You specified that your number is in the additional code. For further conversion, you need to get a direct number code. Therefore, let\'s perform the conversion from additional code to direct code.
to do this, first perform the conversion from the additional code to the reverse by subtracting 1 bit, then get the direct code by inverting all the bits except the signed one.
| | | | | | | | | |
| 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | дополнительный код |
| | | | | | - | 1 | -1 bit |
| 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | обратный код |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | прямой код |
got It:10100101
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 left
let\'s make a direct translation from binary to post-binary like this:
0101001012 = 010 100 101 = 010(=2) 100(=4) 101(=5) = 2458
answer: 110110112 = 2458
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
0∙28+1∙27+0∙26+1∙25+0∙24+0∙23+1∙22+0∙21+1∙20 = 0∙256+1∙128+0∙64+1∙32+0∙16+0∙8+1∙4+0∙2+1∙1 = 0+128+0+32+0+0+4+0+1 = 16510
got It: 0101001012 =16510
Translate the number 16510 в octal like this:
the Integer part of the number is divided by the base of the new number system:
| 165 | 8 | | |
| -160 | 20 | 8 | |
| 5 | -16 | 2 | |
| 4 | | |
 |
the result of the conversion was:
16510 = 2458
answer: 110110112 = 2458