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:
4Fc7.b516 = 4 F c 7. b 5 = 4(=0100) F(=1111) c(=1100) 7(=0111). b(=1011) 5(=0101) = 100111111000111.101101012
answer: 4Fc7.b516 = 100111111000111.101101012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
4∙163+15∙162+12∙161+7∙160+11∙16-1+5∙16-2 = 4∙4096+15∙256+12∙16+7∙1+11∙0.0625+5∙0.00390625 = 16384+3840+192+7+0.6875+0.01953125 = 20423.7070312510
got It: 4Fc7.b516 =20423.7070312510
Translate the number 20423.7070312510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
20423 | 2 | | | | | | | | | | | | | | |
-20422 | 10211 | 2 | | | | | | | | | | | | | |
1 | -10210 | 5105 | 2 | | | | | | | | | | | | |
| 1 | -5104 | 2552 | 2 | | | | | | | | | | | |
| | 1 | -2552 | 1276 | 2 | | | | | | | | | | |
| | | 0 | -1276 | 638 | 2 | | | | | | | | | |
| | | | 0 | -638 | 319 | 2 | | | | | | | | |
| | | | | 0 | -318 | 159 | 2 | | | | | | | |
| | | | | | 1 | -158 | 79 | 2 | | | | | | |
| | | | | | | 1 | -78 | 39 | 2 | | | | | |
| | | | | | | | 1 | -38 | 19 | 2 | | | | |
| | | | | | | | | 1 | -18 | 9 | 2 | | | |
| | | | | | | | | | 1 | -8 | 4 | 2 | | |
| | | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | | 0 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 70703125*2 |
1 | .41406*2 |
0 | .82813*2 |
1 | .65625*2 |
1 | .3125*2 |
0 | .625*2 |
1 | .25*2 |
0 | .5*2 |
1 | .0*2 |
the result of the conversion was:
20423.7070312510 = 100111111000111.101101012
answer: 4Fc7.b516 = 100111111000111.101101012