-0.000 000 000 000 014 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 37(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
-0.000 000 000 000 014 37(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:

|-0.000 000 000 000 014 37| = 0.000 000 000 000 014 37


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: 0.000 000 000 000 014 37.

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) 0.000 000 000 000 014 37 × 2 = 0 + 0.000 000 000 000 028 74;
  • 2) 0.000 000 000 000 028 74 × 2 = 0 + 0.000 000 000 000 057 48;
  • 3) 0.000 000 000 000 057 48 × 2 = 0 + 0.000 000 000 000 114 96;
  • 4) 0.000 000 000 000 114 96 × 2 = 0 + 0.000 000 000 000 229 92;
  • 5) 0.000 000 000 000 229 92 × 2 = 0 + 0.000 000 000 000 459 84;
  • 6) 0.000 000 000 000 459 84 × 2 = 0 + 0.000 000 000 000 919 68;
  • 7) 0.000 000 000 000 919 68 × 2 = 0 + 0.000 000 000 001 839 36;
  • 8) 0.000 000 000 001 839 36 × 2 = 0 + 0.000 000 000 003 678 72;
  • 9) 0.000 000 000 003 678 72 × 2 = 0 + 0.000 000 000 007 357 44;
  • 10) 0.000 000 000 007 357 44 × 2 = 0 + 0.000 000 000 014 714 88;
  • 11) 0.000 000 000 014 714 88 × 2 = 0 + 0.000 000 000 029 429 76;
  • 12) 0.000 000 000 029 429 76 × 2 = 0 + 0.000 000 000 058 859 52;
  • 13) 0.000 000 000 058 859 52 × 2 = 0 + 0.000 000 000 117 719 04;
  • 14) 0.000 000 000 117 719 04 × 2 = 0 + 0.000 000 000 235 438 08;
  • 15) 0.000 000 000 235 438 08 × 2 = 0 + 0.000 000 000 470 876 16;
  • 16) 0.000 000 000 470 876 16 × 2 = 0 + 0.000 000 000 941 752 32;
  • 17) 0.000 000 000 941 752 32 × 2 = 0 + 0.000 000 001 883 504 64;
  • 18) 0.000 000 001 883 504 64 × 2 = 0 + 0.000 000 003 767 009 28;
  • 19) 0.000 000 003 767 009 28 × 2 = 0 + 0.000 000 007 534 018 56;
  • 20) 0.000 000 007 534 018 56 × 2 = 0 + 0.000 000 015 068 037 12;
  • 21) 0.000 000 015 068 037 12 × 2 = 0 + 0.000 000 030 136 074 24;
  • 22) 0.000 000 030 136 074 24 × 2 = 0 + 0.000 000 060 272 148 48;
  • 23) 0.000 000 060 272 148 48 × 2 = 0 + 0.000 000 120 544 296 96;
  • 24) 0.000 000 120 544 296 96 × 2 = 0 + 0.000 000 241 088 593 92;
  • 25) 0.000 000 241 088 593 92 × 2 = 0 + 0.000 000 482 177 187 84;
  • 26) 0.000 000 482 177 187 84 × 2 = 0 + 0.000 000 964 354 375 68;
  • 27) 0.000 000 964 354 375 68 × 2 = 0 + 0.000 001 928 708 751 36;
  • 28) 0.000 001 928 708 751 36 × 2 = 0 + 0.000 003 857 417 502 72;
  • 29) 0.000 003 857 417 502 72 × 2 = 0 + 0.000 007 714 835 005 44;
  • 30) 0.000 007 714 835 005 44 × 2 = 0 + 0.000 015 429 670 010 88;
  • 31) 0.000 015 429 670 010 88 × 2 = 0 + 0.000 030 859 340 021 76;
  • 32) 0.000 030 859 340 021 76 × 2 = 0 + 0.000 061 718 680 043 52;
  • 33) 0.000 061 718 680 043 52 × 2 = 0 + 0.000 123 437 360 087 04;
  • 34) 0.000 123 437 360 087 04 × 2 = 0 + 0.000 246 874 720 174 08;
  • 35) 0.000 246 874 720 174 08 × 2 = 0 + 0.000 493 749 440 348 16;
  • 36) 0.000 493 749 440 348 16 × 2 = 0 + 0.000 987 498 880 696 32;
  • 37) 0.000 987 498 880 696 32 × 2 = 0 + 0.001 974 997 761 392 64;
  • 38) 0.001 974 997 761 392 64 × 2 = 0 + 0.003 949 995 522 785 28;
  • 39) 0.003 949 995 522 785 28 × 2 = 0 + 0.007 899 991 045 570 56;
  • 40) 0.007 899 991 045 570 56 × 2 = 0 + 0.015 799 982 091 141 12;
  • 41) 0.015 799 982 091 141 12 × 2 = 0 + 0.031 599 964 182 282 24;
  • 42) 0.031 599 964 182 282 24 × 2 = 0 + 0.063 199 928 364 564 48;
  • 43) 0.063 199 928 364 564 48 × 2 = 0 + 0.126 399 856 729 128 96;
  • 44) 0.126 399 856 729 128 96 × 2 = 0 + 0.252 799 713 458 257 92;
  • 45) 0.252 799 713 458 257 92 × 2 = 0 + 0.505 599 426 916 515 84;
  • 46) 0.505 599 426 916 515 84 × 2 = 1 + 0.011 198 853 833 031 68;
  • 47) 0.011 198 853 833 031 68 × 2 = 0 + 0.022 397 707 666 063 36;
  • 48) 0.022 397 707 666 063 36 × 2 = 0 + 0.044 795 415 332 126 72;
  • 49) 0.044 795 415 332 126 72 × 2 = 0 + 0.089 590 830 664 253 44;
  • 50) 0.089 590 830 664 253 44 × 2 = 0 + 0.179 181 661 328 506 88;
  • 51) 0.179 181 661 328 506 88 × 2 = 0 + 0.358 363 322 657 013 76;
  • 52) 0.358 363 322 657 013 76 × 2 = 0 + 0.716 726 645 314 027 52;
  • 53) 0.716 726 645 314 027 52 × 2 = 1 + 0.433 453 290 628 055 04;
  • 54) 0.433 453 290 628 055 04 × 2 = 0 + 0.866 906 581 256 110 08;
  • 55) 0.866 906 581 256 110 08 × 2 = 1 + 0.733 813 162 512 220 16;
  • 56) 0.733 813 162 512 220 16 × 2 = 1 + 0.467 626 325 024 440 32;
  • 57) 0.467 626 325 024 440 32 × 2 = 0 + 0.935 252 650 048 880 64;
  • 58) 0.935 252 650 048 880 64 × 2 = 1 + 0.870 505 300 097 761 28;
  • 59) 0.870 505 300 097 761 28 × 2 = 1 + 0.741 010 600 195 522 56;
  • 60) 0.741 010 600 195 522 56 × 2 = 1 + 0.482 021 200 391 045 12;
  • 61) 0.482 021 200 391 045 12 × 2 = 0 + 0.964 042 400 782 090 24;
  • 62) 0.964 042 400 782 090 24 × 2 = 1 + 0.928 084 801 564 180 48;
  • 63) 0.928 084 801 564 180 48 × 2 = 1 + 0.856 169 603 128 360 96;
  • 64) 0.856 169 603 128 360 96 × 2 = 1 + 0.712 339 206 256 721 92;
  • 65) 0.712 339 206 256 721 92 × 2 = 1 + 0.424 678 412 513 443 84;
  • 66) 0.424 678 412 513 443 84 × 2 = 0 + 0.849 356 825 026 887 68;
  • 67) 0.849 356 825 026 887 68 × 2 = 1 + 0.698 713 650 053 775 36;
  • 68) 0.698 713 650 053 775 36 × 2 = 1 + 0.397 427 300 107 550 72;
  • 69) 0.397 427 300 107 550 72 × 2 = 0 + 0.794 854 600 215 101 44;
  • 70) 0.794 854 600 215 101 44 × 2 = 1 + 0.589 709 200 430 202 88;
  • 71) 0.589 709 200 430 202 88 × 2 = 1 + 0.179 418 400 860 405 76;
  • 72) 0.179 418 400 860 405 76 × 2 = 0 + 0.358 836 801 720 811 52;
  • 73) 0.358 836 801 720 811 52 × 2 = 0 + 0.717 673 603 441 623 04;
  • 74) 0.717 673 603 441 623 04 × 2 = 1 + 0.435 347 206 883 246 08;
  • 75) 0.435 347 206 883 246 08 × 2 = 0 + 0.870 694 413 766 492 16;
  • 76) 0.870 694 413 766 492 16 × 2 = 1 + 0.741 388 827 532 984 32;
  • 77) 0.741 388 827 532 984 32 × 2 = 1 + 0.482 777 655 065 968 64;
  • 78) 0.482 777 655 065 968 64 × 2 = 0 + 0.965 555 310 131 937 28;
  • 79) 0.965 555 310 131 937 28 × 2 = 1 + 0.931 110 620 263 874 56;
  • 80) 0.931 110 620 263 874 56 × 2 = 1 + 0.862 221 240 527 749 12;
  • 81) 0.862 221 240 527 749 12 × 2 = 1 + 0.724 442 481 055 498 24;
  • 82) 0.724 442 481 055 498 24 × 2 = 1 + 0.448 884 962 110 996 48;
  • 83) 0.448 884 962 110 996 48 × 2 = 0 + 0.897 769 924 221 992 96;
  • 84) 0.897 769 924 221 992 96 × 2 = 1 + 0.795 539 848 443 985 92;
  • 85) 0.795 539 848 443 985 92 × 2 = 1 + 0.591 079 696 887 971 84;
  • 86) 0.591 079 696 887 971 84 × 2 = 1 + 0.182 159 393 775 943 68;
  • 87) 0.182 159 393 775 943 68 × 2 = 0 + 0.364 318 787 551 887 36;
  • 88) 0.364 318 787 551 887 36 × 2 = 0 + 0.728 637 575 103 774 72;
  • 89) 0.728 637 575 103 774 72 × 2 = 1 + 0.457 275 150 207 549 44;
  • 90) 0.457 275 150 207 549 44 × 2 = 0 + 0.914 550 300 415 098 88;
  • 91) 0.914 550 300 415 098 88 × 2 = 1 + 0.829 100 600 830 197 76;
  • 92) 0.829 100 600 830 197 76 × 2 = 1 + 0.658 201 201 660 395 52;
  • 93) 0.658 201 201 660 395 52 × 2 = 1 + 0.316 402 403 320 791 04;
  • 94) 0.316 402 403 320 791 04 × 2 = 0 + 0.632 804 806 641 582 08;
  • 95) 0.632 804 806 641 582 08 × 2 = 1 + 0.265 609 613 283 164 16;
  • 96) 0.265 609 613 283 164 16 × 2 = 0 + 0.531 219 226 566 328 32;
  • 97) 0.531 219 226 566 328 32 × 2 = 1 + 0.062 438 453 132 656 64;
  • 98) 0.062 438 453 132 656 64 × 2 = 0 + 0.124 876 906 265 313 28;

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:


0.000 000 000 000 014 37(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0111 0111 1011 0110 0101 1011 1101 1100 1011 1010 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 37(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0111 0111 1011 0110 0101 1011 1101 1100 1011 1010 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 014 37(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0111 0111 1011 0110 0101 1011 1101 1100 1011 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0111 0111 1011 0110 0101 1011 1101 1100 1011 1010 10(2) × 20 =

1.0000 0010 1101 1101 1110 1101 1001 0110 1111 0111 0010 1110 1010(2) × 2-46


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): -46

Mantissa (not normalized):
1.0000 0010 1101 1101 1110 1101 1001 0110 1111 0111 0010 1110 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 0000 0010 1101 1101 1110 1101 1001 0110 1111 0111 0010 1110 1010 =

0000 0010 1101 1101 1110 1101 1001 0110 1111 0111 0010 1110 1010


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 0001

Mantissa (52 bits) =
0000 0010 1101 1101 1110 1101 1001 0110 1111 0111 0010 1110 1010


Decimal number -0.000 000 000 000 014 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1101 1101 1110 1101 1001 0110 1111 0111 0010 1110 1010

» Convert decimal -0.000 000 000 000 014 22 to 64 bit double precision IEEE 754 binary floating point representation standard

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» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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