This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
Fill in the number with missing zeros on the right
let\'s make a direct translation from binary to post-binary like this:
110001.1101002 = 110 001. 110 100 = 110(=6) 001(=1). 110(=6) 100(=4) = 61.648
answer: 110001.11012 = 61.648
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙25+1∙24+0∙23+0∙22+0∙21+1∙20+1∙2-1+1∙2-2+0∙2-3+1∙2-4+0∙2-5+0∙2-6 = 1∙32+1∙16+0∙8+0∙4+0∙2+1∙1+1∙0.5+1∙0.25+0∙0.125+1∙0.0625+0∙0.03125+0∙0.015625 = 32+16+0+0+0+1+0.5+0.25+0+0.0625+0+0 = 49.812510
got It: 110001.1101002 =49.812510
Translate the number 49.812510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
49 | 8 | |
-48 | 6 | |
1 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 8125*8 |
6 | .5*8 |
4 | .0*8 |
the result of the conversion was:
49.812510 = 61.648
answer: 110001.11012 = 61.648