0.009 234 567 808 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.009 234 567 808(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
0.009 234 567 808(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: 0.009 234 567 808.

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) 0.009 234 567 808 × 2 = 0 + 0.018 469 135 616;
  • 2) 0.018 469 135 616 × 2 = 0 + 0.036 938 271 232;
  • 3) 0.036 938 271 232 × 2 = 0 + 0.073 876 542 464;
  • 4) 0.073 876 542 464 × 2 = 0 + 0.147 753 084 928;
  • 5) 0.147 753 084 928 × 2 = 0 + 0.295 506 169 856;
  • 6) 0.295 506 169 856 × 2 = 0 + 0.591 012 339 712;
  • 7) 0.591 012 339 712 × 2 = 1 + 0.182 024 679 424;
  • 8) 0.182 024 679 424 × 2 = 0 + 0.364 049 358 848;
  • 9) 0.364 049 358 848 × 2 = 0 + 0.728 098 717 696;
  • 10) 0.728 098 717 696 × 2 = 1 + 0.456 197 435 392;
  • 11) 0.456 197 435 392 × 2 = 0 + 0.912 394 870 784;
  • 12) 0.912 394 870 784 × 2 = 1 + 0.824 789 741 568;
  • 13) 0.824 789 741 568 × 2 = 1 + 0.649 579 483 136;
  • 14) 0.649 579 483 136 × 2 = 1 + 0.299 158 966 272;
  • 15) 0.299 158 966 272 × 2 = 0 + 0.598 317 932 544;
  • 16) 0.598 317 932 544 × 2 = 1 + 0.196 635 865 088;
  • 17) 0.196 635 865 088 × 2 = 0 + 0.393 271 730 176;
  • 18) 0.393 271 730 176 × 2 = 0 + 0.786 543 460 352;
  • 19) 0.786 543 460 352 × 2 = 1 + 0.573 086 920 704;
  • 20) 0.573 086 920 704 × 2 = 1 + 0.146 173 841 408;
  • 21) 0.146 173 841 408 × 2 = 0 + 0.292 347 682 816;
  • 22) 0.292 347 682 816 × 2 = 0 + 0.584 695 365 632;
  • 23) 0.584 695 365 632 × 2 = 1 + 0.169 390 731 264;
  • 24) 0.169 390 731 264 × 2 = 0 + 0.338 781 462 528;
  • 25) 0.338 781 462 528 × 2 = 0 + 0.677 562 925 056;
  • 26) 0.677 562 925 056 × 2 = 1 + 0.355 125 850 112;
  • 27) 0.355 125 850 112 × 2 = 0 + 0.710 251 700 224;
  • 28) 0.710 251 700 224 × 2 = 1 + 0.420 503 400 448;
  • 29) 0.420 503 400 448 × 2 = 0 + 0.841 006 800 896;
  • 30) 0.841 006 800 896 × 2 = 1 + 0.682 013 601 792;
  • 31) 0.682 013 601 792 × 2 = 1 + 0.364 027 203 584;
  • 32) 0.364 027 203 584 × 2 = 0 + 0.728 054 407 168;
  • 33) 0.728 054 407 168 × 2 = 1 + 0.456 108 814 336;
  • 34) 0.456 108 814 336 × 2 = 0 + 0.912 217 628 672;
  • 35) 0.912 217 628 672 × 2 = 1 + 0.824 435 257 344;
  • 36) 0.824 435 257 344 × 2 = 1 + 0.648 870 514 688;
  • 37) 0.648 870 514 688 × 2 = 1 + 0.297 741 029 376;
  • 38) 0.297 741 029 376 × 2 = 0 + 0.595 482 058 752;
  • 39) 0.595 482 058 752 × 2 = 1 + 0.190 964 117 504;
  • 40) 0.190 964 117 504 × 2 = 0 + 0.381 928 235 008;
  • 41) 0.381 928 235 008 × 2 = 0 + 0.763 856 470 016;
  • 42) 0.763 856 470 016 × 2 = 1 + 0.527 712 940 032;
  • 43) 0.527 712 940 032 × 2 = 1 + 0.055 425 880 064;
  • 44) 0.055 425 880 064 × 2 = 0 + 0.110 851 760 128;
  • 45) 0.110 851 760 128 × 2 = 0 + 0.221 703 520 256;
  • 46) 0.221 703 520 256 × 2 = 0 + 0.443 407 040 512;
  • 47) 0.443 407 040 512 × 2 = 0 + 0.886 814 081 024;
  • 48) 0.886 814 081 024 × 2 = 1 + 0.773 628 162 048;
  • 49) 0.773 628 162 048 × 2 = 1 + 0.547 256 324 096;
  • 50) 0.547 256 324 096 × 2 = 1 + 0.094 512 648 192;
  • 51) 0.094 512 648 192 × 2 = 0 + 0.189 025 296 384;
  • 52) 0.189 025 296 384 × 2 = 0 + 0.378 050 592 768;
  • 53) 0.378 050 592 768 × 2 = 0 + 0.756 101 185 536;
  • 54) 0.756 101 185 536 × 2 = 1 + 0.512 202 371 072;
  • 55) 0.512 202 371 072 × 2 = 1 + 0.024 404 742 144;
  • 56) 0.024 404 742 144 × 2 = 0 + 0.048 809 484 288;
  • 57) 0.048 809 484 288 × 2 = 0 + 0.097 618 968 576;
  • 58) 0.097 618 968 576 × 2 = 0 + 0.195 237 937 152;
  • 59) 0.195 237 937 152 × 2 = 0 + 0.390 475 874 304;

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:


0.009 234 567 808(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1010 0110 0001 1100 0110 000(2)

5. Positive number before normalization:

0.009 234 567 808(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1010 0110 0001 1100 0110 000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.009 234 567 808(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1010 0110 0001 1100 0110 000(2) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1010 0110 0001 1100 0110 000(2) × 20 =

1.0010 1110 1001 1001 0010 1011 0101 1101 0011 0000 1110 0011 0000(2) × 2-7


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): -7

Mantissa (not normalized):
1.0010 1110 1001 1001 0010 1011 0101 1101 0011 0000 1110 0011 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 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;

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) =

1016(10) =

011 1111 1000(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. 0010 1110 1001 1001 0010 1011 0101 1101 0011 0000 1110 0011 0000 =

0010 1110 1001 1001 0010 1011 0101 1101 0011 0000 1110 0011 0000


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 1000

Mantissa (52 bits) =
0010 1110 1001 1001 0010 1011 0101 1101 0011 0000 1110 0011 0000


Decimal number 0.009 234 567 808 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1000 - 0010 1110 1001 1001 0010 1011 0101 1101 0011 0000 1110 0011 0000

» Convert decimal 0.009 234 567 725 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 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - 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