64bit IEEE 754: Decimal -> Double Precision Floating Point Binary: 0.000 000 000 931 322 55 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.000 000 000 931 322 55(10) converted and written in 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: 0.000 000 000 931 322 55.

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) 0.000 000 000 931 322 55 × 2 = 0 + 0.000 000 001 862 645 1;
  • 2) 0.000 000 001 862 645 1 × 2 = 0 + 0.000 000 003 725 290 2;
  • 3) 0.000 000 003 725 290 2 × 2 = 0 + 0.000 000 007 450 580 4;
  • 4) 0.000 000 007 450 580 4 × 2 = 0 + 0.000 000 014 901 160 8;
  • 5) 0.000 000 014 901 160 8 × 2 = 0 + 0.000 000 029 802 321 6;
  • 6) 0.000 000 029 802 321 6 × 2 = 0 + 0.000 000 059 604 643 2;
  • 7) 0.000 000 059 604 643 2 × 2 = 0 + 0.000 000 119 209 286 4;
  • 8) 0.000 000 119 209 286 4 × 2 = 0 + 0.000 000 238 418 572 8;
  • 9) 0.000 000 238 418 572 8 × 2 = 0 + 0.000 000 476 837 145 6;
  • 10) 0.000 000 476 837 145 6 × 2 = 0 + 0.000 000 953 674 291 2;
  • 11) 0.000 000 953 674 291 2 × 2 = 0 + 0.000 001 907 348 582 4;
  • 12) 0.000 001 907 348 582 4 × 2 = 0 + 0.000 003 814 697 164 8;
  • 13) 0.000 003 814 697 164 8 × 2 = 0 + 0.000 007 629 394 329 6;
  • 14) 0.000 007 629 394 329 6 × 2 = 0 + 0.000 015 258 788 659 2;
  • 15) 0.000 015 258 788 659 2 × 2 = 0 + 0.000 030 517 577 318 4;
  • 16) 0.000 030 517 577 318 4 × 2 = 0 + 0.000 061 035 154 636 8;
  • 17) 0.000 061 035 154 636 8 × 2 = 0 + 0.000 122 070 309 273 6;
  • 18) 0.000 122 070 309 273 6 × 2 = 0 + 0.000 244 140 618 547 2;
  • 19) 0.000 244 140 618 547 2 × 2 = 0 + 0.000 488 281 237 094 4;
  • 20) 0.000 488 281 237 094 4 × 2 = 0 + 0.000 976 562 474 188 8;
  • 21) 0.000 976 562 474 188 8 × 2 = 0 + 0.001 953 124 948 377 6;
  • 22) 0.001 953 124 948 377 6 × 2 = 0 + 0.003 906 249 896 755 2;
  • 23) 0.003 906 249 896 755 2 × 2 = 0 + 0.007 812 499 793 510 4;
  • 24) 0.007 812 499 793 510 4 × 2 = 0 + 0.015 624 999 587 020 8;
  • 25) 0.015 624 999 587 020 8 × 2 = 0 + 0.031 249 999 174 041 6;
  • 26) 0.031 249 999 174 041 6 × 2 = 0 + 0.062 499 998 348 083 2;
  • 27) 0.062 499 998 348 083 2 × 2 = 0 + 0.124 999 996 696 166 4;
  • 28) 0.124 999 996 696 166 4 × 2 = 0 + 0.249 999 993 392 332 8;
  • 29) 0.249 999 993 392 332 8 × 2 = 0 + 0.499 999 986 784 665 6;
  • 30) 0.499 999 986 784 665 6 × 2 = 0 + 0.999 999 973 569 331 2;
  • 31) 0.999 999 973 569 331 2 × 2 = 1 + 0.999 999 947 138 662 4;
  • 32) 0.999 999 947 138 662 4 × 2 = 1 + 0.999 999 894 277 324 8;
  • 33) 0.999 999 894 277 324 8 × 2 = 1 + 0.999 999 788 554 649 6;
  • 34) 0.999 999 788 554 649 6 × 2 = 1 + 0.999 999 577 109 299 2;
  • 35) 0.999 999 577 109 299 2 × 2 = 1 + 0.999 999 154 218 598 4;
  • 36) 0.999 999 154 218 598 4 × 2 = 1 + 0.999 998 308 437 196 8;
  • 37) 0.999 998 308 437 196 8 × 2 = 1 + 0.999 996 616 874 393 6;
  • 38) 0.999 996 616 874 393 6 × 2 = 1 + 0.999 993 233 748 787 2;
  • 39) 0.999 993 233 748 787 2 × 2 = 1 + 0.999 986 467 497 574 4;
  • 40) 0.999 986 467 497 574 4 × 2 = 1 + 0.999 972 934 995 148 8;
  • 41) 0.999 972 934 995 148 8 × 2 = 1 + 0.999 945 869 990 297 6;
  • 42) 0.999 945 869 990 297 6 × 2 = 1 + 0.999 891 739 980 595 2;
  • 43) 0.999 891 739 980 595 2 × 2 = 1 + 0.999 783 479 961 190 4;
  • 44) 0.999 783 479 961 190 4 × 2 = 1 + 0.999 566 959 922 380 8;
  • 45) 0.999 566 959 922 380 8 × 2 = 1 + 0.999 133 919 844 761 6;
  • 46) 0.999 133 919 844 761 6 × 2 = 1 + 0.998 267 839 689 523 2;
  • 47) 0.998 267 839 689 523 2 × 2 = 1 + 0.996 535 679 379 046 4;
  • 48) 0.996 535 679 379 046 4 × 2 = 1 + 0.993 071 358 758 092 8;
  • 49) 0.993 071 358 758 092 8 × 2 = 1 + 0.986 142 717 516 185 6;
  • 50) 0.986 142 717 516 185 6 × 2 = 1 + 0.972 285 435 032 371 2;
  • 51) 0.972 285 435 032 371 2 × 2 = 1 + 0.944 570 870 064 742 4;
  • 52) 0.944 570 870 064 742 4 × 2 = 1 + 0.889 141 740 129 484 8;
  • 53) 0.889 141 740 129 484 8 × 2 = 1 + 0.778 283 480 258 969 6;
  • 54) 0.778 283 480 258 969 6 × 2 = 1 + 0.556 566 960 517 939 2;
  • 55) 0.556 566 960 517 939 2 × 2 = 1 + 0.113 133 921 035 878 4;
  • 56) 0.113 133 921 035 878 4 × 2 = 0 + 0.226 267 842 071 756 8;
  • 57) 0.226 267 842 071 756 8 × 2 = 0 + 0.452 535 684 143 513 6;
  • 58) 0.452 535 684 143 513 6 × 2 = 0 + 0.905 071 368 287 027 2;
  • 59) 0.905 071 368 287 027 2 × 2 = 1 + 0.810 142 736 574 054 4;
  • 60) 0.810 142 736 574 054 4 × 2 = 1 + 0.620 285 473 148 108 8;
  • 61) 0.620 285 473 148 108 8 × 2 = 1 + 0.240 570 946 296 217 6;
  • 62) 0.240 570 946 296 217 6 × 2 = 0 + 0.481 141 892 592 435 2;
  • 63) 0.481 141 892 592 435 2 × 2 = 0 + 0.962 283 785 184 870 4;
  • 64) 0.962 283 785 184 870 4 × 2 = 1 + 0.924 567 570 369 740 8;
  • 65) 0.924 567 570 369 740 8 × 2 = 1 + 0.849 135 140 739 481 6;
  • 66) 0.849 135 140 739 481 6 × 2 = 1 + 0.698 270 281 478 963 2;
  • 67) 0.698 270 281 478 963 2 × 2 = 1 + 0.396 540 562 957 926 4;
  • 68) 0.396 540 562 957 926 4 × 2 = 0 + 0.793 081 125 915 852 8;
  • 69) 0.793 081 125 915 852 8 × 2 = 1 + 0.586 162 251 831 705 6;
  • 70) 0.586 162 251 831 705 6 × 2 = 1 + 0.172 324 503 663 411 2;
  • 71) 0.172 324 503 663 411 2 × 2 = 0 + 0.344 649 007 326 822 4;
  • 72) 0.344 649 007 326 822 4 × 2 = 0 + 0.689 298 014 653 644 8;
  • 73) 0.689 298 014 653 644 8 × 2 = 1 + 0.378 596 029 307 289 6;
  • 74) 0.378 596 029 307 289 6 × 2 = 0 + 0.757 192 058 614 579 2;
  • 75) 0.757 192 058 614 579 2 × 2 = 1 + 0.514 384 117 229 158 4;
  • 76) 0.514 384 117 229 158 4 × 2 = 1 + 0.028 768 234 458 316 8;
  • 77) 0.028 768 234 458 316 8 × 2 = 0 + 0.057 536 468 916 633 6;
  • 78) 0.057 536 468 916 633 6 × 2 = 0 + 0.115 072 937 833 267 2;
  • 79) 0.115 072 937 833 267 2 × 2 = 0 + 0.230 145 875 666 534 4;
  • 80) 0.230 145 875 666 534 4 × 2 = 0 + 0.460 291 751 333 068 8;
  • 81) 0.460 291 751 333 068 8 × 2 = 0 + 0.920 583 502 666 137 6;
  • 82) 0.920 583 502 666 137 6 × 2 = 1 + 0.841 167 005 332 275 2;
  • 83) 0.841 167 005 332 275 2 × 2 = 1 + 0.682 334 010 664 550 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...)


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:


0.000 000 000 931 322 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1110 0011 1001 1110 1100 1011 0000 011(2)


5. Positive number before normalization:

0.000 000 000 931 322 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1110 0011 1001 1110 1100 1011 0000 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 931 322 55(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1110 0011 1001 1110 1100 1011 0000 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1110 0011 1001 1110 1100 1011 0000 011(2) × 20 =

1.1111 1111 1111 1111 1111 1111 0001 1100 1111 0110 0101 1000 0011(2) × 2-31


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): -31

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 0001 1100 1111 0110 0101 1000 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(11-1) - 1 =

(-31 + 1 023)(10) =

992(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;
  • 992 ÷ 2 = 496 + 0;
  • 496 ÷ 2 = 248 + 0;
  • 248 ÷ 2 = 124 + 0;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

992(10) =

011 1110 0000(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. 1111 1111 1111 1111 1111 1111 0001 1100 1111 0110 0101 1000 0011 =

1111 1111 1111 1111 1111 1111 0001 1100 1111 0110 0101 1000 0011


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 1110 0000

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 0001 1100 1111 0110 0101 1000 0011


The base ten decimal number 0.000 000 000 931 322 55 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1110 0000 - 1111 1111 1111 1111 1111 1111 0001 1100 1111 0110 0101 1000 0011

(64 bits IEEE 754)

  • Sign (1 bit):

    • 0

      63

  • Exponent (11 bits):

    • 0

      62

    • 1

      61

    • 1

      60

    • 1

      59

    • 1

      58

    • 1

      57

    • 0

      56

    • 0

      55

    • 0

      54

    • 0

      53

    • 0

      52

  • Mantissa (52 bits):

    • 1

      51

    • 1

      50

    • 1

      49

    • 1

      48

    • 1

      47

    • 1

      46

    • 1

      45

    • 1

      44

    • 1

      43

    • 1

      42

    • 1

      41

    • 1

      40

    • 1

      39

    • 1

      38

    • 1

      37

    • 1

      36

    • 1

      35

    • 1

      34

    • 1

      33

    • 1

      32

    • 1

      31

    • 1

      30

    • 1

      29

    • 1

      28

    • 0

      27

    • 0

      26

    • 0

      25

    • 1

      24

    • 1

      23

    • 1

      22

    • 0

      21

    • 0

      20

    • 1

      19

    • 1

      18

    • 1

      17

    • 1

      16

    • 0

      15

    • 1

      14

    • 1

      13

    • 0

      12

    • 0

      11

    • 1

      10

    • 0

      9

    • 1

      8

    • 1

      7

    • 0

      6

    • 0

      5

    • 0

      4

    • 0

      3

    • 0

      2

    • 1

      1

    • 1

      0

Number 0.000 000 000 931 322 54 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Number 0.000 000 000 931 322 56 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Convert to 64 bit double precision IEEE 754 binary floating point representation standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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