-0.000 000 000 010 068 461 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 010 068 461 09(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
-0.000 000 000 010 068 461 09(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:

|-0.000 000 000 010 068 461 09| = 0.000 000 000 010 068 461 09


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: 0.000 000 000 010 068 461 09.

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) 0.000 000 000 010 068 461 09 × 2 = 0 + 0.000 000 000 020 136 922 18;
  • 2) 0.000 000 000 020 136 922 18 × 2 = 0 + 0.000 000 000 040 273 844 36;
  • 3) 0.000 000 000 040 273 844 36 × 2 = 0 + 0.000 000 000 080 547 688 72;
  • 4) 0.000 000 000 080 547 688 72 × 2 = 0 + 0.000 000 000 161 095 377 44;
  • 5) 0.000 000 000 161 095 377 44 × 2 = 0 + 0.000 000 000 322 190 754 88;
  • 6) 0.000 000 000 322 190 754 88 × 2 = 0 + 0.000 000 000 644 381 509 76;
  • 7) 0.000 000 000 644 381 509 76 × 2 = 0 + 0.000 000 001 288 763 019 52;
  • 8) 0.000 000 001 288 763 019 52 × 2 = 0 + 0.000 000 002 577 526 039 04;
  • 9) 0.000 000 002 577 526 039 04 × 2 = 0 + 0.000 000 005 155 052 078 08;
  • 10) 0.000 000 005 155 052 078 08 × 2 = 0 + 0.000 000 010 310 104 156 16;
  • 11) 0.000 000 010 310 104 156 16 × 2 = 0 + 0.000 000 020 620 208 312 32;
  • 12) 0.000 000 020 620 208 312 32 × 2 = 0 + 0.000 000 041 240 416 624 64;
  • 13) 0.000 000 041 240 416 624 64 × 2 = 0 + 0.000 000 082 480 833 249 28;
  • 14) 0.000 000 082 480 833 249 28 × 2 = 0 + 0.000 000 164 961 666 498 56;
  • 15) 0.000 000 164 961 666 498 56 × 2 = 0 + 0.000 000 329 923 332 997 12;
  • 16) 0.000 000 329 923 332 997 12 × 2 = 0 + 0.000 000 659 846 665 994 24;
  • 17) 0.000 000 659 846 665 994 24 × 2 = 0 + 0.000 001 319 693 331 988 48;
  • 18) 0.000 001 319 693 331 988 48 × 2 = 0 + 0.000 002 639 386 663 976 96;
  • 19) 0.000 002 639 386 663 976 96 × 2 = 0 + 0.000 005 278 773 327 953 92;
  • 20) 0.000 005 278 773 327 953 92 × 2 = 0 + 0.000 010 557 546 655 907 84;
  • 21) 0.000 010 557 546 655 907 84 × 2 = 0 + 0.000 021 115 093 311 815 68;
  • 22) 0.000 021 115 093 311 815 68 × 2 = 0 + 0.000 042 230 186 623 631 36;
  • 23) 0.000 042 230 186 623 631 36 × 2 = 0 + 0.000 084 460 373 247 262 72;
  • 24) 0.000 084 460 373 247 262 72 × 2 = 0 + 0.000 168 920 746 494 525 44;
  • 25) 0.000 168 920 746 494 525 44 × 2 = 0 + 0.000 337 841 492 989 050 88;
  • 26) 0.000 337 841 492 989 050 88 × 2 = 0 + 0.000 675 682 985 978 101 76;
  • 27) 0.000 675 682 985 978 101 76 × 2 = 0 + 0.001 351 365 971 956 203 52;
  • 28) 0.001 351 365 971 956 203 52 × 2 = 0 + 0.002 702 731 943 912 407 04;
  • 29) 0.002 702 731 943 912 407 04 × 2 = 0 + 0.005 405 463 887 824 814 08;
  • 30) 0.005 405 463 887 824 814 08 × 2 = 0 + 0.010 810 927 775 649 628 16;
  • 31) 0.010 810 927 775 649 628 16 × 2 = 0 + 0.021 621 855 551 299 256 32;
  • 32) 0.021 621 855 551 299 256 32 × 2 = 0 + 0.043 243 711 102 598 512 64;
  • 33) 0.043 243 711 102 598 512 64 × 2 = 0 + 0.086 487 422 205 197 025 28;
  • 34) 0.086 487 422 205 197 025 28 × 2 = 0 + 0.172 974 844 410 394 050 56;
  • 35) 0.172 974 844 410 394 050 56 × 2 = 0 + 0.345 949 688 820 788 101 12;
  • 36) 0.345 949 688 820 788 101 12 × 2 = 0 + 0.691 899 377 641 576 202 24;
  • 37) 0.691 899 377 641 576 202 24 × 2 = 1 + 0.383 798 755 283 152 404 48;
  • 38) 0.383 798 755 283 152 404 48 × 2 = 0 + 0.767 597 510 566 304 808 96;
  • 39) 0.767 597 510 566 304 808 96 × 2 = 1 + 0.535 195 021 132 609 617 92;
  • 40) 0.535 195 021 132 609 617 92 × 2 = 1 + 0.070 390 042 265 219 235 84;
  • 41) 0.070 390 042 265 219 235 84 × 2 = 0 + 0.140 780 084 530 438 471 68;
  • 42) 0.140 780 084 530 438 471 68 × 2 = 0 + 0.281 560 169 060 876 943 36;
  • 43) 0.281 560 169 060 876 943 36 × 2 = 0 + 0.563 120 338 121 753 886 72;
  • 44) 0.563 120 338 121 753 886 72 × 2 = 1 + 0.126 240 676 243 507 773 44;
  • 45) 0.126 240 676 243 507 773 44 × 2 = 0 + 0.252 481 352 487 015 546 88;
  • 46) 0.252 481 352 487 015 546 88 × 2 = 0 + 0.504 962 704 974 031 093 76;
  • 47) 0.504 962 704 974 031 093 76 × 2 = 1 + 0.009 925 409 948 062 187 52;
  • 48) 0.009 925 409 948 062 187 52 × 2 = 0 + 0.019 850 819 896 124 375 04;
  • 49) 0.019 850 819 896 124 375 04 × 2 = 0 + 0.039 701 639 792 248 750 08;
  • 50) 0.039 701 639 792 248 750 08 × 2 = 0 + 0.079 403 279 584 497 500 16;
  • 51) 0.079 403 279 584 497 500 16 × 2 = 0 + 0.158 806 559 168 995 000 32;
  • 52) 0.158 806 559 168 995 000 32 × 2 = 0 + 0.317 613 118 337 990 000 64;
  • 53) 0.317 613 118 337 990 000 64 × 2 = 0 + 0.635 226 236 675 980 001 28;
  • 54) 0.635 226 236 675 980 001 28 × 2 = 1 + 0.270 452 473 351 960 002 56;
  • 55) 0.270 452 473 351 960 002 56 × 2 = 0 + 0.540 904 946 703 920 005 12;
  • 56) 0.540 904 946 703 920 005 12 × 2 = 1 + 0.081 809 893 407 840 010 24;
  • 57) 0.081 809 893 407 840 010 24 × 2 = 0 + 0.163 619 786 815 680 020 48;
  • 58) 0.163 619 786 815 680 020 48 × 2 = 0 + 0.327 239 573 631 360 040 96;
  • 59) 0.327 239 573 631 360 040 96 × 2 = 0 + 0.654 479 147 262 720 081 92;
  • 60) 0.654 479 147 262 720 081 92 × 2 = 1 + 0.308 958 294 525 440 163 84;
  • 61) 0.308 958 294 525 440 163 84 × 2 = 0 + 0.617 916 589 050 880 327 68;
  • 62) 0.617 916 589 050 880 327 68 × 2 = 1 + 0.235 833 178 101 760 655 36;
  • 63) 0.235 833 178 101 760 655 36 × 2 = 0 + 0.471 666 356 203 521 310 72;
  • 64) 0.471 666 356 203 521 310 72 × 2 = 0 + 0.943 332 712 407 042 621 44;
  • 65) 0.943 332 712 407 042 621 44 × 2 = 1 + 0.886 665 424 814 085 242 88;
  • 66) 0.886 665 424 814 085 242 88 × 2 = 1 + 0.773 330 849 628 170 485 76;
  • 67) 0.773 330 849 628 170 485 76 × 2 = 1 + 0.546 661 699 256 340 971 52;
  • 68) 0.546 661 699 256 340 971 52 × 2 = 1 + 0.093 323 398 512 681 943 04;
  • 69) 0.093 323 398 512 681 943 04 × 2 = 0 + 0.186 646 797 025 363 886 08;
  • 70) 0.186 646 797 025 363 886 08 × 2 = 0 + 0.373 293 594 050 727 772 16;
  • 71) 0.373 293 594 050 727 772 16 × 2 = 0 + 0.746 587 188 101 455 544 32;
  • 72) 0.746 587 188 101 455 544 32 × 2 = 1 + 0.493 174 376 202 911 088 64;
  • 73) 0.493 174 376 202 911 088 64 × 2 = 0 + 0.986 348 752 405 822 177 28;
  • 74) 0.986 348 752 405 822 177 28 × 2 = 1 + 0.972 697 504 811 644 354 56;
  • 75) 0.972 697 504 811 644 354 56 × 2 = 1 + 0.945 395 009 623 288 709 12;
  • 76) 0.945 395 009 623 288 709 12 × 2 = 1 + 0.890 790 019 246 577 418 24;
  • 77) 0.890 790 019 246 577 418 24 × 2 = 1 + 0.781 580 038 493 154 836 48;
  • 78) 0.781 580 038 493 154 836 48 × 2 = 1 + 0.563 160 076 986 309 672 96;
  • 79) 0.563 160 076 986 309 672 96 × 2 = 1 + 0.126 320 153 972 619 345 92;
  • 80) 0.126 320 153 972 619 345 92 × 2 = 0 + 0.252 640 307 945 238 691 84;
  • 81) 0.252 640 307 945 238 691 84 × 2 = 0 + 0.505 280 615 890 477 383 68;
  • 82) 0.505 280 615 890 477 383 68 × 2 = 1 + 0.010 561 231 780 954 767 36;
  • 83) 0.010 561 231 780 954 767 36 × 2 = 0 + 0.021 122 463 561 909 534 72;
  • 84) 0.021 122 463 561 909 534 72 × 2 = 0 + 0.042 244 927 123 819 069 44;
  • 85) 0.042 244 927 123 819 069 44 × 2 = 0 + 0.084 489 854 247 638 138 88;
  • 86) 0.084 489 854 247 638 138 88 × 2 = 0 + 0.168 979 708 495 276 277 76;
  • 87) 0.168 979 708 495 276 277 76 × 2 = 0 + 0.337 959 416 990 552 555 52;
  • 88) 0.337 959 416 990 552 555 52 × 2 = 0 + 0.675 918 833 981 105 111 04;
  • 89) 0.675 918 833 981 105 111 04 × 2 = 1 + 0.351 837 667 962 210 222 08;

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:


0.000 000 000 010 068 461 09(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0100 1111 0001 0111 1110 0100 0000 1(2)

6. Positive number before normalization:

0.000 000 000 010 068 461 09(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0100 1111 0001 0111 1110 0100 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:


0.000 000 000 010 068 461 09(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0100 1111 0001 0111 1110 0100 0000 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0010 0000 0101 0001 0100 1111 0001 0111 1110 0100 0000 1(2) × 20 =

1.0110 0010 0100 0000 1010 0010 1001 1110 0010 1111 1100 1000 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 0010 1001 1110 0010 1111 1100 1000 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 0010 1001 1110 0010 1111 1100 1000 0001 =

0110 0010 0100 0000 1010 0010 1001 1110 0010 1111 1100 1000 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 0010 1001 1110 0010 1111 1100 1000 0001


Decimal number -0.000 000 000 010 068 461 09 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0100 0000 1010 0010 1001 1110 0010 1111 1100 1000 0001

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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