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:
285A.25716 = 2 8 5 A. 2 5 7 = 2(=0010) 8(=1000) 5(=0101) A(=1010). 2(=0010) 5(=0101) 7(=0111) = 10100001011010.0010010101112
the Final answer: 285A.25716 = 10100001011010.0010010101112
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
2∙163+8∙162+5∙161+10∙160+2∙16-1+5∙16-2+7∙16-3 = 2∙4096+8∙256+5∙16+10∙1+2∙0.0625+5∙0.00390625+7∙0.000244140625 = 8192+2048+80+10+0.125+0.01953125+0.001708984375 = 10330.14624023437510
got It: 285A.25716 =10330.14624023437510
Translate the number 10330.14624023437510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
10330 | 2 | | | | | | | | | | | | | |
-10330 | 5165 | 2 | | | | | | | | | | | | |
0 | -5164 | 2582 | 2 | | | | | | | | | | | |
| 1 | -2582 | 1291 | 2 | | | | | | | | | | |
| | 0 | -1290 | 645 | 2 | | | | | | | | | |
| | | 1 | -644 | 322 | 2 | | | | | | | | |
| | | | 1 | -322 | 161 | 2 | | | | | | | |
| | | | | 0 | -160 | 80 | 2 | | | | | | |
| | | | | | 1 | -80 | 40 | 2 | | | | | |
| | | | | | | 0 | -40 | 20 | 2 | | | | |
| | | | | | | | 0 | -20 | 10 | 2 | | | |
| | | | | | | | | 0 | -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. | 146240234375*2 |
0 | .29248*2 |
0 | .58496*2 |
1 | .16992*2 |
0 | .33984*2 |
0 | .67969*2 |
1 | .35938*2 |
0 | .71875*2 |
1 | .4375*2 |
0 | .875*2 |
1 | .75*2 |
the result of the conversion was:
10330.14624023437510 = 10100001011010.00100101012
the Final answer: 285A.25716 = 10100001011010.00100101012