-0.000 000 000 000 176 56 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 56(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 56(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 56| = 0.000 000 000 000 176 56


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 56.

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 56 × 2 = 0 + 0.000 000 000 000 353 12;
  • 2) 0.000 000 000 000 353 12 × 2 = 0 + 0.000 000 000 000 706 24;
  • 3) 0.000 000 000 000 706 24 × 2 = 0 + 0.000 000 000 001 412 48;
  • 4) 0.000 000 000 001 412 48 × 2 = 0 + 0.000 000 000 002 824 96;
  • 5) 0.000 000 000 002 824 96 × 2 = 0 + 0.000 000 000 005 649 92;
  • 6) 0.000 000 000 005 649 92 × 2 = 0 + 0.000 000 000 011 299 84;
  • 7) 0.000 000 000 011 299 84 × 2 = 0 + 0.000 000 000 022 599 68;
  • 8) 0.000 000 000 022 599 68 × 2 = 0 + 0.000 000 000 045 199 36;
  • 9) 0.000 000 000 045 199 36 × 2 = 0 + 0.000 000 000 090 398 72;
  • 10) 0.000 000 000 090 398 72 × 2 = 0 + 0.000 000 000 180 797 44;
  • 11) 0.000 000 000 180 797 44 × 2 = 0 + 0.000 000 000 361 594 88;
  • 12) 0.000 000 000 361 594 88 × 2 = 0 + 0.000 000 000 723 189 76;
  • 13) 0.000 000 000 723 189 76 × 2 = 0 + 0.000 000 001 446 379 52;
  • 14) 0.000 000 001 446 379 52 × 2 = 0 + 0.000 000 002 892 759 04;
  • 15) 0.000 000 002 892 759 04 × 2 = 0 + 0.000 000 005 785 518 08;
  • 16) 0.000 000 005 785 518 08 × 2 = 0 + 0.000 000 011 571 036 16;
  • 17) 0.000 000 011 571 036 16 × 2 = 0 + 0.000 000 023 142 072 32;
  • 18) 0.000 000 023 142 072 32 × 2 = 0 + 0.000 000 046 284 144 64;
  • 19) 0.000 000 046 284 144 64 × 2 = 0 + 0.000 000 092 568 289 28;
  • 20) 0.000 000 092 568 289 28 × 2 = 0 + 0.000 000 185 136 578 56;
  • 21) 0.000 000 185 136 578 56 × 2 = 0 + 0.000 000 370 273 157 12;
  • 22) 0.000 000 370 273 157 12 × 2 = 0 + 0.000 000 740 546 314 24;
  • 23) 0.000 000 740 546 314 24 × 2 = 0 + 0.000 001 481 092 628 48;
  • 24) 0.000 001 481 092 628 48 × 2 = 0 + 0.000 002 962 185 256 96;
  • 25) 0.000 002 962 185 256 96 × 2 = 0 + 0.000 005 924 370 513 92;
  • 26) 0.000 005 924 370 513 92 × 2 = 0 + 0.000 011 848 741 027 84;
  • 27) 0.000 011 848 741 027 84 × 2 = 0 + 0.000 023 697 482 055 68;
  • 28) 0.000 023 697 482 055 68 × 2 = 0 + 0.000 047 394 964 111 36;
  • 29) 0.000 047 394 964 111 36 × 2 = 0 + 0.000 094 789 928 222 72;
  • 30) 0.000 094 789 928 222 72 × 2 = 0 + 0.000 189 579 856 445 44;
  • 31) 0.000 189 579 856 445 44 × 2 = 0 + 0.000 379 159 712 890 88;
  • 32) 0.000 379 159 712 890 88 × 2 = 0 + 0.000 758 319 425 781 76;
  • 33) 0.000 758 319 425 781 76 × 2 = 0 + 0.001 516 638 851 563 52;
  • 34) 0.001 516 638 851 563 52 × 2 = 0 + 0.003 033 277 703 127 04;
  • 35) 0.003 033 277 703 127 04 × 2 = 0 + 0.006 066 555 406 254 08;
  • 36) 0.006 066 555 406 254 08 × 2 = 0 + 0.012 133 110 812 508 16;
  • 37) 0.012 133 110 812 508 16 × 2 = 0 + 0.024 266 221 625 016 32;
  • 38) 0.024 266 221 625 016 32 × 2 = 0 + 0.048 532 443 250 032 64;
  • 39) 0.048 532 443 250 032 64 × 2 = 0 + 0.097 064 886 500 065 28;
  • 40) 0.097 064 886 500 065 28 × 2 = 0 + 0.194 129 773 000 130 56;
  • 41) 0.194 129 773 000 130 56 × 2 = 0 + 0.388 259 546 000 261 12;
  • 42) 0.388 259 546 000 261 12 × 2 = 0 + 0.776 519 092 000 522 24;
  • 43) 0.776 519 092 000 522 24 × 2 = 1 + 0.553 038 184 001 044 48;
  • 44) 0.553 038 184 001 044 48 × 2 = 1 + 0.106 076 368 002 088 96;
  • 45) 0.106 076 368 002 088 96 × 2 = 0 + 0.212 152 736 004 177 92;
  • 46) 0.212 152 736 004 177 92 × 2 = 0 + 0.424 305 472 008 355 84;
  • 47) 0.424 305 472 008 355 84 × 2 = 0 + 0.848 610 944 016 711 68;
  • 48) 0.848 610 944 016 711 68 × 2 = 1 + 0.697 221 888 033 423 36;
  • 49) 0.697 221 888 033 423 36 × 2 = 1 + 0.394 443 776 066 846 72;
  • 50) 0.394 443 776 066 846 72 × 2 = 0 + 0.788 887 552 133 693 44;
  • 51) 0.788 887 552 133 693 44 × 2 = 1 + 0.577 775 104 267 386 88;
  • 52) 0.577 775 104 267 386 88 × 2 = 1 + 0.155 550 208 534 773 76;
  • 53) 0.155 550 208 534 773 76 × 2 = 0 + 0.311 100 417 069 547 52;
  • 54) 0.311 100 417 069 547 52 × 2 = 0 + 0.622 200 834 139 095 04;
  • 55) 0.622 200 834 139 095 04 × 2 = 1 + 0.244 401 668 278 190 08;
  • 56) 0.244 401 668 278 190 08 × 2 = 0 + 0.488 803 336 556 380 16;
  • 57) 0.488 803 336 556 380 16 × 2 = 0 + 0.977 606 673 112 760 32;
  • 58) 0.977 606 673 112 760 32 × 2 = 1 + 0.955 213 346 225 520 64;
  • 59) 0.955 213 346 225 520 64 × 2 = 1 + 0.910 426 692 451 041 28;
  • 60) 0.910 426 692 451 041 28 × 2 = 1 + 0.820 853 384 902 082 56;
  • 61) 0.820 853 384 902 082 56 × 2 = 1 + 0.641 706 769 804 165 12;
  • 62) 0.641 706 769 804 165 12 × 2 = 1 + 0.283 413 539 608 330 24;
  • 63) 0.283 413 539 608 330 24 × 2 = 0 + 0.566 827 079 216 660 48;
  • 64) 0.566 827 079 216 660 48 × 2 = 1 + 0.133 654 158 433 320 96;
  • 65) 0.133 654 158 433 320 96 × 2 = 0 + 0.267 308 316 866 641 92;
  • 66) 0.267 308 316 866 641 92 × 2 = 0 + 0.534 616 633 733 283 84;
  • 67) 0.534 616 633 733 283 84 × 2 = 1 + 0.069 233 267 466 567 68;
  • 68) 0.069 233 267 466 567 68 × 2 = 0 + 0.138 466 534 933 135 36;
  • 69) 0.138 466 534 933 135 36 × 2 = 0 + 0.276 933 069 866 270 72;
  • 70) 0.276 933 069 866 270 72 × 2 = 0 + 0.553 866 139 732 541 44;
  • 71) 0.553 866 139 732 541 44 × 2 = 1 + 0.107 732 279 465 082 88;
  • 72) 0.107 732 279 465 082 88 × 2 = 0 + 0.215 464 558 930 165 76;
  • 73) 0.215 464 558 930 165 76 × 2 = 0 + 0.430 929 117 860 331 52;
  • 74) 0.430 929 117 860 331 52 × 2 = 0 + 0.861 858 235 720 663 04;
  • 75) 0.861 858 235 720 663 04 × 2 = 1 + 0.723 716 471 441 326 08;
  • 76) 0.723 716 471 441 326 08 × 2 = 1 + 0.447 432 942 882 652 16;
  • 77) 0.447 432 942 882 652 16 × 2 = 0 + 0.894 865 885 765 304 32;
  • 78) 0.894 865 885 765 304 32 × 2 = 1 + 0.789 731 771 530 608 64;
  • 79) 0.789 731 771 530 608 64 × 2 = 1 + 0.579 463 543 061 217 28;
  • 80) 0.579 463 543 061 217 28 × 2 = 1 + 0.158 927 086 122 434 56;
  • 81) 0.158 927 086 122 434 56 × 2 = 0 + 0.317 854 172 244 869 12;
  • 82) 0.317 854 172 244 869 12 × 2 = 0 + 0.635 708 344 489 738 24;
  • 83) 0.635 708 344 489 738 24 × 2 = 1 + 0.271 416 688 979 476 48;
  • 84) 0.271 416 688 979 476 48 × 2 = 0 + 0.542 833 377 958 952 96;
  • 85) 0.542 833 377 958 952 96 × 2 = 1 + 0.085 666 755 917 905 92;
  • 86) 0.085 666 755 917 905 92 × 2 = 0 + 0.171 333 511 835 811 84;
  • 87) 0.171 333 511 835 811 84 × 2 = 0 + 0.342 667 023 671 623 68;
  • 88) 0.342 667 023 671 623 68 × 2 = 0 + 0.685 334 047 343 247 36;
  • 89) 0.685 334 047 343 247 36 × 2 = 1 + 0.370 668 094 686 494 72;
  • 90) 0.370 668 094 686 494 72 × 2 = 0 + 0.741 336 189 372 989 44;
  • 91) 0.741 336 189 372 989 44 × 2 = 1 + 0.482 672 378 745 978 88;
  • 92) 0.482 672 378 745 978 88 × 2 = 0 + 0.965 344 757 491 957 76;
  • 93) 0.965 344 757 491 957 76 × 2 = 1 + 0.930 689 514 983 915 52;
  • 94) 0.930 689 514 983 915 52 × 2 = 1 + 0.861 379 029 967 831 04;
  • 95) 0.861 379 029 967 831 04 × 2 = 1 + 0.722 758 059 935 662 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:


0.000 000 000 000 176 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0111 1101 0010 0010 0011 0111 0010 1000 1010 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0111 1101 0010 0010 0011 0111 0010 1000 1010 111(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 56(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0111 1101 0010 0010 0011 0111 0010 1000 1010 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0111 1101 0010 0010 0011 0111 0010 1000 1010 111(2) × 20 =

1.1000 1101 1001 0011 1110 1001 0001 0001 1011 1001 0100 0101 0111(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 1001 0011 1110 1001 0001 0001 1011 1001 0100 0101 0111


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 1001 0011 1110 1001 0001 0001 1011 1001 0100 0101 0111 =

1000 1101 1001 0011 1110 1001 0001 0001 1011 1001 0100 0101 0111


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 1001 0011 1110 1001 0001 0001 1011 1001 0100 0101 0111


Decimal number -0.000 000 000 000 176 56 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0011 1110 1001 0001 0001 1011 1001 0100 0101 0111

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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