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:
F516.16 = F 5 1 6. = F(=1111) 5(=0101) 1(=0001) 6(=0110). = 1111010100010110.2
answer: F516.16 = 1111010100010110.2
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
15∙163+5∙162+1∙161+6∙160 = 15∙4096+5∙256+1∙16+6∙1 = 61440+1280+16+6 = 62742.10
got It: F516.16 =62742.10
Translate the number 62742.10 в binary like this:
the Integer part of the number is divided by the base of the new number system:
62742 | 2 | | | | | | | | | | | | | | | |
-62742 | 31371 | 2 | | | | | | | | | | | | | | |
0 | -31370 | 15685 | 2 | | | | | | | | | | | | | |
| 1 | -15684 | 7842 | 2 | | | | | | | | | | | | |
| | 1 | -7842 | 3921 | 2 | | | | | | | | | | | |
| | | 0 | -3920 | 1960 | 2 | | | | | | | | | | |
| | | | 1 | -1960 | 980 | 2 | | | | | | | | | |
| | | | | 0 | -980 | 490 | 2 | | | | | | | | |
| | | | | | 0 | -490 | 245 | 2 | | | | | | | |
| | | | | | | 0 | -244 | 122 | 2 | | | | | | |
| | | | | | | | 1 | -122 | 61 | 2 | | | | | |
| | | | | | | | | 0 | -60 | 30 | 2 | | | | |
| | | | | | | | | | 1 | -30 | 15 | 2 | | | |
| | | | | | | | | | | 0 | -14 | 7 | 2 | | |
| | | | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | | | 1 | -2 | 1 | |
| | | | | | | | | | | | | | 1 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | *2 |
the result of the conversion was:
62742.10 = 1111010100010110.2
answer: F516.16 = 1111010100010110.2