-0.000 000 000 000 028 421 709 399 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 028 421 709 399(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 028 421 709 399(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 028 421 709 399| = 0.000 000 000 000 028 421 709 399


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 028 421 709 399.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 028 421 709 399 × 2 = 0 + 0.000 000 000 000 056 843 418 798;
  • 2) 0.000 000 000 000 056 843 418 798 × 2 = 0 + 0.000 000 000 000 113 686 837 596;
  • 3) 0.000 000 000 000 113 686 837 596 × 2 = 0 + 0.000 000 000 000 227 373 675 192;
  • 4) 0.000 000 000 000 227 373 675 192 × 2 = 0 + 0.000 000 000 000 454 747 350 384;
  • 5) 0.000 000 000 000 454 747 350 384 × 2 = 0 + 0.000 000 000 000 909 494 700 768;
  • 6) 0.000 000 000 000 909 494 700 768 × 2 = 0 + 0.000 000 000 001 818 989 401 536;
  • 7) 0.000 000 000 001 818 989 401 536 × 2 = 0 + 0.000 000 000 003 637 978 803 072;
  • 8) 0.000 000 000 003 637 978 803 072 × 2 = 0 + 0.000 000 000 007 275 957 606 144;
  • 9) 0.000 000 000 007 275 957 606 144 × 2 = 0 + 0.000 000 000 014 551 915 212 288;
  • 10) 0.000 000 000 014 551 915 212 288 × 2 = 0 + 0.000 000 000 029 103 830 424 576;
  • 11) 0.000 000 000 029 103 830 424 576 × 2 = 0 + 0.000 000 000 058 207 660 849 152;
  • 12) 0.000 000 000 058 207 660 849 152 × 2 = 0 + 0.000 000 000 116 415 321 698 304;
  • 13) 0.000 000 000 116 415 321 698 304 × 2 = 0 + 0.000 000 000 232 830 643 396 608;
  • 14) 0.000 000 000 232 830 643 396 608 × 2 = 0 + 0.000 000 000 465 661 286 793 216;
  • 15) 0.000 000 000 465 661 286 793 216 × 2 = 0 + 0.000 000 000 931 322 573 586 432;
  • 16) 0.000 000 000 931 322 573 586 432 × 2 = 0 + 0.000 000 001 862 645 147 172 864;
  • 17) 0.000 000 001 862 645 147 172 864 × 2 = 0 + 0.000 000 003 725 290 294 345 728;
  • 18) 0.000 000 003 725 290 294 345 728 × 2 = 0 + 0.000 000 007 450 580 588 691 456;
  • 19) 0.000 000 007 450 580 588 691 456 × 2 = 0 + 0.000 000 014 901 161 177 382 912;
  • 20) 0.000 000 014 901 161 177 382 912 × 2 = 0 + 0.000 000 029 802 322 354 765 824;
  • 21) 0.000 000 029 802 322 354 765 824 × 2 = 0 + 0.000 000 059 604 644 709 531 648;
  • 22) 0.000 000 059 604 644 709 531 648 × 2 = 0 + 0.000 000 119 209 289 419 063 296;
  • 23) 0.000 000 119 209 289 419 063 296 × 2 = 0 + 0.000 000 238 418 578 838 126 592;
  • 24) 0.000 000 238 418 578 838 126 592 × 2 = 0 + 0.000 000 476 837 157 676 253 184;
  • 25) 0.000 000 476 837 157 676 253 184 × 2 = 0 + 0.000 000 953 674 315 352 506 368;
  • 26) 0.000 000 953 674 315 352 506 368 × 2 = 0 + 0.000 001 907 348 630 705 012 736;
  • 27) 0.000 001 907 348 630 705 012 736 × 2 = 0 + 0.000 003 814 697 261 410 025 472;
  • 28) 0.000 003 814 697 261 410 025 472 × 2 = 0 + 0.000 007 629 394 522 820 050 944;
  • 29) 0.000 007 629 394 522 820 050 944 × 2 = 0 + 0.000 015 258 789 045 640 101 888;
  • 30) 0.000 015 258 789 045 640 101 888 × 2 = 0 + 0.000 030 517 578 091 280 203 776;
  • 31) 0.000 030 517 578 091 280 203 776 × 2 = 0 + 0.000 061 035 156 182 560 407 552;
  • 32) 0.000 061 035 156 182 560 407 552 × 2 = 0 + 0.000 122 070 312 365 120 815 104;
  • 33) 0.000 122 070 312 365 120 815 104 × 2 = 0 + 0.000 244 140 624 730 241 630 208;
  • 34) 0.000 244 140 624 730 241 630 208 × 2 = 0 + 0.000 488 281 249 460 483 260 416;
  • 35) 0.000 488 281 249 460 483 260 416 × 2 = 0 + 0.000 976 562 498 920 966 520 832;
  • 36) 0.000 976 562 498 920 966 520 832 × 2 = 0 + 0.001 953 124 997 841 933 041 664;
  • 37) 0.001 953 124 997 841 933 041 664 × 2 = 0 + 0.003 906 249 995 683 866 083 328;
  • 38) 0.003 906 249 995 683 866 083 328 × 2 = 0 + 0.007 812 499 991 367 732 166 656;
  • 39) 0.007 812 499 991 367 732 166 656 × 2 = 0 + 0.015 624 999 982 735 464 333 312;
  • 40) 0.015 624 999 982 735 464 333 312 × 2 = 0 + 0.031 249 999 965 470 928 666 624;
  • 41) 0.031 249 999 965 470 928 666 624 × 2 = 0 + 0.062 499 999 930 941 857 333 248;
  • 42) 0.062 499 999 930 941 857 333 248 × 2 = 0 + 0.124 999 999 861 883 714 666 496;
  • 43) 0.124 999 999 861 883 714 666 496 × 2 = 0 + 0.249 999 999 723 767 429 332 992;
  • 44) 0.249 999 999 723 767 429 332 992 × 2 = 0 + 0.499 999 999 447 534 858 665 984;
  • 45) 0.499 999 999 447 534 858 665 984 × 2 = 0 + 0.999 999 998 895 069 717 331 968;
  • 46) 0.999 999 998 895 069 717 331 968 × 2 = 1 + 0.999 999 997 790 139 434 663 936;
  • 47) 0.999 999 997 790 139 434 663 936 × 2 = 1 + 0.999 999 995 580 278 869 327 872;
  • 48) 0.999 999 995 580 278 869 327 872 × 2 = 1 + 0.999 999 991 160 557 738 655 744;
  • 49) 0.999 999 991 160 557 738 655 744 × 2 = 1 + 0.999 999 982 321 115 477 311 488;
  • 50) 0.999 999 982 321 115 477 311 488 × 2 = 1 + 0.999 999 964 642 230 954 622 976;
  • 51) 0.999 999 964 642 230 954 622 976 × 2 = 1 + 0.999 999 929 284 461 909 245 952;
  • 52) 0.999 999 929 284 461 909 245 952 × 2 = 1 + 0.999 999 858 568 923 818 491 904;
  • 53) 0.999 999 858 568 923 818 491 904 × 2 = 1 + 0.999 999 717 137 847 636 983 808;
  • 54) 0.999 999 717 137 847 636 983 808 × 2 = 1 + 0.999 999 434 275 695 273 967 616;
  • 55) 0.999 999 434 275 695 273 967 616 × 2 = 1 + 0.999 998 868 551 390 547 935 232;
  • 56) 0.999 998 868 551 390 547 935 232 × 2 = 1 + 0.999 997 737 102 781 095 870 464;
  • 57) 0.999 997 737 102 781 095 870 464 × 2 = 1 + 0.999 995 474 205 562 191 740 928;
  • 58) 0.999 995 474 205 562 191 740 928 × 2 = 1 + 0.999 990 948 411 124 383 481 856;
  • 59) 0.999 990 948 411 124 383 481 856 × 2 = 1 + 0.999 981 896 822 248 766 963 712;
  • 60) 0.999 981 896 822 248 766 963 712 × 2 = 1 + 0.999 963 793 644 497 533 927 424;
  • 61) 0.999 963 793 644 497 533 927 424 × 2 = 1 + 0.999 927 587 288 995 067 854 848;
  • 62) 0.999 927 587 288 995 067 854 848 × 2 = 1 + 0.999 855 174 577 990 135 709 696;
  • 63) 0.999 855 174 577 990 135 709 696 × 2 = 1 + 0.999 710 349 155 980 271 419 392;
  • 64) 0.999 710 349 155 980 271 419 392 × 2 = 1 + 0.999 420 698 311 960 542 838 784;
  • 65) 0.999 420 698 311 960 542 838 784 × 2 = 1 + 0.998 841 396 623 921 085 677 568;
  • 66) 0.998 841 396 623 921 085 677 568 × 2 = 1 + 0.997 682 793 247 842 171 355 136;
  • 67) 0.997 682 793 247 842 171 355 136 × 2 = 1 + 0.995 365 586 495 684 342 710 272;
  • 68) 0.995 365 586 495 684 342 710 272 × 2 = 1 + 0.990 731 172 991 368 685 420 544;
  • 69) 0.990 731 172 991 368 685 420 544 × 2 = 1 + 0.981 462 345 982 737 370 841 088;
  • 70) 0.981 462 345 982 737 370 841 088 × 2 = 1 + 0.962 924 691 965 474 741 682 176;
  • 71) 0.962 924 691 965 474 741 682 176 × 2 = 1 + 0.925 849 383 930 949 483 364 352;
  • 72) 0.925 849 383 930 949 483 364 352 × 2 = 1 + 0.851 698 767 861 898 966 728 704;
  • 73) 0.851 698 767 861 898 966 728 704 × 2 = 1 + 0.703 397 535 723 797 933 457 408;
  • 74) 0.703 397 535 723 797 933 457 408 × 2 = 1 + 0.406 795 071 447 595 866 914 816;
  • 75) 0.406 795 071 447 595 866 914 816 × 2 = 0 + 0.813 590 142 895 191 733 829 632;
  • 76) 0.813 590 142 895 191 733 829 632 × 2 = 1 + 0.627 180 285 790 383 467 659 264;
  • 77) 0.627 180 285 790 383 467 659 264 × 2 = 1 + 0.254 360 571 580 766 935 318 528;
  • 78) 0.254 360 571 580 766 935 318 528 × 2 = 0 + 0.508 721 143 161 533 870 637 056;
  • 79) 0.508 721 143 161 533 870 637 056 × 2 = 1 + 0.017 442 286 323 067 741 274 112;
  • 80) 0.017 442 286 323 067 741 274 112 × 2 = 0 + 0.034 884 572 646 135 482 548 224;
  • 81) 0.034 884 572 646 135 482 548 224 × 2 = 0 + 0.069 769 145 292 270 965 096 448;
  • 82) 0.069 769 145 292 270 965 096 448 × 2 = 0 + 0.139 538 290 584 541 930 192 896;
  • 83) 0.139 538 290 584 541 930 192 896 × 2 = 0 + 0.279 076 581 169 083 860 385 792;
  • 84) 0.279 076 581 169 083 860 385 792 × 2 = 0 + 0.558 153 162 338 167 720 771 584;
  • 85) 0.558 153 162 338 167 720 771 584 × 2 = 1 + 0.116 306 324 676 335 441 543 168;
  • 86) 0.116 306 324 676 335 441 543 168 × 2 = 0 + 0.232 612 649 352 670 883 086 336;
  • 87) 0.232 612 649 352 670 883 086 336 × 2 = 0 + 0.465 225 298 705 341 766 172 672;
  • 88) 0.465 225 298 705 341 766 172 672 × 2 = 0 + 0.930 450 597 410 683 532 345 344;
  • 89) 0.930 450 597 410 683 532 345 344 × 2 = 1 + 0.860 901 194 821 367 064 690 688;
  • 90) 0.860 901 194 821 367 064 690 688 × 2 = 1 + 0.721 802 389 642 734 129 381 376;
  • 91) 0.721 802 389 642 734 129 381 376 × 2 = 1 + 0.443 604 779 285 468 258 762 752;
  • 92) 0.443 604 779 285 468 258 762 752 × 2 = 0 + 0.887 209 558 570 936 517 525 504;
  • 93) 0.887 209 558 570 936 517 525 504 × 2 = 1 + 0.774 419 117 141 873 035 051 008;
  • 94) 0.774 419 117 141 873 035 051 008 × 2 = 1 + 0.548 838 234 283 746 070 102 016;
  • 95) 0.548 838 234 283 746 070 102 016 × 2 = 1 + 0.097 676 468 567 492 140 204 032;
  • 96) 0.097 676 468 567 492 140 204 032 × 2 = 0 + 0.195 352 937 134 984 280 408 064;
  • 97) 0.195 352 937 134 984 280 408 064 × 2 = 0 + 0.390 705 874 269 968 560 816 128;
  • 98) 0.390 705 874 269 968 560 816 128 × 2 = 0 + 0.781 411 748 539 937 121 632 256;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 028 421 709 399(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1101 1010 0000 1000 1110 1110 00(2)

6. Positive number before normalization:

0.000 000 000 000 028 421 709 399(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1101 1010 0000 1000 1110 1110 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 028 421 709 399(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1101 1010 0000 1000 1110 1110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1101 1010 0000 1000 1110 1110 00(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 0110 1000 0010 0011 1011 1000(2) × 2-46


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -46

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 0110 1000 0010 0011 1011 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

977(10) =

011 1101 0001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1111 1111 1111 1111 1111 1111 1111 0110 1000 0010 0011 1011 1000 =

1111 1111 1111 1111 1111 1111 1111 0110 1000 0010 0011 1011 1000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1101 0001

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1111 0110 1000 0010 0011 1011 1000


Decimal number -0.000 000 000 000 028 421 709 399 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 1111 1111 1111 1111 1111 1111 1111 0110 1000 0010 0011 1011 1000

» Convert decimal -0.000 000 000 000 028 421 709 332 to 64 bit double precision IEEE 754 binary floating point representation standard

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100