-0.104 999 999 998 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.104 999 999 998(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
-0.104 999 999 998(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:

|-0.104 999 999 998| = 0.104 999 999 998


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: 0.104 999 999 998.

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) 0.104 999 999 998 × 2 = 0 + 0.209 999 999 996;
  • 2) 0.209 999 999 996 × 2 = 0 + 0.419 999 999 992;
  • 3) 0.419 999 999 992 × 2 = 0 + 0.839 999 999 984;
  • 4) 0.839 999 999 984 × 2 = 1 + 0.679 999 999 968;
  • 5) 0.679 999 999 968 × 2 = 1 + 0.359 999 999 936;
  • 6) 0.359 999 999 936 × 2 = 0 + 0.719 999 999 872;
  • 7) 0.719 999 999 872 × 2 = 1 + 0.439 999 999 744;
  • 8) 0.439 999 999 744 × 2 = 0 + 0.879 999 999 488;
  • 9) 0.879 999 999 488 × 2 = 1 + 0.759 999 998 976;
  • 10) 0.759 999 998 976 × 2 = 1 + 0.519 999 997 952;
  • 11) 0.519 999 997 952 × 2 = 1 + 0.039 999 995 904;
  • 12) 0.039 999 995 904 × 2 = 0 + 0.079 999 991 808;
  • 13) 0.079 999 991 808 × 2 = 0 + 0.159 999 983 616;
  • 14) 0.159 999 983 616 × 2 = 0 + 0.319 999 967 232;
  • 15) 0.319 999 967 232 × 2 = 0 + 0.639 999 934 464;
  • 16) 0.639 999 934 464 × 2 = 1 + 0.279 999 868 928;
  • 17) 0.279 999 868 928 × 2 = 0 + 0.559 999 737 856;
  • 18) 0.559 999 737 856 × 2 = 1 + 0.119 999 475 712;
  • 19) 0.119 999 475 712 × 2 = 0 + 0.239 998 951 424;
  • 20) 0.239 998 951 424 × 2 = 0 + 0.479 997 902 848;
  • 21) 0.479 997 902 848 × 2 = 0 + 0.959 995 805 696;
  • 22) 0.959 995 805 696 × 2 = 1 + 0.919 991 611 392;
  • 23) 0.919 991 611 392 × 2 = 1 + 0.839 983 222 784;
  • 24) 0.839 983 222 784 × 2 = 1 + 0.679 966 445 568;
  • 25) 0.679 966 445 568 × 2 = 1 + 0.359 932 891 136;
  • 26) 0.359 932 891 136 × 2 = 0 + 0.719 865 782 272;
  • 27) 0.719 865 782 272 × 2 = 1 + 0.439 731 564 544;
  • 28) 0.439 731 564 544 × 2 = 0 + 0.879 463 129 088;
  • 29) 0.879 463 129 088 × 2 = 1 + 0.758 926 258 176;
  • 30) 0.758 926 258 176 × 2 = 1 + 0.517 852 516 352;
  • 31) 0.517 852 516 352 × 2 = 1 + 0.035 705 032 704;
  • 32) 0.035 705 032 704 × 2 = 0 + 0.071 410 065 408;
  • 33) 0.071 410 065 408 × 2 = 0 + 0.142 820 130 816;
  • 34) 0.142 820 130 816 × 2 = 0 + 0.285 640 261 632;
  • 35) 0.285 640 261 632 × 2 = 0 + 0.571 280 523 264;
  • 36) 0.571 280 523 264 × 2 = 1 + 0.142 561 046 528;
  • 37) 0.142 561 046 528 × 2 = 0 + 0.285 122 093 056;
  • 38) 0.285 122 093 056 × 2 = 0 + 0.570 244 186 112;
  • 39) 0.570 244 186 112 × 2 = 1 + 0.140 488 372 224;
  • 40) 0.140 488 372 224 × 2 = 0 + 0.280 976 744 448;
  • 41) 0.280 976 744 448 × 2 = 0 + 0.561 953 488 896;
  • 42) 0.561 953 488 896 × 2 = 1 + 0.123 906 977 792;
  • 43) 0.123 906 977 792 × 2 = 0 + 0.247 813 955 584;
  • 44) 0.247 813 955 584 × 2 = 0 + 0.495 627 911 168;
  • 45) 0.495 627 911 168 × 2 = 0 + 0.991 255 822 336;
  • 46) 0.991 255 822 336 × 2 = 1 + 0.982 511 644 672;
  • 47) 0.982 511 644 672 × 2 = 1 + 0.965 023 289 344;
  • 48) 0.965 023 289 344 × 2 = 1 + 0.930 046 578 688;
  • 49) 0.930 046 578 688 × 2 = 1 + 0.860 093 157 376;
  • 50) 0.860 093 157 376 × 2 = 1 + 0.720 186 314 752;
  • 51) 0.720 186 314 752 × 2 = 1 + 0.440 372 629 504;
  • 52) 0.440 372 629 504 × 2 = 0 + 0.880 745 259 008;
  • 53) 0.880 745 259 008 × 2 = 1 + 0.761 490 518 016;
  • 54) 0.761 490 518 016 × 2 = 1 + 0.522 981 036 032;
  • 55) 0.522 981 036 032 × 2 = 1 + 0.045 962 072 064;
  • 56) 0.045 962 072 064 × 2 = 0 + 0.091 924 144 128;

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:


0.104 999 999 998(10) =

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

6. Positive number before normalization:

0.104 999 999 998(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 0010 0100 0111 1110 1110(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:


0.104 999 999 998(10) =

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

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

1.1010 1110 0001 0100 0111 1010 1110 0001 0010 0100 0111 1110 1110(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 0001 0010 0100 0111 1110 1110


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 0001 0010 0100 0111 1110 1110 =

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


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 0001 0010 0100 0111 1110 1110


Decimal number -0.104 999 999 998 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

Share

» Convert decimal -0.104 999 999 921 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: The Attention Economy - The Battlefield For Your Focus. Gemini Notebook YouTube Video of 7.50 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