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:
34C9.2A16 = 3 4 C 9. 2 A = 3(=0011) 4(=0100) C(=1100) 9(=1001). 2(=0010) A(=1010) = 11010011001001.00101012
the Final answer: 34C9.2A16 = 11010011001001.00101012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
3∙163+4∙162+12∙161+9∙160+2∙16-1+10∙16-2 = 3∙4096+4∙256+12∙16+9∙1+2∙0.0625+10∙0.00390625 = 12288+1024+192+9+0.125+0.0390625 = 13513.164062510
got It: 34C9.2A16 =13513.164062510
Translate the number 13513.164062510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
13513 | 2 | | | | | | | | | | | | | |
-13512 | 6756 | 2 | | | | | | | | | | | | |
1 | -6756 | 3378 | 2 | | | | | | | | | | | |
| 0 | -3378 | 1689 | 2 | | | | | | | | | | |
| | 0 | -1688 | 844 | 2 | | | | | | | | | |
| | | 1 | -844 | 422 | 2 | | | | | | | | |
| | | | 0 | -422 | 211 | 2 | | | | | | | |
| | | | | 0 | -210 | 105 | 2 | | | | | | |
| | | | | | 1 | -104 | 52 | 2 | | | | | |
| | | | | | | 1 | -52 | 26 | 2 | | | | |
| | | | | | | | 0 | -26 | 13 | 2 | | | |
| | | | | | | | | 0 | -12 | 6 | 2 | | |
| | | | | | | | | | 1 | -6 | 3 | 2 | |
| | | | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | | | 1 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 1640625*2 |
0 | .32813*2 |
0 | .65625*2 |
1 | .3125*2 |
0 | .625*2 |
1 | .25*2 |
0 | .5*2 |
1 | .0*2 |
the result of the conversion was:
13513.164062510 = 11010011001001.00101012
the Final answer: 34C9.2A16 = 11010011001001.00101012