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 | 0 | 1 | 0 | 1 | twos-complement |
| | | | | | - | 1 | -1 bit |
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | ones complement |
| 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | direct code |
got It:10101011
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 binary to hexadecimal like this:
101010112 = 1010 1011 = 1010(=A) 1011(=B) = AB16
the Final answer: 101010112 = AB16
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
1∙27+0∙26+1∙25+0∙24+1∙23+0∙22+1∙21+1∙20 = 1∙128+0∙64+1∙32+0∙16+1∙8+0∙4+1∙2+1∙1 = 128+0+32+0+8+0+2+1 = 17110
got It: 101010112 =17110
Translate the number 17110 в hexadecimal like this:
the Integer part of the number is divided by the base of the new number system:
| 171 | 16 | |
| -160 | A | |
| B | | |
 |
the result of the conversion was:
17110 = AB16
the Final answer: 101010112 = AB16