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:
AD92.BCA16 = A D 9 2. B C A = A(=1010) D(=1101) 9(=1001) 2(=0010). B(=1011) C(=1100) A(=1010) = 1010110110010010.101111001012
the Final answer: AD92.BCA16 = 1010110110010010.101111001012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙163+13∙162+9∙161+2∙160+11∙16-1+12∙16-2+10∙16-3 = 10∙4096+13∙256+9∙16+2∙1+11∙0.0625+12∙0.00390625+10∙0.000244140625 = 40960+3328+144+2+0.6875+0.046875+0.00244140625 = 44434.7368164062510
got It: AD92.BCA16 =44434.7368164062510
Translate the number 44434.7368164062510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
44434 | 2 | | | | | | | | | | | | | | | |
-44434 | 22217 | 2 | | | | | | | | | | | | | | |
0 | -22216 | 11108 | 2 | | | | | | | | | | | | | |
| 1 | -11108 | 5554 | 2 | | | | | | | | | | | | |
| | 0 | -5554 | 2777 | 2 | | | | | | | | | | | |
| | | 0 | -2776 | 1388 | 2 | | | | | | | | | | |
| | | | 1 | -1388 | 694 | 2 | | | | | | | | | |
| | | | | 0 | -694 | 347 | 2 | | | | | | | | |
| | | | | | 0 | -346 | 173 | 2 | | | | | | | |
| | | | | | | 1 | -172 | 86 | 2 | | | | | | |
| | | | | | | | 1 | -86 | 43 | 2 | | | | | |
| | | | | | | | | 0 | -42 | 21 | 2 | | | | |
| | | | | | | | | | 1 | -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. | 73681640625*2 |
1 | .47363*2 |
0 | .94727*2 |
1 | .89453*2 |
1 | .78906*2 |
1 | .57813*2 |
1 | .15625*2 |
0 | .3125*2 |
0 | .625*2 |
1 | .25*2 |
0 | .5*2 |
the result of the conversion was:
44434.7368164062510 = 1010110110010010.10111100102
the Final answer: AD92.BCA16 = 1010110110010010.10111100102