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:
2C6B.F216 = 2 C 6 B. F 2 = 2(=0010) C(=1100) 6(=0110) B(=1011). F(=1111) 2(=0010) = 10110001101011.11110012
the Final answer: 2C6B.F216 = 10110001101011.11110012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+12∙162+6∙161+11∙160+15∙16-1+2∙16-2 = 2∙4096+12∙256+6∙16+11∙1+15∙0.0625+2∙0.00390625 = 8192+3072+96+11+0.9375+0.0078125 = 11371.945312510
got It: 2C6B.F216 =11371.945312510
Translate the number 11371.945312510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
11371 | 2 | | | | | | | | | | | | | |
-11370 | 5685 | 2 | | | | | | | | | | | | |
1 | -5684 | 2842 | 2 | | | | | | | | | | | |
| 1 | -2842 | 1421 | 2 | | | | | | | | | | |
| | 0 | -1420 | 710 | 2 | | | | | | | | | |
| | | 1 | -710 | 355 | 2 | | | | | | | | |
| | | | 0 | -354 | 177 | 2 | | | | | | | |
| | | | | 1 | -176 | 88 | 2 | | | | | | |
| | | | | | 1 | -88 | 44 | 2 | | | | | |
| | | | | | | 0 | -44 | 22 | 2 | | | | |
| | | | | | | | 0 | -22 | 11 | 2 | | | |
| | | | | | | | | 0 | -10 | 5 | 2 | | |
| | | | | | | | | | 1 | -4 | 2 | 2 | |
| | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | 0 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 9453125*2 |
1 | .89063*2 |
1 | .78125*2 |
1 | .5625*2 |
1 | .125*2 |
0 | .25*2 |
0 | .5*2 |
1 | .0*2 |
the result of the conversion was:
11371.945312510 = 10110001101011.11110012
the Final answer: 2C6B.F216 = 10110001101011.11110012