0.092 290 816 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.092 290 816(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
0.092 290 816(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: 0.092 290 816.

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) 0.092 290 816 × 2 = 0 + 0.184 581 632;
  • 2) 0.184 581 632 × 2 = 0 + 0.369 163 264;
  • 3) 0.369 163 264 × 2 = 0 + 0.738 326 528;
  • 4) 0.738 326 528 × 2 = 1 + 0.476 653 056;
  • 5) 0.476 653 056 × 2 = 0 + 0.953 306 112;
  • 6) 0.953 306 112 × 2 = 1 + 0.906 612 224;
  • 7) 0.906 612 224 × 2 = 1 + 0.813 224 448;
  • 8) 0.813 224 448 × 2 = 1 + 0.626 448 896;
  • 9) 0.626 448 896 × 2 = 1 + 0.252 897 792;
  • 10) 0.252 897 792 × 2 = 0 + 0.505 795 584;
  • 11) 0.505 795 584 × 2 = 1 + 0.011 591 168;
  • 12) 0.011 591 168 × 2 = 0 + 0.023 182 336;
  • 13) 0.023 182 336 × 2 = 0 + 0.046 364 672;
  • 14) 0.046 364 672 × 2 = 0 + 0.092 729 344;
  • 15) 0.092 729 344 × 2 = 0 + 0.185 458 688;
  • 16) 0.185 458 688 × 2 = 0 + 0.370 917 376;
  • 17) 0.370 917 376 × 2 = 0 + 0.741 834 752;
  • 18) 0.741 834 752 × 2 = 1 + 0.483 669 504;
  • 19) 0.483 669 504 × 2 = 0 + 0.967 339 008;
  • 20) 0.967 339 008 × 2 = 1 + 0.934 678 016;
  • 21) 0.934 678 016 × 2 = 1 + 0.869 356 032;
  • 22) 0.869 356 032 × 2 = 1 + 0.738 712 064;
  • 23) 0.738 712 064 × 2 = 1 + 0.477 424 128;
  • 24) 0.477 424 128 × 2 = 0 + 0.954 848 256;
  • 25) 0.954 848 256 × 2 = 1 + 0.909 696 512;
  • 26) 0.909 696 512 × 2 = 1 + 0.819 393 024;
  • 27) 0.819 393 024 × 2 = 1 + 0.638 786 048;
  • 28) 0.638 786 048 × 2 = 1 + 0.277 572 096;
  • 29) 0.277 572 096 × 2 = 0 + 0.555 144 192;
  • 30) 0.555 144 192 × 2 = 1 + 0.110 288 384;
  • 31) 0.110 288 384 × 2 = 0 + 0.220 576 768;
  • 32) 0.220 576 768 × 2 = 0 + 0.441 153 536;
  • 33) 0.441 153 536 × 2 = 0 + 0.882 307 072;
  • 34) 0.882 307 072 × 2 = 1 + 0.764 614 144;
  • 35) 0.764 614 144 × 2 = 1 + 0.529 228 288;
  • 36) 0.529 228 288 × 2 = 1 + 0.058 456 576;
  • 37) 0.058 456 576 × 2 = 0 + 0.116 913 152;
  • 38) 0.116 913 152 × 2 = 0 + 0.233 826 304;
  • 39) 0.233 826 304 × 2 = 0 + 0.467 652 608;
  • 40) 0.467 652 608 × 2 = 0 + 0.935 305 216;
  • 41) 0.935 305 216 × 2 = 1 + 0.870 610 432;
  • 42) 0.870 610 432 × 2 = 1 + 0.741 220 864;
  • 43) 0.741 220 864 × 2 = 1 + 0.482 441 728;
  • 44) 0.482 441 728 × 2 = 0 + 0.964 883 456;
  • 45) 0.964 883 456 × 2 = 1 + 0.929 766 912;
  • 46) 0.929 766 912 × 2 = 1 + 0.859 533 824;
  • 47) 0.859 533 824 × 2 = 1 + 0.719 067 648;
  • 48) 0.719 067 648 × 2 = 1 + 0.438 135 296;
  • 49) 0.438 135 296 × 2 = 0 + 0.876 270 592;
  • 50) 0.876 270 592 × 2 = 1 + 0.752 541 184;
  • 51) 0.752 541 184 × 2 = 1 + 0.505 082 368;
  • 52) 0.505 082 368 × 2 = 1 + 0.010 164 736;
  • 53) 0.010 164 736 × 2 = 0 + 0.020 329 472;
  • 54) 0.020 329 472 × 2 = 0 + 0.040 658 944;
  • 55) 0.040 658 944 × 2 = 0 + 0.081 317 888;
  • 56) 0.081 317 888 × 2 = 0 + 0.162 635 776;

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:


0.092 290 816(10) =

0.0001 0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000(2)

5. Positive number before normalization:

0.092 290 816(10) =

0.0001 0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000(2)

6. 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:


0.092 290 816(10) =

0.0001 0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000(2) =

0.0001 0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000(2) × 20 =

1.0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000(2) × 2-4


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): -4

Mantissa (not normalized):
1.0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000


8. 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)


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;
  • 1 019 ÷ 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;

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) =

1019(10) =

011 1111 1011(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. 0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000 =

0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 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 1011

Mantissa (52 bits) =
0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000


Decimal number 0.092 290 816 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 0111 1010 0000 0101 1110 1111 0100 0111 0000 1110 1111 0111 0000

» Convert decimal 0.092 290 836 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