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:
A23.C0916 = A 2 3. C 0 9 = A(=1010) 2(=0010) 3(=0011). C(=1100) 0(=0000) 9(=1001) = 101000100011.1100000010012
the Final answer: A23.C0916 = 101000100011.1100000010012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
10∙162+2∙161+3∙160+12∙16-1+0∙16-2+9∙16-3 = 10∙256+2∙16+3∙1+12∙0.0625+0∙0.00390625+9∙0.000244140625 = 2560+32+3+0.75+0+0.002197265625 = 2595.75219726562510
got It: A23.C0916 =2595.75219726562510
Translate the number 2595.75219726562510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
2595 | 2 | | | | | | | | | | | |
-2594 | 1297 | 2 | | | | | | | | | | |
1 | -1296 | 648 | 2 | | | | | | | | | |
| 1 | -648 | 324 | 2 | | | | | | | | |
| | 0 | -324 | 162 | 2 | | | | | | | |
| | | 0 | -162 | 81 | 2 | | | | | | |
| | | | 0 | -80 | 40 | 2 | | | | | |
| | | | | 1 | -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. | 752197265625*2 |
1 | .50439*2 |
1 | .00879*2 |
0 | .01758*2 |
0 | .03516*2 |
0 | .07031*2 |
0 | .14063*2 |
0 | .28125*2 |
0 | .5625*2 |
1 | .125*2 |
0 | .25*2 |
the result of the conversion was:
2595.75219726562510 = 101000100011.11000000102
the Final answer: A23.C0916 = 101000100011.11000000102