-0.000 792 035 636 43 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 792 035 636 43(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
-0.000 792 035 636 43(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:

|-0.000 792 035 636 43| = 0.000 792 035 636 43


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: 0.000 792 035 636 43.

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) 0.000 792 035 636 43 × 2 = 0 + 0.001 584 071 272 86;
  • 2) 0.001 584 071 272 86 × 2 = 0 + 0.003 168 142 545 72;
  • 3) 0.003 168 142 545 72 × 2 = 0 + 0.006 336 285 091 44;
  • 4) 0.006 336 285 091 44 × 2 = 0 + 0.012 672 570 182 88;
  • 5) 0.012 672 570 182 88 × 2 = 0 + 0.025 345 140 365 76;
  • 6) 0.025 345 140 365 76 × 2 = 0 + 0.050 690 280 731 52;
  • 7) 0.050 690 280 731 52 × 2 = 0 + 0.101 380 561 463 04;
  • 8) 0.101 380 561 463 04 × 2 = 0 + 0.202 761 122 926 08;
  • 9) 0.202 761 122 926 08 × 2 = 0 + 0.405 522 245 852 16;
  • 10) 0.405 522 245 852 16 × 2 = 0 + 0.811 044 491 704 32;
  • 11) 0.811 044 491 704 32 × 2 = 1 + 0.622 088 983 408 64;
  • 12) 0.622 088 983 408 64 × 2 = 1 + 0.244 177 966 817 28;
  • 13) 0.244 177 966 817 28 × 2 = 0 + 0.488 355 933 634 56;
  • 14) 0.488 355 933 634 56 × 2 = 0 + 0.976 711 867 269 12;
  • 15) 0.976 711 867 269 12 × 2 = 1 + 0.953 423 734 538 24;
  • 16) 0.953 423 734 538 24 × 2 = 1 + 0.906 847 469 076 48;
  • 17) 0.906 847 469 076 48 × 2 = 1 + 0.813 694 938 152 96;
  • 18) 0.813 694 938 152 96 × 2 = 1 + 0.627 389 876 305 92;
  • 19) 0.627 389 876 305 92 × 2 = 1 + 0.254 779 752 611 84;
  • 20) 0.254 779 752 611 84 × 2 = 0 + 0.509 559 505 223 68;
  • 21) 0.509 559 505 223 68 × 2 = 1 + 0.019 119 010 447 36;
  • 22) 0.019 119 010 447 36 × 2 = 0 + 0.038 238 020 894 72;
  • 23) 0.038 238 020 894 72 × 2 = 0 + 0.076 476 041 789 44;
  • 24) 0.076 476 041 789 44 × 2 = 0 + 0.152 952 083 578 88;
  • 25) 0.152 952 083 578 88 × 2 = 0 + 0.305 904 167 157 76;
  • 26) 0.305 904 167 157 76 × 2 = 0 + 0.611 808 334 315 52;
  • 27) 0.611 808 334 315 52 × 2 = 1 + 0.223 616 668 631 04;
  • 28) 0.223 616 668 631 04 × 2 = 0 + 0.447 233 337 262 08;
  • 29) 0.447 233 337 262 08 × 2 = 0 + 0.894 466 674 524 16;
  • 30) 0.894 466 674 524 16 × 2 = 1 + 0.788 933 349 048 32;
  • 31) 0.788 933 349 048 32 × 2 = 1 + 0.577 866 698 096 64;
  • 32) 0.577 866 698 096 64 × 2 = 1 + 0.155 733 396 193 28;
  • 33) 0.155 733 396 193 28 × 2 = 0 + 0.311 466 792 386 56;
  • 34) 0.311 466 792 386 56 × 2 = 0 + 0.622 933 584 773 12;
  • 35) 0.622 933 584 773 12 × 2 = 1 + 0.245 867 169 546 24;
  • 36) 0.245 867 169 546 24 × 2 = 0 + 0.491 734 339 092 48;
  • 37) 0.491 734 339 092 48 × 2 = 0 + 0.983 468 678 184 96;
  • 38) 0.983 468 678 184 96 × 2 = 1 + 0.966 937 356 369 92;
  • 39) 0.966 937 356 369 92 × 2 = 1 + 0.933 874 712 739 84;
  • 40) 0.933 874 712 739 84 × 2 = 1 + 0.867 749 425 479 68;
  • 41) 0.867 749 425 479 68 × 2 = 1 + 0.735 498 850 959 36;
  • 42) 0.735 498 850 959 36 × 2 = 1 + 0.470 997 701 918 72;
  • 43) 0.470 997 701 918 72 × 2 = 0 + 0.941 995 403 837 44;
  • 44) 0.941 995 403 837 44 × 2 = 1 + 0.883 990 807 674 88;
  • 45) 0.883 990 807 674 88 × 2 = 1 + 0.767 981 615 349 76;
  • 46) 0.767 981 615 349 76 × 2 = 1 + 0.535 963 230 699 52;
  • 47) 0.535 963 230 699 52 × 2 = 1 + 0.071 926 461 399 04;
  • 48) 0.071 926 461 399 04 × 2 = 0 + 0.143 852 922 798 08;
  • 49) 0.143 852 922 798 08 × 2 = 0 + 0.287 705 845 596 16;
  • 50) 0.287 705 845 596 16 × 2 = 0 + 0.575 411 691 192 32;
  • 51) 0.575 411 691 192 32 × 2 = 1 + 0.150 823 382 384 64;
  • 52) 0.150 823 382 384 64 × 2 = 0 + 0.301 646 764 769 28;
  • 53) 0.301 646 764 769 28 × 2 = 0 + 0.603 293 529 538 56;
  • 54) 0.603 293 529 538 56 × 2 = 1 + 0.206 587 059 077 12;
  • 55) 0.206 587 059 077 12 × 2 = 0 + 0.413 174 118 154 24;
  • 56) 0.413 174 118 154 24 × 2 = 0 + 0.826 348 236 308 48;
  • 57) 0.826 348 236 308 48 × 2 = 1 + 0.652 696 472 616 96;
  • 58) 0.652 696 472 616 96 × 2 = 1 + 0.305 392 945 233 92;
  • 59) 0.305 392 945 233 92 × 2 = 0 + 0.610 785 890 467 84;
  • 60) 0.610 785 890 467 84 × 2 = 1 + 0.221 571 780 935 68;
  • 61) 0.221 571 780 935 68 × 2 = 0 + 0.443 143 561 871 36;
  • 62) 0.443 143 561 871 36 × 2 = 0 + 0.886 287 123 742 72;
  • 63) 0.886 287 123 742 72 × 2 = 1 + 0.772 574 247 485 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:


0.000 792 035 636 43(10) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1110 0010 0100 1101 001(2)

6. Positive number before normalization:

0.000 792 035 636 43(10) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1110 0010 0100 1101 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 792 035 636 43(10) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1110 0010 0100 1101 001(2) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1110 0010 0100 1101 001(2) × 20 =

1.1001 1111 0100 0001 0011 1001 0011 1110 1111 0001 0010 0110 1001(2) × 2-11


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): -11

Mantissa (not normalized):
1.1001 1111 0100 0001 0011 1001 0011 1110 1111 0001 0010 0110 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1012(10) =

011 1111 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. 1001 1111 0100 0001 0011 1001 0011 1110 1111 0001 0010 0110 1001 =

1001 1111 0100 0001 0011 1001 0011 1110 1111 0001 0010 0110 1001


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 1111 0100

Mantissa (52 bits) =
1001 1111 0100 0001 0011 1001 0011 1110 1111 0001 0010 0110 1001


Decimal number -0.000 792 035 636 43 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0100 - 1001 1111 0100 0001 0011 1001 0011 1110 1111 0001 0010 0110 1001

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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