-0.001 038 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.001 038 71(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
-0.001 038 71(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:

|-0.001 038 71| = 0.001 038 71


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: 0.001 038 71.

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) 0.001 038 71 × 2 = 0 + 0.002 077 42;
  • 2) 0.002 077 42 × 2 = 0 + 0.004 154 84;
  • 3) 0.004 154 84 × 2 = 0 + 0.008 309 68;
  • 4) 0.008 309 68 × 2 = 0 + 0.016 619 36;
  • 5) 0.016 619 36 × 2 = 0 + 0.033 238 72;
  • 6) 0.033 238 72 × 2 = 0 + 0.066 477 44;
  • 7) 0.066 477 44 × 2 = 0 + 0.132 954 88;
  • 8) 0.132 954 88 × 2 = 0 + 0.265 909 76;
  • 9) 0.265 909 76 × 2 = 0 + 0.531 819 52;
  • 10) 0.531 819 52 × 2 = 1 + 0.063 639 04;
  • 11) 0.063 639 04 × 2 = 0 + 0.127 278 08;
  • 12) 0.127 278 08 × 2 = 0 + 0.254 556 16;
  • 13) 0.254 556 16 × 2 = 0 + 0.509 112 32;
  • 14) 0.509 112 32 × 2 = 1 + 0.018 224 64;
  • 15) 0.018 224 64 × 2 = 0 + 0.036 449 28;
  • 16) 0.036 449 28 × 2 = 0 + 0.072 898 56;
  • 17) 0.072 898 56 × 2 = 0 + 0.145 797 12;
  • 18) 0.145 797 12 × 2 = 0 + 0.291 594 24;
  • 19) 0.291 594 24 × 2 = 0 + 0.583 188 48;
  • 20) 0.583 188 48 × 2 = 1 + 0.166 376 96;
  • 21) 0.166 376 96 × 2 = 0 + 0.332 753 92;
  • 22) 0.332 753 92 × 2 = 0 + 0.665 507 84;
  • 23) 0.665 507 84 × 2 = 1 + 0.331 015 68;
  • 24) 0.331 015 68 × 2 = 0 + 0.662 031 36;
  • 25) 0.662 031 36 × 2 = 1 + 0.324 062 72;
  • 26) 0.324 062 72 × 2 = 0 + 0.648 125 44;
  • 27) 0.648 125 44 × 2 = 1 + 0.296 250 88;
  • 28) 0.296 250 88 × 2 = 0 + 0.592 501 76;
  • 29) 0.592 501 76 × 2 = 1 + 0.185 003 52;
  • 30) 0.185 003 52 × 2 = 0 + 0.370 007 04;
  • 31) 0.370 007 04 × 2 = 0 + 0.740 014 08;
  • 32) 0.740 014 08 × 2 = 1 + 0.480 028 16;
  • 33) 0.480 028 16 × 2 = 0 + 0.960 056 32;
  • 34) 0.960 056 32 × 2 = 1 + 0.920 112 64;
  • 35) 0.920 112 64 × 2 = 1 + 0.840 225 28;
  • 36) 0.840 225 28 × 2 = 1 + 0.680 450 56;
  • 37) 0.680 450 56 × 2 = 1 + 0.360 901 12;
  • 38) 0.360 901 12 × 2 = 0 + 0.721 802 24;
  • 39) 0.721 802 24 × 2 = 1 + 0.443 604 48;
  • 40) 0.443 604 48 × 2 = 0 + 0.887 208 96;
  • 41) 0.887 208 96 × 2 = 1 + 0.774 417 92;
  • 42) 0.774 417 92 × 2 = 1 + 0.548 835 84;
  • 43) 0.548 835 84 × 2 = 1 + 0.097 671 68;
  • 44) 0.097 671 68 × 2 = 0 + 0.195 343 36;
  • 45) 0.195 343 36 × 2 = 0 + 0.390 686 72;
  • 46) 0.390 686 72 × 2 = 0 + 0.781 373 44;
  • 47) 0.781 373 44 × 2 = 1 + 0.562 746 88;
  • 48) 0.562 746 88 × 2 = 1 + 0.125 493 76;
  • 49) 0.125 493 76 × 2 = 0 + 0.250 987 52;
  • 50) 0.250 987 52 × 2 = 0 + 0.501 975 04;
  • 51) 0.501 975 04 × 2 = 1 + 0.003 950 08;
  • 52) 0.003 950 08 × 2 = 0 + 0.007 900 16;
  • 53) 0.007 900 16 × 2 = 0 + 0.015 800 32;
  • 54) 0.015 800 32 × 2 = 0 + 0.031 600 64;
  • 55) 0.031 600 64 × 2 = 0 + 0.063 201 28;
  • 56) 0.063 201 28 × 2 = 0 + 0.126 402 56;
  • 57) 0.126 402 56 × 2 = 0 + 0.252 805 12;
  • 58) 0.252 805 12 × 2 = 0 + 0.505 610 24;
  • 59) 0.505 610 24 × 2 = 1 + 0.011 220 48;
  • 60) 0.011 220 48 × 2 = 0 + 0.022 440 96;
  • 61) 0.022 440 96 × 2 = 0 + 0.044 881 92;
  • 62) 0.044 881 92 × 2 = 0 + 0.089 763 84;

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:


0.001 038 71(10) =

0.0000 0000 0100 0100 0001 0010 1010 1001 0111 1010 1110 0011 0010 0000 0010 00(2)

6. Positive number before normalization:

0.001 038 71(10) =

0.0000 0000 0100 0100 0001 0010 1010 1001 0111 1010 1110 0011 0010 0000 0010 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 038 71(10) =

0.0000 0000 0100 0100 0001 0010 1010 1001 0111 1010 1110 0011 0010 0000 0010 00(2) =

0.0000 0000 0100 0100 0001 0010 1010 1001 0111 1010 1110 0011 0010 0000 0010 00(2) × 20 =

1.0001 0000 0100 1010 1010 0101 1110 1011 1000 1100 1000 0000 1000(2) × 2-10


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): -10

Mantissa (not normalized):
1.0001 0000 0100 1010 1010 0101 1110 1011 1000 1100 1000 0000 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(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;
  • 1 013 ÷ 2 = 506 + 1;
  • 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) =

1013(10) =

011 1111 0101(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. 0001 0000 0100 1010 1010 0101 1110 1011 1000 1100 1000 0000 1000 =

0001 0000 0100 1010 1010 0101 1110 1011 1000 1100 1000 0000 1000


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 0101

Mantissa (52 bits) =
0001 0000 0100 1010 1010 0101 1110 1011 1000 1100 1000 0000 1000


Decimal number -0.001 038 71 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0101 - 0001 0000 0100 1010 1010 0101 1110 1011 1000 1100 1000 0000 1000

Share

» Convert decimal -0.001 038 69 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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