-0.000 000 000 000 177 44 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 177 44(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
-0.000 000 000 000 177 44(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:

|-0.000 000 000 000 177 44| = 0.000 000 000 000 177 44


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: 0.000 000 000 000 177 44.

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) 0.000 000 000 000 177 44 × 2 = 0 + 0.000 000 000 000 354 88;
  • 2) 0.000 000 000 000 354 88 × 2 = 0 + 0.000 000 000 000 709 76;
  • 3) 0.000 000 000 000 709 76 × 2 = 0 + 0.000 000 000 001 419 52;
  • 4) 0.000 000 000 001 419 52 × 2 = 0 + 0.000 000 000 002 839 04;
  • 5) 0.000 000 000 002 839 04 × 2 = 0 + 0.000 000 000 005 678 08;
  • 6) 0.000 000 000 005 678 08 × 2 = 0 + 0.000 000 000 011 356 16;
  • 7) 0.000 000 000 011 356 16 × 2 = 0 + 0.000 000 000 022 712 32;
  • 8) 0.000 000 000 022 712 32 × 2 = 0 + 0.000 000 000 045 424 64;
  • 9) 0.000 000 000 045 424 64 × 2 = 0 + 0.000 000 000 090 849 28;
  • 10) 0.000 000 000 090 849 28 × 2 = 0 + 0.000 000 000 181 698 56;
  • 11) 0.000 000 000 181 698 56 × 2 = 0 + 0.000 000 000 363 397 12;
  • 12) 0.000 000 000 363 397 12 × 2 = 0 + 0.000 000 000 726 794 24;
  • 13) 0.000 000 000 726 794 24 × 2 = 0 + 0.000 000 001 453 588 48;
  • 14) 0.000 000 001 453 588 48 × 2 = 0 + 0.000 000 002 907 176 96;
  • 15) 0.000 000 002 907 176 96 × 2 = 0 + 0.000 000 005 814 353 92;
  • 16) 0.000 000 005 814 353 92 × 2 = 0 + 0.000 000 011 628 707 84;
  • 17) 0.000 000 011 628 707 84 × 2 = 0 + 0.000 000 023 257 415 68;
  • 18) 0.000 000 023 257 415 68 × 2 = 0 + 0.000 000 046 514 831 36;
  • 19) 0.000 000 046 514 831 36 × 2 = 0 + 0.000 000 093 029 662 72;
  • 20) 0.000 000 093 029 662 72 × 2 = 0 + 0.000 000 186 059 325 44;
  • 21) 0.000 000 186 059 325 44 × 2 = 0 + 0.000 000 372 118 650 88;
  • 22) 0.000 000 372 118 650 88 × 2 = 0 + 0.000 000 744 237 301 76;
  • 23) 0.000 000 744 237 301 76 × 2 = 0 + 0.000 001 488 474 603 52;
  • 24) 0.000 001 488 474 603 52 × 2 = 0 + 0.000 002 976 949 207 04;
  • 25) 0.000 002 976 949 207 04 × 2 = 0 + 0.000 005 953 898 414 08;
  • 26) 0.000 005 953 898 414 08 × 2 = 0 + 0.000 011 907 796 828 16;
  • 27) 0.000 011 907 796 828 16 × 2 = 0 + 0.000 023 815 593 656 32;
  • 28) 0.000 023 815 593 656 32 × 2 = 0 + 0.000 047 631 187 312 64;
  • 29) 0.000 047 631 187 312 64 × 2 = 0 + 0.000 095 262 374 625 28;
  • 30) 0.000 095 262 374 625 28 × 2 = 0 + 0.000 190 524 749 250 56;
  • 31) 0.000 190 524 749 250 56 × 2 = 0 + 0.000 381 049 498 501 12;
  • 32) 0.000 381 049 498 501 12 × 2 = 0 + 0.000 762 098 997 002 24;
  • 33) 0.000 762 098 997 002 24 × 2 = 0 + 0.001 524 197 994 004 48;
  • 34) 0.001 524 197 994 004 48 × 2 = 0 + 0.003 048 395 988 008 96;
  • 35) 0.003 048 395 988 008 96 × 2 = 0 + 0.006 096 791 976 017 92;
  • 36) 0.006 096 791 976 017 92 × 2 = 0 + 0.012 193 583 952 035 84;
  • 37) 0.012 193 583 952 035 84 × 2 = 0 + 0.024 387 167 904 071 68;
  • 38) 0.024 387 167 904 071 68 × 2 = 0 + 0.048 774 335 808 143 36;
  • 39) 0.048 774 335 808 143 36 × 2 = 0 + 0.097 548 671 616 286 72;
  • 40) 0.097 548 671 616 286 72 × 2 = 0 + 0.195 097 343 232 573 44;
  • 41) 0.195 097 343 232 573 44 × 2 = 0 + 0.390 194 686 465 146 88;
  • 42) 0.390 194 686 465 146 88 × 2 = 0 + 0.780 389 372 930 293 76;
  • 43) 0.780 389 372 930 293 76 × 2 = 1 + 0.560 778 745 860 587 52;
  • 44) 0.560 778 745 860 587 52 × 2 = 1 + 0.121 557 491 721 175 04;
  • 45) 0.121 557 491 721 175 04 × 2 = 0 + 0.243 114 983 442 350 08;
  • 46) 0.243 114 983 442 350 08 × 2 = 0 + 0.486 229 966 884 700 16;
  • 47) 0.486 229 966 884 700 16 × 2 = 0 + 0.972 459 933 769 400 32;
  • 48) 0.972 459 933 769 400 32 × 2 = 1 + 0.944 919 867 538 800 64;
  • 49) 0.944 919 867 538 800 64 × 2 = 1 + 0.889 839 735 077 601 28;
  • 50) 0.889 839 735 077 601 28 × 2 = 1 + 0.779 679 470 155 202 56;
  • 51) 0.779 679 470 155 202 56 × 2 = 1 + 0.559 358 940 310 405 12;
  • 52) 0.559 358 940 310 405 12 × 2 = 1 + 0.118 717 880 620 810 24;
  • 53) 0.118 717 880 620 810 24 × 2 = 0 + 0.237 435 761 241 620 48;
  • 54) 0.237 435 761 241 620 48 × 2 = 0 + 0.474 871 522 483 240 96;
  • 55) 0.474 871 522 483 240 96 × 2 = 0 + 0.949 743 044 966 481 92;
  • 56) 0.949 743 044 966 481 92 × 2 = 1 + 0.899 486 089 932 963 84;
  • 57) 0.899 486 089 932 963 84 × 2 = 1 + 0.798 972 179 865 927 68;
  • 58) 0.798 972 179 865 927 68 × 2 = 1 + 0.597 944 359 731 855 36;
  • 59) 0.597 944 359 731 855 36 × 2 = 1 + 0.195 888 719 463 710 72;
  • 60) 0.195 888 719 463 710 72 × 2 = 0 + 0.391 777 438 927 421 44;
  • 61) 0.391 777 438 927 421 44 × 2 = 0 + 0.783 554 877 854 842 88;
  • 62) 0.783 554 877 854 842 88 × 2 = 1 + 0.567 109 755 709 685 76;
  • 63) 0.567 109 755 709 685 76 × 2 = 1 + 0.134 219 511 419 371 52;
  • 64) 0.134 219 511 419 371 52 × 2 = 0 + 0.268 439 022 838 743 04;
  • 65) 0.268 439 022 838 743 04 × 2 = 0 + 0.536 878 045 677 486 08;
  • 66) 0.536 878 045 677 486 08 × 2 = 1 + 0.073 756 091 354 972 16;
  • 67) 0.073 756 091 354 972 16 × 2 = 0 + 0.147 512 182 709 944 32;
  • 68) 0.147 512 182 709 944 32 × 2 = 0 + 0.295 024 365 419 888 64;
  • 69) 0.295 024 365 419 888 64 × 2 = 0 + 0.590 048 730 839 777 28;
  • 70) 0.590 048 730 839 777 28 × 2 = 1 + 0.180 097 461 679 554 56;
  • 71) 0.180 097 461 679 554 56 × 2 = 0 + 0.360 194 923 359 109 12;
  • 72) 0.360 194 923 359 109 12 × 2 = 0 + 0.720 389 846 718 218 24;
  • 73) 0.720 389 846 718 218 24 × 2 = 1 + 0.440 779 693 436 436 48;
  • 74) 0.440 779 693 436 436 48 × 2 = 0 + 0.881 559 386 872 872 96;
  • 75) 0.881 559 386 872 872 96 × 2 = 1 + 0.763 118 773 745 745 92;
  • 76) 0.763 118 773 745 745 92 × 2 = 1 + 0.526 237 547 491 491 84;
  • 77) 0.526 237 547 491 491 84 × 2 = 1 + 0.052 475 094 982 983 68;
  • 78) 0.052 475 094 982 983 68 × 2 = 0 + 0.104 950 189 965 967 36;
  • 79) 0.104 950 189 965 967 36 × 2 = 0 + 0.209 900 379 931 934 72;
  • 80) 0.209 900 379 931 934 72 × 2 = 0 + 0.419 800 759 863 869 44;
  • 81) 0.419 800 759 863 869 44 × 2 = 0 + 0.839 601 519 727 738 88;
  • 82) 0.839 601 519 727 738 88 × 2 = 1 + 0.679 203 039 455 477 76;
  • 83) 0.679 203 039 455 477 76 × 2 = 1 + 0.358 406 078 910 955 52;
  • 84) 0.358 406 078 910 955 52 × 2 = 0 + 0.716 812 157 821 911 04;
  • 85) 0.716 812 157 821 911 04 × 2 = 1 + 0.433 624 315 643 822 08;
  • 86) 0.433 624 315 643 822 08 × 2 = 0 + 0.867 248 631 287 644 16;
  • 87) 0.867 248 631 287 644 16 × 2 = 1 + 0.734 497 262 575 288 32;
  • 88) 0.734 497 262 575 288 32 × 2 = 1 + 0.468 994 525 150 576 64;
  • 89) 0.468 994 525 150 576 64 × 2 = 0 + 0.937 989 050 301 153 28;
  • 90) 0.937 989 050 301 153 28 × 2 = 1 + 0.875 978 100 602 306 56;
  • 91) 0.875 978 100 602 306 56 × 2 = 1 + 0.751 956 201 204 613 12;
  • 92) 0.751 956 201 204 613 12 × 2 = 1 + 0.503 912 402 409 226 24;
  • 93) 0.503 912 402 409 226 24 × 2 = 1 + 0.007 824 804 818 452 48;
  • 94) 0.007 824 804 818 452 48 × 2 = 0 + 0.015 649 609 636 904 96;
  • 95) 0.015 649 609 636 904 96 × 2 = 0 + 0.031 299 219 273 809 92;

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:


0.000 000 000 000 177 44(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1111 0001 1110 0110 0100 0100 1011 1000 0110 1011 0111 100(2)

6. Positive number before normalization:

0.000 000 000 000 177 44(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1111 0001 1110 0110 0100 0100 1011 1000 0110 1011 0111 100(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:


0.000 000 000 000 177 44(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1111 0001 1110 0110 0100 0100 1011 1000 0110 1011 0111 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1111 0001 1110 0110 0100 0100 1011 1000 0110 1011 0111 100(2) × 20 =

1.1000 1111 1000 1111 0011 0010 0010 0101 1100 0011 0101 1011 1100(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 1111 1000 1111 0011 0010 0010 0101 1100 0011 0101 1011 1100


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 1111 1000 1111 0011 0010 0010 0101 1100 0011 0101 1011 1100 =

1000 1111 1000 1111 0011 0010 0010 0101 1100 0011 0101 1011 1100


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 1111 1000 1111 0011 0010 0010 0101 1100 0011 0101 1011 1100


Decimal number -0.000 000 000 000 177 44 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1111 1000 1111 0011 0010 0010 0101 1100 0011 0101 1011 1100

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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