Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.000 000 000 010 068 460 71 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.000 000 000 010 068 460 71(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 010 068 460 71| = 0.000 000 000 010 068 460 71


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: 0.000 000 000 010 068 460 71.

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) 0.000 000 000 010 068 460 71 × 2 = 0 + 0.000 000 000 020 136 921 42;
  • 2) 0.000 000 000 020 136 921 42 × 2 = 0 + 0.000 000 000 040 273 842 84;
  • 3) 0.000 000 000 040 273 842 84 × 2 = 0 + 0.000 000 000 080 547 685 68;
  • 4) 0.000 000 000 080 547 685 68 × 2 = 0 + 0.000 000 000 161 095 371 36;
  • 5) 0.000 000 000 161 095 371 36 × 2 = 0 + 0.000 000 000 322 190 742 72;
  • 6) 0.000 000 000 322 190 742 72 × 2 = 0 + 0.000 000 000 644 381 485 44;
  • 7) 0.000 000 000 644 381 485 44 × 2 = 0 + 0.000 000 001 288 762 970 88;
  • 8) 0.000 000 001 288 762 970 88 × 2 = 0 + 0.000 000 002 577 525 941 76;
  • 9) 0.000 000 002 577 525 941 76 × 2 = 0 + 0.000 000 005 155 051 883 52;
  • 10) 0.000 000 005 155 051 883 52 × 2 = 0 + 0.000 000 010 310 103 767 04;
  • 11) 0.000 000 010 310 103 767 04 × 2 = 0 + 0.000 000 020 620 207 534 08;
  • 12) 0.000 000 020 620 207 534 08 × 2 = 0 + 0.000 000 041 240 415 068 16;
  • 13) 0.000 000 041 240 415 068 16 × 2 = 0 + 0.000 000 082 480 830 136 32;
  • 14) 0.000 000 082 480 830 136 32 × 2 = 0 + 0.000 000 164 961 660 272 64;
  • 15) 0.000 000 164 961 660 272 64 × 2 = 0 + 0.000 000 329 923 320 545 28;
  • 16) 0.000 000 329 923 320 545 28 × 2 = 0 + 0.000 000 659 846 641 090 56;
  • 17) 0.000 000 659 846 641 090 56 × 2 = 0 + 0.000 001 319 693 282 181 12;
  • 18) 0.000 001 319 693 282 181 12 × 2 = 0 + 0.000 002 639 386 564 362 24;
  • 19) 0.000 002 639 386 564 362 24 × 2 = 0 + 0.000 005 278 773 128 724 48;
  • 20) 0.000 005 278 773 128 724 48 × 2 = 0 + 0.000 010 557 546 257 448 96;
  • 21) 0.000 010 557 546 257 448 96 × 2 = 0 + 0.000 021 115 092 514 897 92;
  • 22) 0.000 021 115 092 514 897 92 × 2 = 0 + 0.000 042 230 185 029 795 84;
  • 23) 0.000 042 230 185 029 795 84 × 2 = 0 + 0.000 084 460 370 059 591 68;
  • 24) 0.000 084 460 370 059 591 68 × 2 = 0 + 0.000 168 920 740 119 183 36;
  • 25) 0.000 168 920 740 119 183 36 × 2 = 0 + 0.000 337 841 480 238 366 72;
  • 26) 0.000 337 841 480 238 366 72 × 2 = 0 + 0.000 675 682 960 476 733 44;
  • 27) 0.000 675 682 960 476 733 44 × 2 = 0 + 0.001 351 365 920 953 466 88;
  • 28) 0.001 351 365 920 953 466 88 × 2 = 0 + 0.002 702 731 841 906 933 76;
  • 29) 0.002 702 731 841 906 933 76 × 2 = 0 + 0.005 405 463 683 813 867 52;
  • 30) 0.005 405 463 683 813 867 52 × 2 = 0 + 0.010 810 927 367 627 735 04;
  • 31) 0.010 810 927 367 627 735 04 × 2 = 0 + 0.021 621 854 735 255 470 08;
  • 32) 0.021 621 854 735 255 470 08 × 2 = 0 + 0.043 243 709 470 510 940 16;
  • 33) 0.043 243 709 470 510 940 16 × 2 = 0 + 0.086 487 418 941 021 880 32;
  • 34) 0.086 487 418 941 021 880 32 × 2 = 0 + 0.172 974 837 882 043 760 64;
  • 35) 0.172 974 837 882 043 760 64 × 2 = 0 + 0.345 949 675 764 087 521 28;
  • 36) 0.345 949 675 764 087 521 28 × 2 = 0 + 0.691 899 351 528 175 042 56;
  • 37) 0.691 899 351 528 175 042 56 × 2 = 1 + 0.383 798 703 056 350 085 12;
  • 38) 0.383 798 703 056 350 085 12 × 2 = 0 + 0.767 597 406 112 700 170 24;
  • 39) 0.767 597 406 112 700 170 24 × 2 = 1 + 0.535 194 812 225 400 340 48;
  • 40) 0.535 194 812 225 400 340 48 × 2 = 1 + 0.070 389 624 450 800 680 96;
  • 41) 0.070 389 624 450 800 680 96 × 2 = 0 + 0.140 779 248 901 601 361 92;
  • 42) 0.140 779 248 901 601 361 92 × 2 = 0 + 0.281 558 497 803 202 723 84;
  • 43) 0.281 558 497 803 202 723 84 × 2 = 0 + 0.563 116 995 606 405 447 68;
  • 44) 0.563 116 995 606 405 447 68 × 2 = 1 + 0.126 233 991 212 810 895 36;
  • 45) 0.126 233 991 212 810 895 36 × 2 = 0 + 0.252 467 982 425 621 790 72;
  • 46) 0.252 467 982 425 621 790 72 × 2 = 0 + 0.504 935 964 851 243 581 44;
  • 47) 0.504 935 964 851 243 581 44 × 2 = 1 + 0.009 871 929 702 487 162 88;
  • 48) 0.009 871 929 702 487 162 88 × 2 = 0 + 0.019 743 859 404 974 325 76;
  • 49) 0.019 743 859 404 974 325 76 × 2 = 0 + 0.039 487 718 809 948 651 52;
  • 50) 0.039 487 718 809 948 651 52 × 2 = 0 + 0.078 975 437 619 897 303 04;
  • 51) 0.078 975 437 619 897 303 04 × 2 = 0 + 0.157 950 875 239 794 606 08;
  • 52) 0.157 950 875 239 794 606 08 × 2 = 0 + 0.315 901 750 479 589 212 16;
  • 53) 0.315 901 750 479 589 212 16 × 2 = 0 + 0.631 803 500 959 178 424 32;
  • 54) 0.631 803 500 959 178 424 32 × 2 = 1 + 0.263 607 001 918 356 848 64;
  • 55) 0.263 607 001 918 356 848 64 × 2 = 0 + 0.527 214 003 836 713 697 28;
  • 56) 0.527 214 003 836 713 697 28 × 2 = 1 + 0.054 428 007 673 427 394 56;
  • 57) 0.054 428 007 673 427 394 56 × 2 = 0 + 0.108 856 015 346 854 789 12;
  • 58) 0.108 856 015 346 854 789 12 × 2 = 0 + 0.217 712 030 693 709 578 24;
  • 59) 0.217 712 030 693 709 578 24 × 2 = 0 + 0.435 424 061 387 419 156 48;
  • 60) 0.435 424 061 387 419 156 48 × 2 = 0 + 0.870 848 122 774 838 312 96;
  • 61) 0.870 848 122 774 838 312 96 × 2 = 1 + 0.741 696 245 549 676 625 92;
  • 62) 0.741 696 245 549 676 625 92 × 2 = 1 + 0.483 392 491 099 353 251 84;
  • 63) 0.483 392 491 099 353 251 84 × 2 = 0 + 0.966 784 982 198 706 503 68;
  • 64) 0.966 784 982 198 706 503 68 × 2 = 1 + 0.933 569 964 397 413 007 36;
  • 65) 0.933 569 964 397 413 007 36 × 2 = 1 + 0.867 139 928 794 826 014 72;
  • 66) 0.867 139 928 794 826 014 72 × 2 = 1 + 0.734 279 857 589 652 029 44;
  • 67) 0.734 279 857 589 652 029 44 × 2 = 1 + 0.468 559 715 179 304 058 88;
  • 68) 0.468 559 715 179 304 058 88 × 2 = 0 + 0.937 119 430 358 608 117 76;
  • 69) 0.937 119 430 358 608 117 76 × 2 = 1 + 0.874 238 860 717 216 235 52;
  • 70) 0.874 238 860 717 216 235 52 × 2 = 1 + 0.748 477 721 434 432 471 04;
  • 71) 0.748 477 721 434 432 471 04 × 2 = 1 + 0.496 955 442 868 864 942 08;
  • 72) 0.496 955 442 868 864 942 08 × 2 = 0 + 0.993 910 885 737 729 884 16;
  • 73) 0.993 910 885 737 729 884 16 × 2 = 1 + 0.987 821 771 475 459 768 32;
  • 74) 0.987 821 771 475 459 768 32 × 2 = 1 + 0.975 643 542 950 919 536 64;
  • 75) 0.975 643 542 950 919 536 64 × 2 = 1 + 0.951 287 085 901 839 073 28;
  • 76) 0.951 287 085 901 839 073 28 × 2 = 1 + 0.902 574 171 803 678 146 56;
  • 77) 0.902 574 171 803 678 146 56 × 2 = 1 + 0.805 148 343 607 356 293 12;
  • 78) 0.805 148 343 607 356 293 12 × 2 = 1 + 0.610 296 687 214 712 586 24;
  • 79) 0.610 296 687 214 712 586 24 × 2 = 1 + 0.220 593 374 429 425 172 48;
  • 80) 0.220 593 374 429 425 172 48 × 2 = 0 + 0.441 186 748 858 850 344 96;
  • 81) 0.441 186 748 858 850 344 96 × 2 = 0 + 0.882 373 497 717 700 689 92;
  • 82) 0.882 373 497 717 700 689 92 × 2 = 1 + 0.764 746 995 435 401 379 84;
  • 83) 0.764 746 995 435 401 379 84 × 2 = 1 + 0.529 493 990 870 802 759 68;
  • 84) 0.529 493 990 870 802 759 68 × 2 = 1 + 0.058 987 981 741 605 519 36;
  • 85) 0.058 987 981 741 605 519 36 × 2 = 0 + 0.117 975 963 483 211 038 72;
  • 86) 0.117 975 963 483 211 038 72 × 2 = 0 + 0.235 951 926 966 422 077 44;
  • 87) 0.235 951 926 966 422 077 44 × 2 = 0 + 0.471 903 853 932 844 154 88;
  • 88) 0.471 903 853 932 844 154 88 × 2 = 0 + 0.943 807 707 865 688 309 76;
  • 89) 0.943 807 707 865 688 309 76 × 2 = 1 + 0.887 615 415 731 376 619 52;

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:


0.000 000 000 010 068 460 71(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0000 1101 1110 1110 1111 1110 0111 0000 1(2)

6. Positive number before normalization:

0.000 000 000 010 068 460 71(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0000 1101 1110 1110 1111 1110 0111 0000 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 010 068 460 71(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0000 1101 1110 1110 1111 1110 0111 0000 1(2) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0000 1101 1110 1110 1111 1110 0111 0000 1(2) × 20 =


1.0110 0010 0100 0000 1010 0001 1011 1101 1101 1111 1100 1110 0001(2) × 2-37


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): -37


Mantissa (not normalized):
1.0110 0010 0100 0000 1010 0001 1011 1101 1101 1111 1100 1110 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


-37 + 2(11-1) - 1 =


(-37 + 1 023)(10) =


986(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;
  • 986 ÷ 2 = 493 + 0;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 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) =


986(10) =


011 1101 1010(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. 0110 0010 0100 0000 1010 0001 1011 1101 1101 1111 1100 1110 0001 =


0110 0010 0100 0000 1010 0001 1011 1101 1101 1111 1100 1110 0001


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 1010


Mantissa (52 bits) =
0110 0010 0100 0000 1010 0001 1011 1101 1101 1111 1100 1110 0001


The base ten decimal number -0.000 000 000 010 068 460 71 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0100 0000 1010 0001 1011 1101 1101 1111 1100 1110 0001

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