-0.000 000 000 000 014 385 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 385(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
-0.000 000 000 000 014 385(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:

|-0.000 000 000 000 014 385| = 0.000 000 000 000 014 385


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: 0.000 000 000 000 014 385.

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) 0.000 000 000 000 014 385 × 2 = 0 + 0.000 000 000 000 028 77;
  • 2) 0.000 000 000 000 028 77 × 2 = 0 + 0.000 000 000 000 057 54;
  • 3) 0.000 000 000 000 057 54 × 2 = 0 + 0.000 000 000 000 115 08;
  • 4) 0.000 000 000 000 115 08 × 2 = 0 + 0.000 000 000 000 230 16;
  • 5) 0.000 000 000 000 230 16 × 2 = 0 + 0.000 000 000 000 460 32;
  • 6) 0.000 000 000 000 460 32 × 2 = 0 + 0.000 000 000 000 920 64;
  • 7) 0.000 000 000 000 920 64 × 2 = 0 + 0.000 000 000 001 841 28;
  • 8) 0.000 000 000 001 841 28 × 2 = 0 + 0.000 000 000 003 682 56;
  • 9) 0.000 000 000 003 682 56 × 2 = 0 + 0.000 000 000 007 365 12;
  • 10) 0.000 000 000 007 365 12 × 2 = 0 + 0.000 000 000 014 730 24;
  • 11) 0.000 000 000 014 730 24 × 2 = 0 + 0.000 000 000 029 460 48;
  • 12) 0.000 000 000 029 460 48 × 2 = 0 + 0.000 000 000 058 920 96;
  • 13) 0.000 000 000 058 920 96 × 2 = 0 + 0.000 000 000 117 841 92;
  • 14) 0.000 000 000 117 841 92 × 2 = 0 + 0.000 000 000 235 683 84;
  • 15) 0.000 000 000 235 683 84 × 2 = 0 + 0.000 000 000 471 367 68;
  • 16) 0.000 000 000 471 367 68 × 2 = 0 + 0.000 000 000 942 735 36;
  • 17) 0.000 000 000 942 735 36 × 2 = 0 + 0.000 000 001 885 470 72;
  • 18) 0.000 000 001 885 470 72 × 2 = 0 + 0.000 000 003 770 941 44;
  • 19) 0.000 000 003 770 941 44 × 2 = 0 + 0.000 000 007 541 882 88;
  • 20) 0.000 000 007 541 882 88 × 2 = 0 + 0.000 000 015 083 765 76;
  • 21) 0.000 000 015 083 765 76 × 2 = 0 + 0.000 000 030 167 531 52;
  • 22) 0.000 000 030 167 531 52 × 2 = 0 + 0.000 000 060 335 063 04;
  • 23) 0.000 000 060 335 063 04 × 2 = 0 + 0.000 000 120 670 126 08;
  • 24) 0.000 000 120 670 126 08 × 2 = 0 + 0.000 000 241 340 252 16;
  • 25) 0.000 000 241 340 252 16 × 2 = 0 + 0.000 000 482 680 504 32;
  • 26) 0.000 000 482 680 504 32 × 2 = 0 + 0.000 000 965 361 008 64;
  • 27) 0.000 000 965 361 008 64 × 2 = 0 + 0.000 001 930 722 017 28;
  • 28) 0.000 001 930 722 017 28 × 2 = 0 + 0.000 003 861 444 034 56;
  • 29) 0.000 003 861 444 034 56 × 2 = 0 + 0.000 007 722 888 069 12;
  • 30) 0.000 007 722 888 069 12 × 2 = 0 + 0.000 015 445 776 138 24;
  • 31) 0.000 015 445 776 138 24 × 2 = 0 + 0.000 030 891 552 276 48;
  • 32) 0.000 030 891 552 276 48 × 2 = 0 + 0.000 061 783 104 552 96;
  • 33) 0.000 061 783 104 552 96 × 2 = 0 + 0.000 123 566 209 105 92;
  • 34) 0.000 123 566 209 105 92 × 2 = 0 + 0.000 247 132 418 211 84;
  • 35) 0.000 247 132 418 211 84 × 2 = 0 + 0.000 494 264 836 423 68;
  • 36) 0.000 494 264 836 423 68 × 2 = 0 + 0.000 988 529 672 847 36;
  • 37) 0.000 988 529 672 847 36 × 2 = 0 + 0.001 977 059 345 694 72;
  • 38) 0.001 977 059 345 694 72 × 2 = 0 + 0.003 954 118 691 389 44;
  • 39) 0.003 954 118 691 389 44 × 2 = 0 + 0.007 908 237 382 778 88;
  • 40) 0.007 908 237 382 778 88 × 2 = 0 + 0.015 816 474 765 557 76;
  • 41) 0.015 816 474 765 557 76 × 2 = 0 + 0.031 632 949 531 115 52;
  • 42) 0.031 632 949 531 115 52 × 2 = 0 + 0.063 265 899 062 231 04;
  • 43) 0.063 265 899 062 231 04 × 2 = 0 + 0.126 531 798 124 462 08;
  • 44) 0.126 531 798 124 462 08 × 2 = 0 + 0.253 063 596 248 924 16;
  • 45) 0.253 063 596 248 924 16 × 2 = 0 + 0.506 127 192 497 848 32;
  • 46) 0.506 127 192 497 848 32 × 2 = 1 + 0.012 254 384 995 696 64;
  • 47) 0.012 254 384 995 696 64 × 2 = 0 + 0.024 508 769 991 393 28;
  • 48) 0.024 508 769 991 393 28 × 2 = 0 + 0.049 017 539 982 786 56;
  • 49) 0.049 017 539 982 786 56 × 2 = 0 + 0.098 035 079 965 573 12;
  • 50) 0.098 035 079 965 573 12 × 2 = 0 + 0.196 070 159 931 146 24;
  • 51) 0.196 070 159 931 146 24 × 2 = 0 + 0.392 140 319 862 292 48;
  • 52) 0.392 140 319 862 292 48 × 2 = 0 + 0.784 280 639 724 584 96;
  • 53) 0.784 280 639 724 584 96 × 2 = 1 + 0.568 561 279 449 169 92;
  • 54) 0.568 561 279 449 169 92 × 2 = 1 + 0.137 122 558 898 339 84;
  • 55) 0.137 122 558 898 339 84 × 2 = 0 + 0.274 245 117 796 679 68;
  • 56) 0.274 245 117 796 679 68 × 2 = 0 + 0.548 490 235 593 359 36;
  • 57) 0.548 490 235 593 359 36 × 2 = 1 + 0.096 980 471 186 718 72;
  • 58) 0.096 980 471 186 718 72 × 2 = 0 + 0.193 960 942 373 437 44;
  • 59) 0.193 960 942 373 437 44 × 2 = 0 + 0.387 921 884 746 874 88;
  • 60) 0.387 921 884 746 874 88 × 2 = 0 + 0.775 843 769 493 749 76;
  • 61) 0.775 843 769 493 749 76 × 2 = 1 + 0.551 687 538 987 499 52;
  • 62) 0.551 687 538 987 499 52 × 2 = 1 + 0.103 375 077 974 999 04;
  • 63) 0.103 375 077 974 999 04 × 2 = 0 + 0.206 750 155 949 998 08;
  • 64) 0.206 750 155 949 998 08 × 2 = 0 + 0.413 500 311 899 996 16;
  • 65) 0.413 500 311 899 996 16 × 2 = 0 + 0.827 000 623 799 992 32;
  • 66) 0.827 000 623 799 992 32 × 2 = 1 + 0.654 001 247 599 984 64;
  • 67) 0.654 001 247 599 984 64 × 2 = 1 + 0.308 002 495 199 969 28;
  • 68) 0.308 002 495 199 969 28 × 2 = 0 + 0.616 004 990 399 938 56;
  • 69) 0.616 004 990 399 938 56 × 2 = 1 + 0.232 009 980 799 877 12;
  • 70) 0.232 009 980 799 877 12 × 2 = 0 + 0.464 019 961 599 754 24;
  • 71) 0.464 019 961 599 754 24 × 2 = 0 + 0.928 039 923 199 508 48;
  • 72) 0.928 039 923 199 508 48 × 2 = 1 + 0.856 079 846 399 016 96;
  • 73) 0.856 079 846 399 016 96 × 2 = 1 + 0.712 159 692 798 033 92;
  • 74) 0.712 159 692 798 033 92 × 2 = 1 + 0.424 319 385 596 067 84;
  • 75) 0.424 319 385 596 067 84 × 2 = 0 + 0.848 638 771 192 135 68;
  • 76) 0.848 638 771 192 135 68 × 2 = 1 + 0.697 277 542 384 271 36;
  • 77) 0.697 277 542 384 271 36 × 2 = 1 + 0.394 555 084 768 542 72;
  • 78) 0.394 555 084 768 542 72 × 2 = 0 + 0.789 110 169 537 085 44;
  • 79) 0.789 110 169 537 085 44 × 2 = 1 + 0.578 220 339 074 170 88;
  • 80) 0.578 220 339 074 170 88 × 2 = 1 + 0.156 440 678 148 341 76;
  • 81) 0.156 440 678 148 341 76 × 2 = 0 + 0.312 881 356 296 683 52;
  • 82) 0.312 881 356 296 683 52 × 2 = 0 + 0.625 762 712 593 367 04;
  • 83) 0.625 762 712 593 367 04 × 2 = 1 + 0.251 525 425 186 734 08;
  • 84) 0.251 525 425 186 734 08 × 2 = 0 + 0.503 050 850 373 468 16;
  • 85) 0.503 050 850 373 468 16 × 2 = 1 + 0.006 101 700 746 936 32;
  • 86) 0.006 101 700 746 936 32 × 2 = 0 + 0.012 203 401 493 872 64;
  • 87) 0.012 203 401 493 872 64 × 2 = 0 + 0.024 406 802 987 745 28;
  • 88) 0.024 406 802 987 745 28 × 2 = 0 + 0.048 813 605 975 490 56;
  • 89) 0.048 813 605 975 490 56 × 2 = 0 + 0.097 627 211 950 981 12;
  • 90) 0.097 627 211 950 981 12 × 2 = 0 + 0.195 254 423 901 962 24;
  • 91) 0.195 254 423 901 962 24 × 2 = 0 + 0.390 508 847 803 924 48;
  • 92) 0.390 508 847 803 924 48 × 2 = 0 + 0.781 017 695 607 848 96;
  • 93) 0.781 017 695 607 848 96 × 2 = 1 + 0.562 035 391 215 697 92;
  • 94) 0.562 035 391 215 697 92 × 2 = 1 + 0.124 070 782 431 395 84;
  • 95) 0.124 070 782 431 395 84 × 2 = 0 + 0.248 141 564 862 791 68;
  • 96) 0.248 141 564 862 791 68 × 2 = 0 + 0.496 283 129 725 583 36;
  • 97) 0.496 283 129 725 583 36 × 2 = 0 + 0.992 566 259 451 166 72;
  • 98) 0.992 566 259 451 166 72 × 2 = 1 + 0.985 132 518 902 333 44;

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:


0.000 000 000 000 014 385(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1000 1100 0110 1001 1101 1011 0010 1000 0000 1100 01(2)

6. Positive number before normalization:

0.000 000 000 000 014 385(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1000 1100 0110 1001 1101 1011 0010 1000 0000 1100 01(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:


0.000 000 000 000 014 385(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1000 1100 0110 1001 1101 1011 0010 1000 0000 1100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1000 1100 0110 1001 1101 1011 0010 1000 0000 1100 01(2) × 20 =

1.0000 0011 0010 0011 0001 1010 0111 0110 1100 1010 0000 0011 0001(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 0011 0010 0011 0001 1010 0111 0110 1100 1010 0000 0011 0001


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 0011 0010 0011 0001 1010 0111 0110 1100 1010 0000 0011 0001 =

0000 0011 0010 0011 0001 1010 0111 0110 1100 1010 0000 0011 0001


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 0011 0010 0011 0001 1010 0111 0110 1100 1010 0000 0011 0001


Decimal number -0.000 000 000 000 014 385 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0010 0011 0001 1010 0111 0110 1100 1010 0000 0011 0001

» Convert decimal -0.000 000 000 000 014 41 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - 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