-81 869.602 684 384 910 389 780 86 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -81 869.602 684 384 910 389 780 86(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
-81 869.602 684 384 910 389 780 86(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:

|-81 869.602 684 384 910 389 780 86| = 81 869.602 684 384 910 389 780 86


2. First, convert to binary (in base 2) the integer part: 81 869.
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;
  • 81 869 ÷ 2 = 40 934 + 1;
  • 40 934 ÷ 2 = 20 467 + 0;
  • 20 467 ÷ 2 = 10 233 + 1;
  • 10 233 ÷ 2 = 5 116 + 1;
  • 5 116 ÷ 2 = 2 558 + 0;
  • 2 558 ÷ 2 = 1 279 + 0;
  • 1 279 ÷ 2 = 639 + 1;
  • 639 ÷ 2 = 319 + 1;
  • 319 ÷ 2 = 159 + 1;
  • 159 ÷ 2 = 79 + 1;
  • 79 ÷ 2 = 39 + 1;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

81 869(10) =

1 0011 1111 1100 1101(2)


4. Convert to binary (base 2) the fractional part: 0.602 684 384 910 389 780 86.

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) 0.602 684 384 910 389 780 86 × 2 = 1 + 0.205 368 769 820 779 561 72;
  • 2) 0.205 368 769 820 779 561 72 × 2 = 0 + 0.410 737 539 641 559 123 44;
  • 3) 0.410 737 539 641 559 123 44 × 2 = 0 + 0.821 475 079 283 118 246 88;
  • 4) 0.821 475 079 283 118 246 88 × 2 = 1 + 0.642 950 158 566 236 493 76;
  • 5) 0.642 950 158 566 236 493 76 × 2 = 1 + 0.285 900 317 132 472 987 52;
  • 6) 0.285 900 317 132 472 987 52 × 2 = 0 + 0.571 800 634 264 945 975 04;
  • 7) 0.571 800 634 264 945 975 04 × 2 = 1 + 0.143 601 268 529 891 950 08;
  • 8) 0.143 601 268 529 891 950 08 × 2 = 0 + 0.287 202 537 059 783 900 16;
  • 9) 0.287 202 537 059 783 900 16 × 2 = 0 + 0.574 405 074 119 567 800 32;
  • 10) 0.574 405 074 119 567 800 32 × 2 = 1 + 0.148 810 148 239 135 600 64;
  • 11) 0.148 810 148 239 135 600 64 × 2 = 0 + 0.297 620 296 478 271 201 28;
  • 12) 0.297 620 296 478 271 201 28 × 2 = 0 + 0.595 240 592 956 542 402 56;
  • 13) 0.595 240 592 956 542 402 56 × 2 = 1 + 0.190 481 185 913 084 805 12;
  • 14) 0.190 481 185 913 084 805 12 × 2 = 0 + 0.380 962 371 826 169 610 24;
  • 15) 0.380 962 371 826 169 610 24 × 2 = 0 + 0.761 924 743 652 339 220 48;
  • 16) 0.761 924 743 652 339 220 48 × 2 = 1 + 0.523 849 487 304 678 440 96;
  • 17) 0.523 849 487 304 678 440 96 × 2 = 1 + 0.047 698 974 609 356 881 92;
  • 18) 0.047 698 974 609 356 881 92 × 2 = 0 + 0.095 397 949 218 713 763 84;
  • 19) 0.095 397 949 218 713 763 84 × 2 = 0 + 0.190 795 898 437 427 527 68;
  • 20) 0.190 795 898 437 427 527 68 × 2 = 0 + 0.381 591 796 874 855 055 36;
  • 21) 0.381 591 796 874 855 055 36 × 2 = 0 + 0.763 183 593 749 710 110 72;
  • 22) 0.763 183 593 749 710 110 72 × 2 = 1 + 0.526 367 187 499 420 221 44;
  • 23) 0.526 367 187 499 420 221 44 × 2 = 1 + 0.052 734 374 998 840 442 88;
  • 24) 0.052 734 374 998 840 442 88 × 2 = 0 + 0.105 468 749 997 680 885 76;
  • 25) 0.105 468 749 997 680 885 76 × 2 = 0 + 0.210 937 499 995 361 771 52;
  • 26) 0.210 937 499 995 361 771 52 × 2 = 0 + 0.421 874 999 990 723 543 04;
  • 27) 0.421 874 999 990 723 543 04 × 2 = 0 + 0.843 749 999 981 447 086 08;
  • 28) 0.843 749 999 981 447 086 08 × 2 = 1 + 0.687 499 999 962 894 172 16;
  • 29) 0.687 499 999 962 894 172 16 × 2 = 1 + 0.374 999 999 925 788 344 32;
  • 30) 0.374 999 999 925 788 344 32 × 2 = 0 + 0.749 999 999 851 576 688 64;
  • 31) 0.749 999 999 851 576 688 64 × 2 = 1 + 0.499 999 999 703 153 377 28;
  • 32) 0.499 999 999 703 153 377 28 × 2 = 0 + 0.999 999 999 406 306 754 56;
  • 33) 0.999 999 999 406 306 754 56 × 2 = 1 + 0.999 999 998 812 613 509 12;
  • 34) 0.999 999 998 812 613 509 12 × 2 = 1 + 0.999 999 997 625 227 018 24;
  • 35) 0.999 999 997 625 227 018 24 × 2 = 1 + 0.999 999 995 250 454 036 48;
  • 36) 0.999 999 995 250 454 036 48 × 2 = 1 + 0.999 999 990 500 908 072 96;
  • 37) 0.999 999 990 500 908 072 96 × 2 = 1 + 0.999 999 981 001 816 145 92;
  • 38) 0.999 999 981 001 816 145 92 × 2 = 1 + 0.999 999 962 003 632 291 84;
  • 39) 0.999 999 962 003 632 291 84 × 2 = 1 + 0.999 999 924 007 264 583 68;
  • 40) 0.999 999 924 007 264 583 68 × 2 = 1 + 0.999 999 848 014 529 167 36;
  • 41) 0.999 999 848 014 529 167 36 × 2 = 1 + 0.999 999 696 029 058 334 72;
  • 42) 0.999 999 696 029 058 334 72 × 2 = 1 + 0.999 999 392 058 116 669 44;
  • 43) 0.999 999 392 058 116 669 44 × 2 = 1 + 0.999 998 784 116 233 338 88;
  • 44) 0.999 998 784 116 233 338 88 × 2 = 1 + 0.999 997 568 232 466 677 76;
  • 45) 0.999 997 568 232 466 677 76 × 2 = 1 + 0.999 995 136 464 933 355 52;
  • 46) 0.999 995 136 464 933 355 52 × 2 = 1 + 0.999 990 272 929 866 711 04;
  • 47) 0.999 990 272 929 866 711 04 × 2 = 1 + 0.999 980 545 859 733 422 08;
  • 48) 0.999 980 545 859 733 422 08 × 2 = 1 + 0.999 961 091 719 466 844 16;
  • 49) 0.999 961 091 719 466 844 16 × 2 = 1 + 0.999 922 183 438 933 688 32;
  • 50) 0.999 922 183 438 933 688 32 × 2 = 1 + 0.999 844 366 877 867 376 64;
  • 51) 0.999 844 366 877 867 376 64 × 2 = 1 + 0.999 688 733 755 734 753 28;
  • 52) 0.999 688 733 755 734 753 28 × 2 = 1 + 0.999 377 467 511 469 506 56;
  • 53) 0.999 377 467 511 469 506 56 × 2 = 1 + 0.998 754 935 022 939 013 12;

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:


0.602 684 384 910 389 780 86(10) =

0.1001 1010 0100 1001 1000 0110 0001 1010 1111 1111 1111 1111 1111 1(2)

6. Positive number before normalization:

81 869.602 684 384 910 389 780 86(10) =

1 0011 1111 1100 1101.1001 1010 0100 1001 1000 0110 0001 1010 1111 1111 1111 1111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the left, so that only one non zero digit remains to the left of it:


81 869.602 684 384 910 389 780 86(10) =

1 0011 1111 1100 1101.1001 1010 0100 1001 1000 0110 0001 1010 1111 1111 1111 1111 1111 1(2) =

1 0011 1111 1100 1101.1001 1010 0100 1001 1000 0110 0001 1010 1111 1111 1111 1111 1111 1(2) × 20 =

1.0011 1111 1100 1101 1001 1010 0100 1001 1000 0110 0001 1010 1111 1111 1111 1111 1111 1(2) × 216


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): 16

Mantissa (not normalized):
1.0011 1111 1100 1101 1001 1010 0100 1001 1000 0110 0001 1010 1111 1111 1111 1111 1111 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

16 + 2(11-1) - 1 =

(16 + 1 023)(10) =

1 039(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;
  • 1 039 ÷ 2 = 519 + 1;
  • 519 ÷ 2 = 259 + 1;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1039(10) =

100 0000 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0011 1111 1100 1101 1001 1010 0100 1001 1000 0110 0001 1010 1111 1 1111 1111 1111 1111 =

0011 1111 1100 1101 1001 1010 0100 1001 1000 0110 0001 1010 1111


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) =
100 0000 1111

Mantissa (52 bits) =
0011 1111 1100 1101 1001 1010 0100 1001 1000 0110 0001 1010 1111


Decimal number -81 869.602 684 384 910 389 780 86 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 1111 - 0011 1111 1100 1101 1001 1010 0100 1001 1000 0110 0001 1010 1111

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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