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:
CF3.13B16 = C F 3. 1 3 B = C(=1100) F(=1111) 3(=0011). 1(=0001) 3(=0011) B(=1011) = 110011110011.0001001110112
the Final answer: CF3.13B16 = 110011110011.0001001110112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
12∙162+15∙161+3∙160+1∙16-1+3∙16-2+11∙16-3 = 12∙256+15∙16+3∙1+1∙0.0625+3∙0.00390625+11∙0.000244140625 = 3072+240+3+0.0625+0.01171875+0.002685546875 = 3315.07690429687510
got It: CF3.13B16 =3315.07690429687510
Translate the number 3315.07690429687510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
3315 | 2 | | | | | | | | | | | |
-3314 | 1657 | 2 | | | | | | | | | | |
1 | -1656 | 828 | 2 | | | | | | | | | |
| 1 | -828 | 414 | 2 | | | | | | | | |
| | 0 | -414 | 207 | 2 | | | | | | | |
| | | 0 | -206 | 103 | 2 | | | | | | |
| | | | 1 | -102 | 51 | 2 | | | | | |
| | | | | 1 | -50 | 25 | 2 | | | | |
| | | | | | 1 | -24 | 12 | 2 | | | |
| | | | | | | 1 | -12 | 6 | 2 | | |
| | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | 1 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 076904296875*2 |
0 | .15381*2 |
0 | .30762*2 |
0 | .61523*2 |
1 | .23047*2 |
0 | .46094*2 |
0 | .92188*2 |
1 | .84375*2 |
1 | .6875*2 |
1 | .375*2 |
0 | .75*2 |
the result of the conversion was:
3315.07690429687510 = 110011110011.00010011102
the Final answer: CF3.13B16 = 110011110011.00010011102