August 2003
Intermediate to advanced
672 pages
16h 23m
English
In order to convert decimals to binaries, we reverse the process outlined in Section G.1 for converting a binary to a decimal.
| 4110 = | 20×2 | + 1 | Dividing 41 by 2, gives the quotient 20 and remainder 1. |
| 2010 = | 10×2 | + 0 | We again divide the current quotient 20 by 2. |
| 1010 = | 5×2 | + 0 | |
| 510 = | 2×2 | + 1 | We repeat this procedure until ... |
| 210 = | 1×2 | + 0 | |
| 110 = | 0×2 | + 1 | ... the quotient is 0. |
| 4110 = | 1010012 | ||
The divisor used in the steps above is the base of the target number system (binary, base 2). The binary value, 1010012, is represented by the remainders, with the last remainder as the left-most bit. Back substitution of the quotient gives the same result:
| 4110 | = (((((0×2 + 1)×2 + 0)×2 + 1)×2 + 0)×2 + 0)×2 + 1 |
| = 1×2 ... |
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