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 hexadecimal to binary like this:
2AC5.D16 = 2 A C 5. D = 2(=0010) A(=1010) C(=1100) 5(=0101). D(=1101) = 10101011000101.11012
the Final answer: 2AC5.D16 = 10101011000101.11012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+10∙162+12∙161+5∙160+13∙16-1 = 2∙4096+10∙256+12∙16+5∙1+13∙0.0625 = 8192+2560+192+5+0.8125 = 10949.812510
got It: 2AC5.D16 =10949.812510
Translate the number 10949.812510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
10949 | 2 | | | | | | | | | | | | | |
-10948 | 5474 | 2 | | | | | | | | | | | | |
1 | -5474 | 2737 | 2 | | | | | | | | | | | |
| 0 | -2736 | 1368 | 2 | | | | | | | | | | |
| | 1 | -1368 | 684 | 2 | | | | | | | | | |
| | | 0 | -684 | 342 | 2 | | | | | | | | |
| | | | 0 | -342 | 171 | 2 | | | | | | | |
| | | | | 0 | -170 | 85 | 2 | | | | | | |
| | | | | | 1 | -84 | 42 | 2 | | | | | |
| | | | | | | 1 | -42 | 21 | 2 | | | | |
| | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | 1 | -10 | 5 | 2 | | |
| | | | | | | | | | 0 | -4 | 2 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 0 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 8125*2 |
1 | .625*2 |
1 | .25*2 |
0 | .5*2 |
1 | .0*2 |
the result of the conversion was:
10949.812510 = 10101011000101.11012
the Final answer: 2AC5.D16 = 10101011000101.11012