. Q 3 = 0 R 3 = 3 LSB+2. Base conversions. Stop when quotient becomes 0 ( 129 ) 10 = ( 323 ) 6 Double-check:

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Base conversions

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  1. Base conversions (129)10 = ( ? )6 129 = Q16 + R1 Q1 = 21 R1 = 3 LSB 21 = Q26 + R2 Q2 = 3 R2 = 3 LSB+1 3 = Q36 + R3 Q3 = 0 R3 = 3 LSB+2

  2. Base conversions • Stop when quotient becomes 0 • ( 129 )10 = ( 323 )6 • Double-check: • 363 + 63 + 3 = 129

  3. Converting fractions • ( 0.125 )10 = ( ? )2 f = 0.1252 = 0.250 I = 0 b-1 = 0 f = 0.2502 = 0.500 I = 0 b-2 = 0 f = 0.5002 = 1.000 I = 1 b-3 = 1 f = 0 algorithm stops( 0.125 )10 = ( 0.001 )2

  4. Converting fractions • ( 0.6 )10 = ( ? )2 f = 0.62 = 1.2 I = 1 b-1 = 1 f = 0.22 = 0.4 I = 0 b-2 = 0 f = 0.42 = 0.8 I = 0 b-3 = 0 f = 0.82 = 1.6 I = 1 b-4 = 1 f = 0.62 = 1.2 I = 1 b-5 = 1 algorithm loops( 0.6 )10 = ( 0.{1001} )2

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