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:
110110.0011002 = 110 110. 001 100 = 110(=6) 110(=6). 001(=1) 100(=4) = 66.148
answer: 110110.001102 = 66.148
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙25+1∙24+0∙23+1∙22+1∙21+0∙20+0∙2-1+0∙2-2+1∙2-3+1∙2-4+0∙2-5+0∙2-6 = 1∙32+1∙16+0∙8+1∙4+1∙2+0∙1+0∙0.5+0∙0.25+1∙0.125+1∙0.0625+0∙0.03125+0∙0.015625 = 32+16+0+4+2+0+0+0+0.125+0.0625+0+0 = 54.187510
got It: 110110.0011002 =54.187510
Translate the number 54.187510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
54 | 8 | |
-48 | 6 | |
6 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 1875*8 |
1 | .5*8 |
4 | .0*8 |
the result of the conversion was:
54.187510 = 66.148
answer: 110110.001102 = 66.148