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

Convert decimal -0.000 000 000 000 014 64(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 64(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 64| = 0.000 000 000 000 014 64


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 64.

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 64 × 2 = 0 + 0.000 000 000 000 029 28;
  • 2) 0.000 000 000 000 029 28 × 2 = 0 + 0.000 000 000 000 058 56;
  • 3) 0.000 000 000 000 058 56 × 2 = 0 + 0.000 000 000 000 117 12;
  • 4) 0.000 000 000 000 117 12 × 2 = 0 + 0.000 000 000 000 234 24;
  • 5) 0.000 000 000 000 234 24 × 2 = 0 + 0.000 000 000 000 468 48;
  • 6) 0.000 000 000 000 468 48 × 2 = 0 + 0.000 000 000 000 936 96;
  • 7) 0.000 000 000 000 936 96 × 2 = 0 + 0.000 000 000 001 873 92;
  • 8) 0.000 000 000 001 873 92 × 2 = 0 + 0.000 000 000 003 747 84;
  • 9) 0.000 000 000 003 747 84 × 2 = 0 + 0.000 000 000 007 495 68;
  • 10) 0.000 000 000 007 495 68 × 2 = 0 + 0.000 000 000 014 991 36;
  • 11) 0.000 000 000 014 991 36 × 2 = 0 + 0.000 000 000 029 982 72;
  • 12) 0.000 000 000 029 982 72 × 2 = 0 + 0.000 000 000 059 965 44;
  • 13) 0.000 000 000 059 965 44 × 2 = 0 + 0.000 000 000 119 930 88;
  • 14) 0.000 000 000 119 930 88 × 2 = 0 + 0.000 000 000 239 861 76;
  • 15) 0.000 000 000 239 861 76 × 2 = 0 + 0.000 000 000 479 723 52;
  • 16) 0.000 000 000 479 723 52 × 2 = 0 + 0.000 000 000 959 447 04;
  • 17) 0.000 000 000 959 447 04 × 2 = 0 + 0.000 000 001 918 894 08;
  • 18) 0.000 000 001 918 894 08 × 2 = 0 + 0.000 000 003 837 788 16;
  • 19) 0.000 000 003 837 788 16 × 2 = 0 + 0.000 000 007 675 576 32;
  • 20) 0.000 000 007 675 576 32 × 2 = 0 + 0.000 000 015 351 152 64;
  • 21) 0.000 000 015 351 152 64 × 2 = 0 + 0.000 000 030 702 305 28;
  • 22) 0.000 000 030 702 305 28 × 2 = 0 + 0.000 000 061 404 610 56;
  • 23) 0.000 000 061 404 610 56 × 2 = 0 + 0.000 000 122 809 221 12;
  • 24) 0.000 000 122 809 221 12 × 2 = 0 + 0.000 000 245 618 442 24;
  • 25) 0.000 000 245 618 442 24 × 2 = 0 + 0.000 000 491 236 884 48;
  • 26) 0.000 000 491 236 884 48 × 2 = 0 + 0.000 000 982 473 768 96;
  • 27) 0.000 000 982 473 768 96 × 2 = 0 + 0.000 001 964 947 537 92;
  • 28) 0.000 001 964 947 537 92 × 2 = 0 + 0.000 003 929 895 075 84;
  • 29) 0.000 003 929 895 075 84 × 2 = 0 + 0.000 007 859 790 151 68;
  • 30) 0.000 007 859 790 151 68 × 2 = 0 + 0.000 015 719 580 303 36;
  • 31) 0.000 015 719 580 303 36 × 2 = 0 + 0.000 031 439 160 606 72;
  • 32) 0.000 031 439 160 606 72 × 2 = 0 + 0.000 062 878 321 213 44;
  • 33) 0.000 062 878 321 213 44 × 2 = 0 + 0.000 125 756 642 426 88;
  • 34) 0.000 125 756 642 426 88 × 2 = 0 + 0.000 251 513 284 853 76;
  • 35) 0.000 251 513 284 853 76 × 2 = 0 + 0.000 503 026 569 707 52;
  • 36) 0.000 503 026 569 707 52 × 2 = 0 + 0.001 006 053 139 415 04;
  • 37) 0.001 006 053 139 415 04 × 2 = 0 + 0.002 012 106 278 830 08;
  • 38) 0.002 012 106 278 830 08 × 2 = 0 + 0.004 024 212 557 660 16;
  • 39) 0.004 024 212 557 660 16 × 2 = 0 + 0.008 048 425 115 320 32;
  • 40) 0.008 048 425 115 320 32 × 2 = 0 + 0.016 096 850 230 640 64;
  • 41) 0.016 096 850 230 640 64 × 2 = 0 + 0.032 193 700 461 281 28;
  • 42) 0.032 193 700 461 281 28 × 2 = 0 + 0.064 387 400 922 562 56;
  • 43) 0.064 387 400 922 562 56 × 2 = 0 + 0.128 774 801 845 125 12;
  • 44) 0.128 774 801 845 125 12 × 2 = 0 + 0.257 549 603 690 250 24;
  • 45) 0.257 549 603 690 250 24 × 2 = 0 + 0.515 099 207 380 500 48;
  • 46) 0.515 099 207 380 500 48 × 2 = 1 + 0.030 198 414 761 000 96;
  • 47) 0.030 198 414 761 000 96 × 2 = 0 + 0.060 396 829 522 001 92;
  • 48) 0.060 396 829 522 001 92 × 2 = 0 + 0.120 793 659 044 003 84;
  • 49) 0.120 793 659 044 003 84 × 2 = 0 + 0.241 587 318 088 007 68;
  • 50) 0.241 587 318 088 007 68 × 2 = 0 + 0.483 174 636 176 015 36;
  • 51) 0.483 174 636 176 015 36 × 2 = 0 + 0.966 349 272 352 030 72;
  • 52) 0.966 349 272 352 030 72 × 2 = 1 + 0.932 698 544 704 061 44;
  • 53) 0.932 698 544 704 061 44 × 2 = 1 + 0.865 397 089 408 122 88;
  • 54) 0.865 397 089 408 122 88 × 2 = 1 + 0.730 794 178 816 245 76;
  • 55) 0.730 794 178 816 245 76 × 2 = 1 + 0.461 588 357 632 491 52;
  • 56) 0.461 588 357 632 491 52 × 2 = 0 + 0.923 176 715 264 983 04;
  • 57) 0.923 176 715 264 983 04 × 2 = 1 + 0.846 353 430 529 966 08;
  • 58) 0.846 353 430 529 966 08 × 2 = 1 + 0.692 706 861 059 932 16;
  • 59) 0.692 706 861 059 932 16 × 2 = 1 + 0.385 413 722 119 864 32;
  • 60) 0.385 413 722 119 864 32 × 2 = 0 + 0.770 827 444 239 728 64;
  • 61) 0.770 827 444 239 728 64 × 2 = 1 + 0.541 654 888 479 457 28;
  • 62) 0.541 654 888 479 457 28 × 2 = 1 + 0.083 309 776 958 914 56;
  • 63) 0.083 309 776 958 914 56 × 2 = 0 + 0.166 619 553 917 829 12;
  • 64) 0.166 619 553 917 829 12 × 2 = 0 + 0.333 239 107 835 658 24;
  • 65) 0.333 239 107 835 658 24 × 2 = 0 + 0.666 478 215 671 316 48;
  • 66) 0.666 478 215 671 316 48 × 2 = 1 + 0.332 956 431 342 632 96;
  • 67) 0.332 956 431 342 632 96 × 2 = 0 + 0.665 912 862 685 265 92;
  • 68) 0.665 912 862 685 265 92 × 2 = 1 + 0.331 825 725 370 531 84;
  • 69) 0.331 825 725 370 531 84 × 2 = 0 + 0.663 651 450 741 063 68;
  • 70) 0.663 651 450 741 063 68 × 2 = 1 + 0.327 302 901 482 127 36;
  • 71) 0.327 302 901 482 127 36 × 2 = 0 + 0.654 605 802 964 254 72;
  • 72) 0.654 605 802 964 254 72 × 2 = 1 + 0.309 211 605 928 509 44;
  • 73) 0.309 211 605 928 509 44 × 2 = 0 + 0.618 423 211 857 018 88;
  • 74) 0.618 423 211 857 018 88 × 2 = 1 + 0.236 846 423 714 037 76;
  • 75) 0.236 846 423 714 037 76 × 2 = 0 + 0.473 692 847 428 075 52;
  • 76) 0.473 692 847 428 075 52 × 2 = 0 + 0.947 385 694 856 151 04;
  • 77) 0.947 385 694 856 151 04 × 2 = 1 + 0.894 771 389 712 302 08;
  • 78) 0.894 771 389 712 302 08 × 2 = 1 + 0.789 542 779 424 604 16;
  • 79) 0.789 542 779 424 604 16 × 2 = 1 + 0.579 085 558 849 208 32;
  • 80) 0.579 085 558 849 208 32 × 2 = 1 + 0.158 171 117 698 416 64;
  • 81) 0.158 171 117 698 416 64 × 2 = 0 + 0.316 342 235 396 833 28;
  • 82) 0.316 342 235 396 833 28 × 2 = 0 + 0.632 684 470 793 666 56;
  • 83) 0.632 684 470 793 666 56 × 2 = 1 + 0.265 368 941 587 333 12;
  • 84) 0.265 368 941 587 333 12 × 2 = 0 + 0.530 737 883 174 666 24;
  • 85) 0.530 737 883 174 666 24 × 2 = 1 + 0.061 475 766 349 332 48;
  • 86) 0.061 475 766 349 332 48 × 2 = 0 + 0.122 951 532 698 664 96;
  • 87) 0.122 951 532 698 664 96 × 2 = 0 + 0.245 903 065 397 329 92;
  • 88) 0.245 903 065 397 329 92 × 2 = 0 + 0.491 806 130 794 659 84;
  • 89) 0.491 806 130 794 659 84 × 2 = 0 + 0.983 612 261 589 319 68;
  • 90) 0.983 612 261 589 319 68 × 2 = 1 + 0.967 224 523 178 639 36;
  • 91) 0.967 224 523 178 639 36 × 2 = 1 + 0.934 449 046 357 278 72;
  • 92) 0.934 449 046 357 278 72 × 2 = 1 + 0.868 898 092 714 557 44;
  • 93) 0.868 898 092 714 557 44 × 2 = 1 + 0.737 796 185 429 114 88;
  • 94) 0.737 796 185 429 114 88 × 2 = 1 + 0.475 592 370 858 229 76;
  • 95) 0.475 592 370 858 229 76 × 2 = 0 + 0.951 184 741 716 459 52;
  • 96) 0.951 184 741 716 459 52 × 2 = 1 + 0.902 369 483 432 919 04;
  • 97) 0.902 369 483 432 919 04 × 2 = 1 + 0.804 738 966 865 838 08;
  • 98) 0.804 738 966 865 838 08 × 2 = 1 + 0.609 477 933 731 676 16;

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 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0001 1110 1110 1100 0101 0101 0100 1111 0010 1000 0111 1101 11(2)

6. Positive number before normalization:

0.000 000 000 000 014 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0001 1110 1110 1100 0101 0101 0100 1111 0010 1000 0111 1101 11(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 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0001 1110 1110 1100 0101 0101 0100 1111 0010 1000 0111 1101 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0001 1110 1110 1100 0101 0101 0100 1111 0010 1000 0111 1101 11(2) × 20 =

1.0000 0111 1011 1011 0001 0101 0101 0011 1100 1010 0001 1111 0111(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 0111 1011 1011 0001 0101 0101 0011 1100 1010 0001 1111 0111


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 0111 1011 1011 0001 0101 0101 0011 1100 1010 0001 1111 0111 =

0000 0111 1011 1011 0001 0101 0101 0011 1100 1010 0001 1111 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 0001

Mantissa (52 bits) =
0000 0111 1011 1011 0001 0101 0101 0011 1100 1010 0001 1111 0111


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

1 - 011 1101 0001 - 0000 0111 1011 1011 0001 0101 0101 0011 1100 1010 0001 1111 0111

» Convert decimal -0.000 000 000 000 015 07 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