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:
1001.10018 = 1 0 0 1. 1 0 0 1 = 1(=001) 0(=000) 0(=000) 1(=001). 1(=001) 0(=000) 0(=000) 1(=001) = 001000000001.0010000000012
answer: 1001.10018 = 1000000001.0010000000012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙83+0∙82+0∙81+1∙80+1∙8-1+0∙8-2+0∙8-3+1∙8-4 = 1∙512+0∙64+0∙8+1∙1+1∙0.125+0∙0.015625+0∙0.001953125+1∙0.000244140625 = 512+0+0+1+0.125+0+0+0.000244140625 = 513.12524414062510
got It: 1001.10018 =513.12524414062510
Translate the number 513.12524414062510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
513 | 2 | | | | | | | | | |
-512 | 256 | 2 | | | | | | | | |
1 | -256 | 128 | 2 | | | | | | | |
| 0 | -128 | 64 | 2 | | | | | | |
| | 0 | -64 | 32 | 2 | | | | | |
| | | 0 | -32 | 16 | 2 | | | | |
| | | | 0 | -16 | 8 | 2 | | | |
| | | | | 0 | -8 | 4 | 2 | | |
| | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | 0 | -2 | 1 | |
| | | | | | | | 0 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 125244140625*2 |
0 | .25049*2 |
0 | .50098*2 |
1 | .00195*2 |
0 | .00391*2 |
0 | .00781*2 |
0 | .01563*2 |
0 | .03125*2 |
0 | .0625*2 |
0 | .125*2 |
0 | .25*2 |
the result of the conversion was:
513.12524414062510 = 1000000001.00100000002
answer: 1001.10018 = 1000000001.00100000002