0.000 000 000 000 021 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 021 35(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
0.000 000 000 000 021 35(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 021 35.

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) 0.000 000 000 000 021 35 × 2 = 0 + 0.000 000 000 000 042 7;
  • 2) 0.000 000 000 000 042 7 × 2 = 0 + 0.000 000 000 000 085 4;
  • 3) 0.000 000 000 000 085 4 × 2 = 0 + 0.000 000 000 000 170 8;
  • 4) 0.000 000 000 000 170 8 × 2 = 0 + 0.000 000 000 000 341 6;
  • 5) 0.000 000 000 000 341 6 × 2 = 0 + 0.000 000 000 000 683 2;
  • 6) 0.000 000 000 000 683 2 × 2 = 0 + 0.000 000 000 001 366 4;
  • 7) 0.000 000 000 001 366 4 × 2 = 0 + 0.000 000 000 002 732 8;
  • 8) 0.000 000 000 002 732 8 × 2 = 0 + 0.000 000 000 005 465 6;
  • 9) 0.000 000 000 005 465 6 × 2 = 0 + 0.000 000 000 010 931 2;
  • 10) 0.000 000 000 010 931 2 × 2 = 0 + 0.000 000 000 021 862 4;
  • 11) 0.000 000 000 021 862 4 × 2 = 0 + 0.000 000 000 043 724 8;
  • 12) 0.000 000 000 043 724 8 × 2 = 0 + 0.000 000 000 087 449 6;
  • 13) 0.000 000 000 087 449 6 × 2 = 0 + 0.000 000 000 174 899 2;
  • 14) 0.000 000 000 174 899 2 × 2 = 0 + 0.000 000 000 349 798 4;
  • 15) 0.000 000 000 349 798 4 × 2 = 0 + 0.000 000 000 699 596 8;
  • 16) 0.000 000 000 699 596 8 × 2 = 0 + 0.000 000 001 399 193 6;
  • 17) 0.000 000 001 399 193 6 × 2 = 0 + 0.000 000 002 798 387 2;
  • 18) 0.000 000 002 798 387 2 × 2 = 0 + 0.000 000 005 596 774 4;
  • 19) 0.000 000 005 596 774 4 × 2 = 0 + 0.000 000 011 193 548 8;
  • 20) 0.000 000 011 193 548 8 × 2 = 0 + 0.000 000 022 387 097 6;
  • 21) 0.000 000 022 387 097 6 × 2 = 0 + 0.000 000 044 774 195 2;
  • 22) 0.000 000 044 774 195 2 × 2 = 0 + 0.000 000 089 548 390 4;
  • 23) 0.000 000 089 548 390 4 × 2 = 0 + 0.000 000 179 096 780 8;
  • 24) 0.000 000 179 096 780 8 × 2 = 0 + 0.000 000 358 193 561 6;
  • 25) 0.000 000 358 193 561 6 × 2 = 0 + 0.000 000 716 387 123 2;
  • 26) 0.000 000 716 387 123 2 × 2 = 0 + 0.000 001 432 774 246 4;
  • 27) 0.000 001 432 774 246 4 × 2 = 0 + 0.000 002 865 548 492 8;
  • 28) 0.000 002 865 548 492 8 × 2 = 0 + 0.000 005 731 096 985 6;
  • 29) 0.000 005 731 096 985 6 × 2 = 0 + 0.000 011 462 193 971 2;
  • 30) 0.000 011 462 193 971 2 × 2 = 0 + 0.000 022 924 387 942 4;
  • 31) 0.000 022 924 387 942 4 × 2 = 0 + 0.000 045 848 775 884 8;
  • 32) 0.000 045 848 775 884 8 × 2 = 0 + 0.000 091 697 551 769 6;
  • 33) 0.000 091 697 551 769 6 × 2 = 0 + 0.000 183 395 103 539 2;
  • 34) 0.000 183 395 103 539 2 × 2 = 0 + 0.000 366 790 207 078 4;
  • 35) 0.000 366 790 207 078 4 × 2 = 0 + 0.000 733 580 414 156 8;
  • 36) 0.000 733 580 414 156 8 × 2 = 0 + 0.001 467 160 828 313 6;
  • 37) 0.001 467 160 828 313 6 × 2 = 0 + 0.002 934 321 656 627 2;
  • 38) 0.002 934 321 656 627 2 × 2 = 0 + 0.005 868 643 313 254 4;
  • 39) 0.005 868 643 313 254 4 × 2 = 0 + 0.011 737 286 626 508 8;
  • 40) 0.011 737 286 626 508 8 × 2 = 0 + 0.023 474 573 253 017 6;
  • 41) 0.023 474 573 253 017 6 × 2 = 0 + 0.046 949 146 506 035 2;
  • 42) 0.046 949 146 506 035 2 × 2 = 0 + 0.093 898 293 012 070 4;
  • 43) 0.093 898 293 012 070 4 × 2 = 0 + 0.187 796 586 024 140 8;
  • 44) 0.187 796 586 024 140 8 × 2 = 0 + 0.375 593 172 048 281 6;
  • 45) 0.375 593 172 048 281 6 × 2 = 0 + 0.751 186 344 096 563 2;
  • 46) 0.751 186 344 096 563 2 × 2 = 1 + 0.502 372 688 193 126 4;
  • 47) 0.502 372 688 193 126 4 × 2 = 1 + 0.004 745 376 386 252 8;
  • 48) 0.004 745 376 386 252 8 × 2 = 0 + 0.009 490 752 772 505 6;
  • 49) 0.009 490 752 772 505 6 × 2 = 0 + 0.018 981 505 545 011 2;
  • 50) 0.018 981 505 545 011 2 × 2 = 0 + 0.037 963 011 090 022 4;
  • 51) 0.037 963 011 090 022 4 × 2 = 0 + 0.075 926 022 180 044 8;
  • 52) 0.075 926 022 180 044 8 × 2 = 0 + 0.151 852 044 360 089 6;
  • 53) 0.151 852 044 360 089 6 × 2 = 0 + 0.303 704 088 720 179 2;
  • 54) 0.303 704 088 720 179 2 × 2 = 0 + 0.607 408 177 440 358 4;
  • 55) 0.607 408 177 440 358 4 × 2 = 1 + 0.214 816 354 880 716 8;
  • 56) 0.214 816 354 880 716 8 × 2 = 0 + 0.429 632 709 761 433 6;
  • 57) 0.429 632 709 761 433 6 × 2 = 0 + 0.859 265 419 522 867 2;
  • 58) 0.859 265 419 522 867 2 × 2 = 1 + 0.718 530 839 045 734 4;
  • 59) 0.718 530 839 045 734 4 × 2 = 1 + 0.437 061 678 091 468 8;
  • 60) 0.437 061 678 091 468 8 × 2 = 0 + 0.874 123 356 182 937 6;
  • 61) 0.874 123 356 182 937 6 × 2 = 1 + 0.748 246 712 365 875 2;
  • 62) 0.748 246 712 365 875 2 × 2 = 1 + 0.496 493 424 731 750 4;
  • 63) 0.496 493 424 731 750 4 × 2 = 0 + 0.992 986 849 463 500 8;
  • 64) 0.992 986 849 463 500 8 × 2 = 1 + 0.985 973 698 927 001 6;
  • 65) 0.985 973 698 927 001 6 × 2 = 1 + 0.971 947 397 854 003 2;
  • 66) 0.971 947 397 854 003 2 × 2 = 1 + 0.943 894 795 708 006 4;
  • 67) 0.943 894 795 708 006 4 × 2 = 1 + 0.887 789 591 416 012 8;
  • 68) 0.887 789 591 416 012 8 × 2 = 1 + 0.775 579 182 832 025 6;
  • 69) 0.775 579 182 832 025 6 × 2 = 1 + 0.551 158 365 664 051 2;
  • 70) 0.551 158 365 664 051 2 × 2 = 1 + 0.102 316 731 328 102 4;
  • 71) 0.102 316 731 328 102 4 × 2 = 0 + 0.204 633 462 656 204 8;
  • 72) 0.204 633 462 656 204 8 × 2 = 0 + 0.409 266 925 312 409 6;
  • 73) 0.409 266 925 312 409 6 × 2 = 0 + 0.818 533 850 624 819 2;
  • 74) 0.818 533 850 624 819 2 × 2 = 1 + 0.637 067 701 249 638 4;
  • 75) 0.637 067 701 249 638 4 × 2 = 1 + 0.274 135 402 499 276 8;
  • 76) 0.274 135 402 499 276 8 × 2 = 0 + 0.548 270 804 998 553 6;
  • 77) 0.548 270 804 998 553 6 × 2 = 1 + 0.096 541 609 997 107 2;
  • 78) 0.096 541 609 997 107 2 × 2 = 0 + 0.193 083 219 994 214 4;
  • 79) 0.193 083 219 994 214 4 × 2 = 0 + 0.386 166 439 988 428 8;
  • 80) 0.386 166 439 988 428 8 × 2 = 0 + 0.772 332 879 976 857 6;
  • 81) 0.772 332 879 976 857 6 × 2 = 1 + 0.544 665 759 953 715 2;
  • 82) 0.544 665 759 953 715 2 × 2 = 1 + 0.089 331 519 907 430 4;
  • 83) 0.089 331 519 907 430 4 × 2 = 0 + 0.178 663 039 814 860 8;
  • 84) 0.178 663 039 814 860 8 × 2 = 0 + 0.357 326 079 629 721 6;
  • 85) 0.357 326 079 629 721 6 × 2 = 0 + 0.714 652 159 259 443 2;
  • 86) 0.714 652 159 259 443 2 × 2 = 1 + 0.429 304 318 518 886 4;
  • 87) 0.429 304 318 518 886 4 × 2 = 0 + 0.858 608 637 037 772 8;
  • 88) 0.858 608 637 037 772 8 × 2 = 1 + 0.717 217 274 075 545 6;
  • 89) 0.717 217 274 075 545 6 × 2 = 1 + 0.434 434 548 151 091 2;
  • 90) 0.434 434 548 151 091 2 × 2 = 0 + 0.868 869 096 302 182 4;
  • 91) 0.868 869 096 302 182 4 × 2 = 1 + 0.737 738 192 604 364 8;
  • 92) 0.737 738 192 604 364 8 × 2 = 1 + 0.475 476 385 208 729 6;
  • 93) 0.475 476 385 208 729 6 × 2 = 0 + 0.950 952 770 417 459 2;
  • 94) 0.950 952 770 417 459 2 × 2 = 1 + 0.901 905 540 834 918 4;
  • 95) 0.901 905 540 834 918 4 × 2 = 1 + 0.803 811 081 669 836 8;
  • 96) 0.803 811 081 669 836 8 × 2 = 1 + 0.607 622 163 339 673 6;
  • 97) 0.607 622 163 339 673 6 × 2 = 1 + 0.215 244 326 679 347 2;
  • 98) 0.215 244 326 679 347 2 × 2 = 0 + 0.430 488 653 358 694 4;

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).


4. 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:


0.000 000 000 000 021 35(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0000 0010 0110 1101 1111 1100 0110 1000 1100 0101 1011 0111 10(2)

5. Positive number before normalization:

0.000 000 000 000 021 35(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0000 0010 0110 1101 1111 1100 0110 1000 1100 0101 1011 0111 10(2)

6. 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:


0.000 000 000 000 021 35(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0000 0010 0110 1101 1111 1100 0110 1000 1100 0101 1011 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 0000 0010 0110 1101 1111 1100 0110 1000 1100 0101 1011 0111 10(2) × 20 =

1.1000 0000 1001 1011 0111 1111 0001 1010 0011 0001 0110 1101 1110(2) × 2-46


7. 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 0 (a positive number)

Exponent (unadjusted): -46

Mantissa (not normalized):
1.1000 0000 1001 1011 0111 1111 0001 1010 0011 0001 0110 1101 1110


8. 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)


9. 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;

10. 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)


11. 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. 1000 0000 1001 1011 0111 1111 0001 1010 0011 0001 0110 1101 1110 =

1000 0000 1001 1011 0111 1111 0001 1010 0011 0001 0110 1101 1110


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

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 0001

Mantissa (52 bits) =
1000 0000 1001 1011 0111 1111 0001 1010 0011 0001 0110 1101 1110


Decimal number 0.000 000 000 000 021 35 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0001 - 1000 0000 1001 1011 0111 1111 0001 1010 0011 0001 0110 1101 1110

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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