CSE 241 Lecture Notes - Lecture 2: Number One Crossbar Switching System, Decimal Mark

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CSE 241
Recitation 2
Example: Converting Arbitrary Bases to Decimal
1.3a (4310)5 -> (?)10
4 x 53 + 3 x 52 + 1 x 51 + 0 x 50
500 + 75 + 5 + 0 = (580)10
(435)6 -> (?)10
4 x 62 + 3 x 61 + 5 x 60
144 + 18 + 5 = (167)10
Example: Determine the base of the numbers such that an equation is true.
(24)b + (17)b = (40)b
(2xb + 4x1) + (1xb + 7x1) = (4xb + 0x1)
2b + 4 + b + 7 = 4b
3b + 11 = 4b
11 = b
Check (convert back to decimal and make sure that it still makes sense):
(22+4) + (11+7) = (44)
26 + 18 = 44
44 = 44
x2 (b + 1)x + (2b + 2) = 0
x = (3)b x = (6)b
In base b: 6 + 3 = 11, 6 x 3 = 22
6 + 3 = 1b + 1 6 x 3 = 2b + 2
8 = b 16 = 2b
8 = b
Example: Converting Decimal Numbers to Binary
(1.125)10 Deal with this in two segments on either side of the decimal point.
(1)10 = (1)2
2 x 0.125
2 x 0.250 0
2 x 0.500 0
2 x 0.000 1 (0.125)10 = (0.001)2
(1.001)2
(123)10
2 | 123
2 | 61 1
2 | 30 1
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