0.000 025 656 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 025 656 91(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
0.000 025 656 91(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: 0.000 025 656 91.

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) 0.000 025 656 91 × 2 = 0 + 0.000 051 313 82;
  • 2) 0.000 051 313 82 × 2 = 0 + 0.000 102 627 64;
  • 3) 0.000 102 627 64 × 2 = 0 + 0.000 205 255 28;
  • 4) 0.000 205 255 28 × 2 = 0 + 0.000 410 510 56;
  • 5) 0.000 410 510 56 × 2 = 0 + 0.000 821 021 12;
  • 6) 0.000 821 021 12 × 2 = 0 + 0.001 642 042 24;
  • 7) 0.001 642 042 24 × 2 = 0 + 0.003 284 084 48;
  • 8) 0.003 284 084 48 × 2 = 0 + 0.006 568 168 96;
  • 9) 0.006 568 168 96 × 2 = 0 + 0.013 136 337 92;
  • 10) 0.013 136 337 92 × 2 = 0 + 0.026 272 675 84;
  • 11) 0.026 272 675 84 × 2 = 0 + 0.052 545 351 68;
  • 12) 0.052 545 351 68 × 2 = 0 + 0.105 090 703 36;
  • 13) 0.105 090 703 36 × 2 = 0 + 0.210 181 406 72;
  • 14) 0.210 181 406 72 × 2 = 0 + 0.420 362 813 44;
  • 15) 0.420 362 813 44 × 2 = 0 + 0.840 725 626 88;
  • 16) 0.840 725 626 88 × 2 = 1 + 0.681 451 253 76;
  • 17) 0.681 451 253 76 × 2 = 1 + 0.362 902 507 52;
  • 18) 0.362 902 507 52 × 2 = 0 + 0.725 805 015 04;
  • 19) 0.725 805 015 04 × 2 = 1 + 0.451 610 030 08;
  • 20) 0.451 610 030 08 × 2 = 0 + 0.903 220 060 16;
  • 21) 0.903 220 060 16 × 2 = 1 + 0.806 440 120 32;
  • 22) 0.806 440 120 32 × 2 = 1 + 0.612 880 240 64;
  • 23) 0.612 880 240 64 × 2 = 1 + 0.225 760 481 28;
  • 24) 0.225 760 481 28 × 2 = 0 + 0.451 520 962 56;
  • 25) 0.451 520 962 56 × 2 = 0 + 0.903 041 925 12;
  • 26) 0.903 041 925 12 × 2 = 1 + 0.806 083 850 24;
  • 27) 0.806 083 850 24 × 2 = 1 + 0.612 167 700 48;
  • 28) 0.612 167 700 48 × 2 = 1 + 0.224 335 400 96;
  • 29) 0.224 335 400 96 × 2 = 0 + 0.448 670 801 92;
  • 30) 0.448 670 801 92 × 2 = 0 + 0.897 341 603 84;
  • 31) 0.897 341 603 84 × 2 = 1 + 0.794 683 207 68;
  • 32) 0.794 683 207 68 × 2 = 1 + 0.589 366 415 36;
  • 33) 0.589 366 415 36 × 2 = 1 + 0.178 732 830 72;
  • 34) 0.178 732 830 72 × 2 = 0 + 0.357 465 661 44;
  • 35) 0.357 465 661 44 × 2 = 0 + 0.714 931 322 88;
  • 36) 0.714 931 322 88 × 2 = 1 + 0.429 862 645 76;
  • 37) 0.429 862 645 76 × 2 = 0 + 0.859 725 291 52;
  • 38) 0.859 725 291 52 × 2 = 1 + 0.719 450 583 04;
  • 39) 0.719 450 583 04 × 2 = 1 + 0.438 901 166 08;
  • 40) 0.438 901 166 08 × 2 = 0 + 0.877 802 332 16;
  • 41) 0.877 802 332 16 × 2 = 1 + 0.755 604 664 32;
  • 42) 0.755 604 664 32 × 2 = 1 + 0.511 209 328 64;
  • 43) 0.511 209 328 64 × 2 = 1 + 0.022 418 657 28;
  • 44) 0.022 418 657 28 × 2 = 0 + 0.044 837 314 56;
  • 45) 0.044 837 314 56 × 2 = 0 + 0.089 674 629 12;
  • 46) 0.089 674 629 12 × 2 = 0 + 0.179 349 258 24;
  • 47) 0.179 349 258 24 × 2 = 0 + 0.358 698 516 48;
  • 48) 0.358 698 516 48 × 2 = 0 + 0.717 397 032 96;
  • 49) 0.717 397 032 96 × 2 = 1 + 0.434 794 065 92;
  • 50) 0.434 794 065 92 × 2 = 0 + 0.869 588 131 84;
  • 51) 0.869 588 131 84 × 2 = 1 + 0.739 176 263 68;
  • 52) 0.739 176 263 68 × 2 = 1 + 0.478 352 527 36;
  • 53) 0.478 352 527 36 × 2 = 0 + 0.956 705 054 72;
  • 54) 0.956 705 054 72 × 2 = 1 + 0.913 410 109 44;
  • 55) 0.913 410 109 44 × 2 = 1 + 0.826 820 218 88;
  • 56) 0.826 820 218 88 × 2 = 1 + 0.653 640 437 76;
  • 57) 0.653 640 437 76 × 2 = 1 + 0.307 280 875 52;
  • 58) 0.307 280 875 52 × 2 = 0 + 0.614 561 751 04;
  • 59) 0.614 561 751 04 × 2 = 1 + 0.229 123 502 08;
  • 60) 0.229 123 502 08 × 2 = 0 + 0.458 247 004 16;
  • 61) 0.458 247 004 16 × 2 = 0 + 0.916 494 008 32;
  • 62) 0.916 494 008 32 × 2 = 1 + 0.832 988 016 64;
  • 63) 0.832 988 016 64 × 2 = 1 + 0.665 976 033 28;
  • 64) 0.665 976 033 28 × 2 = 1 + 0.331 952 066 56;
  • 65) 0.331 952 066 56 × 2 = 0 + 0.663 904 133 12;
  • 66) 0.663 904 133 12 × 2 = 1 + 0.327 808 266 24;
  • 67) 0.327 808 266 24 × 2 = 0 + 0.655 616 532 48;
  • 68) 0.655 616 532 48 × 2 = 1 + 0.311 233 064 96;

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:


0.000 025 656 91(10) =

0.0000 0000 0000 0001 1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101(2)

5. Positive number before normalization:

0.000 025 656 91(10) =

0.0000 0000 0000 0001 1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 025 656 91(10) =

0.0000 0000 0000 0001 1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101(2) =

0.0000 0000 0000 0001 1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101(2) × 20 =

1.1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101(2) × 2-16


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): -16

Mantissa (not normalized):
1.1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-16 + 2(11-1) - 1 =

(-16 + 1 023)(10) =

1 007(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;
  • 1 007 ÷ 2 = 503 + 1;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1007(10) =

011 1110 1111(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. 1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101 =

1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101


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 1110 1111

Mantissa (52 bits) =
1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101


Decimal number 0.000 025 656 91 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1111 - 1010 1110 0111 0011 1001 0110 1110 0000 1011 0111 1010 0111 0101

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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