0.001 207 488 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.001 207 488 57(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
0.001 207 488 57(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.001 207 488 57.

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) 0.001 207 488 57 × 2 = 0 + 0.002 414 977 14;
  • 2) 0.002 414 977 14 × 2 = 0 + 0.004 829 954 28;
  • 3) 0.004 829 954 28 × 2 = 0 + 0.009 659 908 56;
  • 4) 0.009 659 908 56 × 2 = 0 + 0.019 319 817 12;
  • 5) 0.019 319 817 12 × 2 = 0 + 0.038 639 634 24;
  • 6) 0.038 639 634 24 × 2 = 0 + 0.077 279 268 48;
  • 7) 0.077 279 268 48 × 2 = 0 + 0.154 558 536 96;
  • 8) 0.154 558 536 96 × 2 = 0 + 0.309 117 073 92;
  • 9) 0.309 117 073 92 × 2 = 0 + 0.618 234 147 84;
  • 10) 0.618 234 147 84 × 2 = 1 + 0.236 468 295 68;
  • 11) 0.236 468 295 68 × 2 = 0 + 0.472 936 591 36;
  • 12) 0.472 936 591 36 × 2 = 0 + 0.945 873 182 72;
  • 13) 0.945 873 182 72 × 2 = 1 + 0.891 746 365 44;
  • 14) 0.891 746 365 44 × 2 = 1 + 0.783 492 730 88;
  • 15) 0.783 492 730 88 × 2 = 1 + 0.566 985 461 76;
  • 16) 0.566 985 461 76 × 2 = 1 + 0.133 970 923 52;
  • 17) 0.133 970 923 52 × 2 = 0 + 0.267 941 847 04;
  • 18) 0.267 941 847 04 × 2 = 0 + 0.535 883 694 08;
  • 19) 0.535 883 694 08 × 2 = 1 + 0.071 767 388 16;
  • 20) 0.071 767 388 16 × 2 = 0 + 0.143 534 776 32;
  • 21) 0.143 534 776 32 × 2 = 0 + 0.287 069 552 64;
  • 22) 0.287 069 552 64 × 2 = 0 + 0.574 139 105 28;
  • 23) 0.574 139 105 28 × 2 = 1 + 0.148 278 210 56;
  • 24) 0.148 278 210 56 × 2 = 0 + 0.296 556 421 12;
  • 25) 0.296 556 421 12 × 2 = 0 + 0.593 112 842 24;
  • 26) 0.593 112 842 24 × 2 = 1 + 0.186 225 684 48;
  • 27) 0.186 225 684 48 × 2 = 0 + 0.372 451 368 96;
  • 28) 0.372 451 368 96 × 2 = 0 + 0.744 902 737 92;
  • 29) 0.744 902 737 92 × 2 = 1 + 0.489 805 475 84;
  • 30) 0.489 805 475 84 × 2 = 0 + 0.979 610 951 68;
  • 31) 0.979 610 951 68 × 2 = 1 + 0.959 221 903 36;
  • 32) 0.959 221 903 36 × 2 = 1 + 0.918 443 806 72;
  • 33) 0.918 443 806 72 × 2 = 1 + 0.836 887 613 44;
  • 34) 0.836 887 613 44 × 2 = 1 + 0.673 775 226 88;
  • 35) 0.673 775 226 88 × 2 = 1 + 0.347 550 453 76;
  • 36) 0.347 550 453 76 × 2 = 0 + 0.695 100 907 52;
  • 37) 0.695 100 907 52 × 2 = 1 + 0.390 201 815 04;
  • 38) 0.390 201 815 04 × 2 = 0 + 0.780 403 630 08;
  • 39) 0.780 403 630 08 × 2 = 1 + 0.560 807 260 16;
  • 40) 0.560 807 260 16 × 2 = 1 + 0.121 614 520 32;
  • 41) 0.121 614 520 32 × 2 = 0 + 0.243 229 040 64;
  • 42) 0.243 229 040 64 × 2 = 0 + 0.486 458 081 28;
  • 43) 0.486 458 081 28 × 2 = 0 + 0.972 916 162 56;
  • 44) 0.972 916 162 56 × 2 = 1 + 0.945 832 325 12;
  • 45) 0.945 832 325 12 × 2 = 1 + 0.891 664 650 24;
  • 46) 0.891 664 650 24 × 2 = 1 + 0.783 329 300 48;
  • 47) 0.783 329 300 48 × 2 = 1 + 0.566 658 600 96;
  • 48) 0.566 658 600 96 × 2 = 1 + 0.133 317 201 92;
  • 49) 0.133 317 201 92 × 2 = 0 + 0.266 634 403 84;
  • 50) 0.266 634 403 84 × 2 = 0 + 0.533 268 807 68;
  • 51) 0.533 268 807 68 × 2 = 1 + 0.066 537 615 36;
  • 52) 0.066 537 615 36 × 2 = 0 + 0.133 075 230 72;
  • 53) 0.133 075 230 72 × 2 = 0 + 0.266 150 461 44;
  • 54) 0.266 150 461 44 × 2 = 0 + 0.532 300 922 88;
  • 55) 0.532 300 922 88 × 2 = 1 + 0.064 601 845 76;
  • 56) 0.064 601 845 76 × 2 = 0 + 0.129 203 691 52;
  • 57) 0.129 203 691 52 × 2 = 0 + 0.258 407 383 04;
  • 58) 0.258 407 383 04 × 2 = 0 + 0.516 814 766 08;
  • 59) 0.516 814 766 08 × 2 = 1 + 0.033 629 532 16;
  • 60) 0.033 629 532 16 × 2 = 0 + 0.067 259 064 32;
  • 61) 0.067 259 064 32 × 2 = 0 + 0.134 518 128 64;
  • 62) 0.134 518 128 64 × 2 = 0 + 0.269 036 257 28;

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).


4. 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:


0.001 207 488 57(10) =

0.0000 0000 0100 1111 0010 0010 0100 1011 1110 1011 0001 1111 0010 0010 0010 00(2)

5. Positive number before normalization:

0.001 207 488 57(10) =

0.0000 0000 0100 1111 0010 0010 0100 1011 1110 1011 0001 1111 0010 0010 0010 00(2)

6. 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:


0.001 207 488 57(10) =

0.0000 0000 0100 1111 0010 0010 0100 1011 1110 1011 0001 1111 0010 0010 0010 00(2) =

0.0000 0000 0100 1111 0010 0010 0100 1011 1110 1011 0001 1111 0010 0010 0010 00(2) × 20 =

1.0011 1100 1000 1001 0010 1111 1010 1100 0111 1100 1000 1000 1000(2) × 2-10


7. 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 0 (a positive number)

Exponent (unadjusted): -10

Mantissa (not normalized):
1.0011 1100 1000 1001 0010 1111 1010 1100 0111 1100 1000 1000 1000


8. 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)


9. 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;

10. 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)


11. 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. 0011 1100 1000 1001 0010 1111 1010 1100 0111 1100 1000 1000 1000 =

0011 1100 1000 1001 0010 1111 1010 1100 0111 1100 1000 1000 1000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 0101

Mantissa (52 bits) =
0011 1100 1000 1001 0010 1111 1010 1100 0111 1100 1000 1000 1000


Decimal number 0.001 207 488 57 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0011 1100 1000 1001 0010 1111 1010 1100 0111 1100 1000 1000 1000

» Convert decimal 0.001 207 488 42 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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