This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s do a direct translation from octal to binary like this:
4635.078 = 4 6 3 5. 0 7 = 4(=100) 6(=110) 3(=011) 5(=101). 0(=000) 7(=111) = 100110011101.0001112
the Final answer: 4635.078 = 100110011101.0001112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
4∙83+6∙82+3∙81+5∙80+0∙8-1+7∙8-2 = 4∙512+6∙64+3∙8+5∙1+0∙0.125+7∙0.015625 = 2048+384+24+5+0+0.109375 = 2461.10937510
got It: 4635.078 =2461.10937510
Translate the number 2461.10937510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
2461 | 2 | | | | | | | | | | | |
-2460 | 1230 | 2 | | | | | | | | | | |
1 | -1230 | 615 | 2 | | | | | | | | | |
| 0 | -614 | 307 | 2 | | | | | | | | |
| | 1 | -306 | 153 | 2 | | | | | | | |
| | | 1 | -152 | 76 | 2 | | | | | | |
| | | | 1 | -76 | 38 | 2 | | | | | |
| | | | | 0 | -38 | 19 | 2 | | | | |
| | | | | | 0 | -18 | 9 | 2 | | | |
| | | | | | | 1 | -8 | 4 | 2 | | |
| | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | 0 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 109375*2 |
0 | .21875*2 |
0 | .4375*2 |
0 | .875*2 |
1 | .75*2 |
1 | .5*2 |
1 | .0*2 |
the result of the conversion was:
2461.10937510 = 100110011101.0001112
the Final answer: 4635.078 = 100110011101.0001112