-0.000 000 000 000 176 49 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 49(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
-0.000 000 000 000 176 49(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:

|-0.000 000 000 000 176 49| = 0.000 000 000 000 176 49


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: 0.000 000 000 000 176 49.

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) 0.000 000 000 000 176 49 × 2 = 0 + 0.000 000 000 000 352 98;
  • 2) 0.000 000 000 000 352 98 × 2 = 0 + 0.000 000 000 000 705 96;
  • 3) 0.000 000 000 000 705 96 × 2 = 0 + 0.000 000 000 001 411 92;
  • 4) 0.000 000 000 001 411 92 × 2 = 0 + 0.000 000 000 002 823 84;
  • 5) 0.000 000 000 002 823 84 × 2 = 0 + 0.000 000 000 005 647 68;
  • 6) 0.000 000 000 005 647 68 × 2 = 0 + 0.000 000 000 011 295 36;
  • 7) 0.000 000 000 011 295 36 × 2 = 0 + 0.000 000 000 022 590 72;
  • 8) 0.000 000 000 022 590 72 × 2 = 0 + 0.000 000 000 045 181 44;
  • 9) 0.000 000 000 045 181 44 × 2 = 0 + 0.000 000 000 090 362 88;
  • 10) 0.000 000 000 090 362 88 × 2 = 0 + 0.000 000 000 180 725 76;
  • 11) 0.000 000 000 180 725 76 × 2 = 0 + 0.000 000 000 361 451 52;
  • 12) 0.000 000 000 361 451 52 × 2 = 0 + 0.000 000 000 722 903 04;
  • 13) 0.000 000 000 722 903 04 × 2 = 0 + 0.000 000 001 445 806 08;
  • 14) 0.000 000 001 445 806 08 × 2 = 0 + 0.000 000 002 891 612 16;
  • 15) 0.000 000 002 891 612 16 × 2 = 0 + 0.000 000 005 783 224 32;
  • 16) 0.000 000 005 783 224 32 × 2 = 0 + 0.000 000 011 566 448 64;
  • 17) 0.000 000 011 566 448 64 × 2 = 0 + 0.000 000 023 132 897 28;
  • 18) 0.000 000 023 132 897 28 × 2 = 0 + 0.000 000 046 265 794 56;
  • 19) 0.000 000 046 265 794 56 × 2 = 0 + 0.000 000 092 531 589 12;
  • 20) 0.000 000 092 531 589 12 × 2 = 0 + 0.000 000 185 063 178 24;
  • 21) 0.000 000 185 063 178 24 × 2 = 0 + 0.000 000 370 126 356 48;
  • 22) 0.000 000 370 126 356 48 × 2 = 0 + 0.000 000 740 252 712 96;
  • 23) 0.000 000 740 252 712 96 × 2 = 0 + 0.000 001 480 505 425 92;
  • 24) 0.000 001 480 505 425 92 × 2 = 0 + 0.000 002 961 010 851 84;
  • 25) 0.000 002 961 010 851 84 × 2 = 0 + 0.000 005 922 021 703 68;
  • 26) 0.000 005 922 021 703 68 × 2 = 0 + 0.000 011 844 043 407 36;
  • 27) 0.000 011 844 043 407 36 × 2 = 0 + 0.000 023 688 086 814 72;
  • 28) 0.000 023 688 086 814 72 × 2 = 0 + 0.000 047 376 173 629 44;
  • 29) 0.000 047 376 173 629 44 × 2 = 0 + 0.000 094 752 347 258 88;
  • 30) 0.000 094 752 347 258 88 × 2 = 0 + 0.000 189 504 694 517 76;
  • 31) 0.000 189 504 694 517 76 × 2 = 0 + 0.000 379 009 389 035 52;
  • 32) 0.000 379 009 389 035 52 × 2 = 0 + 0.000 758 018 778 071 04;
  • 33) 0.000 758 018 778 071 04 × 2 = 0 + 0.001 516 037 556 142 08;
  • 34) 0.001 516 037 556 142 08 × 2 = 0 + 0.003 032 075 112 284 16;
  • 35) 0.003 032 075 112 284 16 × 2 = 0 + 0.006 064 150 224 568 32;
  • 36) 0.006 064 150 224 568 32 × 2 = 0 + 0.012 128 300 449 136 64;
  • 37) 0.012 128 300 449 136 64 × 2 = 0 + 0.024 256 600 898 273 28;
  • 38) 0.024 256 600 898 273 28 × 2 = 0 + 0.048 513 201 796 546 56;
  • 39) 0.048 513 201 796 546 56 × 2 = 0 + 0.097 026 403 593 093 12;
  • 40) 0.097 026 403 593 093 12 × 2 = 0 + 0.194 052 807 186 186 24;
  • 41) 0.194 052 807 186 186 24 × 2 = 0 + 0.388 105 614 372 372 48;
  • 42) 0.388 105 614 372 372 48 × 2 = 0 + 0.776 211 228 744 744 96;
  • 43) 0.776 211 228 744 744 96 × 2 = 1 + 0.552 422 457 489 489 92;
  • 44) 0.552 422 457 489 489 92 × 2 = 1 + 0.104 844 914 978 979 84;
  • 45) 0.104 844 914 978 979 84 × 2 = 0 + 0.209 689 829 957 959 68;
  • 46) 0.209 689 829 957 959 68 × 2 = 0 + 0.419 379 659 915 919 36;
  • 47) 0.419 379 659 915 919 36 × 2 = 0 + 0.838 759 319 831 838 72;
  • 48) 0.838 759 319 831 838 72 × 2 = 1 + 0.677 518 639 663 677 44;
  • 49) 0.677 518 639 663 677 44 × 2 = 1 + 0.355 037 279 327 354 88;
  • 50) 0.355 037 279 327 354 88 × 2 = 0 + 0.710 074 558 654 709 76;
  • 51) 0.710 074 558 654 709 76 × 2 = 1 + 0.420 149 117 309 419 52;
  • 52) 0.420 149 117 309 419 52 × 2 = 0 + 0.840 298 234 618 839 04;
  • 53) 0.840 298 234 618 839 04 × 2 = 1 + 0.680 596 469 237 678 08;
  • 54) 0.680 596 469 237 678 08 × 2 = 1 + 0.361 192 938 475 356 16;
  • 55) 0.361 192 938 475 356 16 × 2 = 0 + 0.722 385 876 950 712 32;
  • 56) 0.722 385 876 950 712 32 × 2 = 1 + 0.444 771 753 901 424 64;
  • 57) 0.444 771 753 901 424 64 × 2 = 0 + 0.889 543 507 802 849 28;
  • 58) 0.889 543 507 802 849 28 × 2 = 1 + 0.779 087 015 605 698 56;
  • 59) 0.779 087 015 605 698 56 × 2 = 1 + 0.558 174 031 211 397 12;
  • 60) 0.558 174 031 211 397 12 × 2 = 1 + 0.116 348 062 422 794 24;
  • 61) 0.116 348 062 422 794 24 × 2 = 0 + 0.232 696 124 845 588 48;
  • 62) 0.232 696 124 845 588 48 × 2 = 0 + 0.465 392 249 691 176 96;
  • 63) 0.465 392 249 691 176 96 × 2 = 0 + 0.930 784 499 382 353 92;
  • 64) 0.930 784 499 382 353 92 × 2 = 1 + 0.861 568 998 764 707 84;
  • 65) 0.861 568 998 764 707 84 × 2 = 1 + 0.723 137 997 529 415 68;
  • 66) 0.723 137 997 529 415 68 × 2 = 1 + 0.446 275 995 058 831 36;
  • 67) 0.446 275 995 058 831 36 × 2 = 0 + 0.892 551 990 117 662 72;
  • 68) 0.892 551 990 117 662 72 × 2 = 1 + 0.785 103 980 235 325 44;
  • 69) 0.785 103 980 235 325 44 × 2 = 1 + 0.570 207 960 470 650 88;
  • 70) 0.570 207 960 470 650 88 × 2 = 1 + 0.140 415 920 941 301 76;
  • 71) 0.140 415 920 941 301 76 × 2 = 0 + 0.280 831 841 882 603 52;
  • 72) 0.280 831 841 882 603 52 × 2 = 0 + 0.561 663 683 765 207 04;
  • 73) 0.561 663 683 765 207 04 × 2 = 1 + 0.123 327 367 530 414 08;
  • 74) 0.123 327 367 530 414 08 × 2 = 0 + 0.246 654 735 060 828 16;
  • 75) 0.246 654 735 060 828 16 × 2 = 0 + 0.493 309 470 121 656 32;
  • 76) 0.493 309 470 121 656 32 × 2 = 0 + 0.986 618 940 243 312 64;
  • 77) 0.986 618 940 243 312 64 × 2 = 1 + 0.973 237 880 486 625 28;
  • 78) 0.973 237 880 486 625 28 × 2 = 1 + 0.946 475 760 973 250 56;
  • 79) 0.946 475 760 973 250 56 × 2 = 1 + 0.892 951 521 946 501 12;
  • 80) 0.892 951 521 946 501 12 × 2 = 1 + 0.785 903 043 893 002 24;
  • 81) 0.785 903 043 893 002 24 × 2 = 1 + 0.571 806 087 786 004 48;
  • 82) 0.571 806 087 786 004 48 × 2 = 1 + 0.143 612 175 572 008 96;
  • 83) 0.143 612 175 572 008 96 × 2 = 0 + 0.287 224 351 144 017 92;
  • 84) 0.287 224 351 144 017 92 × 2 = 0 + 0.574 448 702 288 035 84;
  • 85) 0.574 448 702 288 035 84 × 2 = 1 + 0.148 897 404 576 071 68;
  • 86) 0.148 897 404 576 071 68 × 2 = 0 + 0.297 794 809 152 143 36;
  • 87) 0.297 794 809 152 143 36 × 2 = 0 + 0.595 589 618 304 286 72;
  • 88) 0.595 589 618 304 286 72 × 2 = 1 + 0.191 179 236 608 573 44;
  • 89) 0.191 179 236 608 573 44 × 2 = 0 + 0.382 358 473 217 146 88;
  • 90) 0.382 358 473 217 146 88 × 2 = 0 + 0.764 716 946 434 293 76;
  • 91) 0.764 716 946 434 293 76 × 2 = 1 + 0.529 433 892 868 587 52;
  • 92) 0.529 433 892 868 587 52 × 2 = 1 + 0.058 867 785 737 175 04;
  • 93) 0.058 867 785 737 175 04 × 2 = 0 + 0.117 735 571 474 350 08;
  • 94) 0.117 735 571 474 350 08 × 2 = 0 + 0.235 471 142 948 700 16;
  • 95) 0.235 471 142 948 700 16 × 2 = 0 + 0.470 942 285 897 400 32;

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:


0.000 000 000 000 176 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1101 0111 0001 1101 1100 1000 1111 1100 1001 0011 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1101 0111 0001 1101 1100 1000 1111 1100 1001 0011 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1101 0111 0001 1101 1100 1000 1111 1100 1001 0011 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1101 0111 0001 1101 1100 1000 1111 1100 1001 0011 000(2) × 20 =

1.1000 1101 0110 1011 1000 1110 1110 0100 0111 1110 0100 1001 1000(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 0110 1011 1000 1110 1110 0100 0111 1110 0100 1001 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 0110 1011 1000 1110 1110 0100 0111 1110 0100 1001 1000 =

1000 1101 0110 1011 1000 1110 1110 0100 0111 1110 0100 1001 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 0100

Mantissa (52 bits) =
1000 1101 0110 1011 1000 1110 1110 0100 0111 1110 0100 1001 1000


Decimal number -0.000 000 000 000 176 49 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 0110 1011 1000 1110 1110 0100 0111 1110 0100 1001 1000

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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