-0.104 999 999 987 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.104 999 999 987 8(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
-0.104 999 999 987 8(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:

|-0.104 999 999 987 8| = 0.104 999 999 987 8


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: 0.104 999 999 987 8.

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) 0.104 999 999 987 8 × 2 = 0 + 0.209 999 999 975 6;
  • 2) 0.209 999 999 975 6 × 2 = 0 + 0.419 999 999 951 2;
  • 3) 0.419 999 999 951 2 × 2 = 0 + 0.839 999 999 902 4;
  • 4) 0.839 999 999 902 4 × 2 = 1 + 0.679 999 999 804 8;
  • 5) 0.679 999 999 804 8 × 2 = 1 + 0.359 999 999 609 6;
  • 6) 0.359 999 999 609 6 × 2 = 0 + 0.719 999 999 219 2;
  • 7) 0.719 999 999 219 2 × 2 = 1 + 0.439 999 998 438 4;
  • 8) 0.439 999 998 438 4 × 2 = 0 + 0.879 999 996 876 8;
  • 9) 0.879 999 996 876 8 × 2 = 1 + 0.759 999 993 753 6;
  • 10) 0.759 999 993 753 6 × 2 = 1 + 0.519 999 987 507 2;
  • 11) 0.519 999 987 507 2 × 2 = 1 + 0.039 999 975 014 4;
  • 12) 0.039 999 975 014 4 × 2 = 0 + 0.079 999 950 028 8;
  • 13) 0.079 999 950 028 8 × 2 = 0 + 0.159 999 900 057 6;
  • 14) 0.159 999 900 057 6 × 2 = 0 + 0.319 999 800 115 2;
  • 15) 0.319 999 800 115 2 × 2 = 0 + 0.639 999 600 230 4;
  • 16) 0.639 999 600 230 4 × 2 = 1 + 0.279 999 200 460 8;
  • 17) 0.279 999 200 460 8 × 2 = 0 + 0.559 998 400 921 6;
  • 18) 0.559 998 400 921 6 × 2 = 1 + 0.119 996 801 843 2;
  • 19) 0.119 996 801 843 2 × 2 = 0 + 0.239 993 603 686 4;
  • 20) 0.239 993 603 686 4 × 2 = 0 + 0.479 987 207 372 8;
  • 21) 0.479 987 207 372 8 × 2 = 0 + 0.959 974 414 745 6;
  • 22) 0.959 974 414 745 6 × 2 = 1 + 0.919 948 829 491 2;
  • 23) 0.919 948 829 491 2 × 2 = 1 + 0.839 897 658 982 4;
  • 24) 0.839 897 658 982 4 × 2 = 1 + 0.679 795 317 964 8;
  • 25) 0.679 795 317 964 8 × 2 = 1 + 0.359 590 635 929 6;
  • 26) 0.359 590 635 929 6 × 2 = 0 + 0.719 181 271 859 2;
  • 27) 0.719 181 271 859 2 × 2 = 1 + 0.438 362 543 718 4;
  • 28) 0.438 362 543 718 4 × 2 = 0 + 0.876 725 087 436 8;
  • 29) 0.876 725 087 436 8 × 2 = 1 + 0.753 450 174 873 6;
  • 30) 0.753 450 174 873 6 × 2 = 1 + 0.506 900 349 747 2;
  • 31) 0.506 900 349 747 2 × 2 = 1 + 0.013 800 699 494 4;
  • 32) 0.013 800 699 494 4 × 2 = 0 + 0.027 601 398 988 8;
  • 33) 0.027 601 398 988 8 × 2 = 0 + 0.055 202 797 977 6;
  • 34) 0.055 202 797 977 6 × 2 = 0 + 0.110 405 595 955 2;
  • 35) 0.110 405 595 955 2 × 2 = 0 + 0.220 811 191 910 4;
  • 36) 0.220 811 191 910 4 × 2 = 0 + 0.441 622 383 820 8;
  • 37) 0.441 622 383 820 8 × 2 = 0 + 0.883 244 767 641 6;
  • 38) 0.883 244 767 641 6 × 2 = 1 + 0.766 489 535 283 2;
  • 39) 0.766 489 535 283 2 × 2 = 1 + 0.532 979 070 566 4;
  • 40) 0.532 979 070 566 4 × 2 = 1 + 0.065 958 141 132 8;
  • 41) 0.065 958 141 132 8 × 2 = 0 + 0.131 916 282 265 6;
  • 42) 0.131 916 282 265 6 × 2 = 0 + 0.263 832 564 531 2;
  • 43) 0.263 832 564 531 2 × 2 = 0 + 0.527 665 129 062 4;
  • 44) 0.527 665 129 062 4 × 2 = 1 + 0.055 330 258 124 8;
  • 45) 0.055 330 258 124 8 × 2 = 0 + 0.110 660 516 249 6;
  • 46) 0.110 660 516 249 6 × 2 = 0 + 0.221 321 032 499 2;
  • 47) 0.221 321 032 499 2 × 2 = 0 + 0.442 642 064 998 4;
  • 48) 0.442 642 064 998 4 × 2 = 0 + 0.885 284 129 996 8;
  • 49) 0.885 284 129 996 8 × 2 = 1 + 0.770 568 259 993 6;
  • 50) 0.770 568 259 993 6 × 2 = 1 + 0.541 136 519 987 2;
  • 51) 0.541 136 519 987 2 × 2 = 1 + 0.082 273 039 974 4;
  • 52) 0.082 273 039 974 4 × 2 = 0 + 0.164 546 079 948 8;
  • 53) 0.164 546 079 948 8 × 2 = 0 + 0.329 092 159 897 6;
  • 54) 0.329 092 159 897 6 × 2 = 0 + 0.658 184 319 795 2;
  • 55) 0.658 184 319 795 2 × 2 = 1 + 0.316 368 639 590 4;
  • 56) 0.316 368 639 590 4 × 2 = 0 + 0.632 737 279 180 8;

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:


0.104 999 999 987 8(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010(2)

6. Positive number before normalization:

0.104 999 999 987 8(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.104 999 999 987 8(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010(2) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010(2) × 20 =

1.1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010(2) × 2-4


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):
1.1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010 =

1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010


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 1011

Mantissa (52 bits) =
1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010


Decimal number -0.104 999 999 987 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 0111 1010 1110 0000 0111 0001 0000 1110 0010

Share

» Convert decimal -0.104 999 999 992 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 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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