0.000 025 657 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 025 657 68(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
0.000 025 657 68(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: 0.000 025 657 68.

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) 0.000 025 657 68 × 2 = 0 + 0.000 051 315 36;
  • 2) 0.000 051 315 36 × 2 = 0 + 0.000 102 630 72;
  • 3) 0.000 102 630 72 × 2 = 0 + 0.000 205 261 44;
  • 4) 0.000 205 261 44 × 2 = 0 + 0.000 410 522 88;
  • 5) 0.000 410 522 88 × 2 = 0 + 0.000 821 045 76;
  • 6) 0.000 821 045 76 × 2 = 0 + 0.001 642 091 52;
  • 7) 0.001 642 091 52 × 2 = 0 + 0.003 284 183 04;
  • 8) 0.003 284 183 04 × 2 = 0 + 0.006 568 366 08;
  • 9) 0.006 568 366 08 × 2 = 0 + 0.013 136 732 16;
  • 10) 0.013 136 732 16 × 2 = 0 + 0.026 273 464 32;
  • 11) 0.026 273 464 32 × 2 = 0 + 0.052 546 928 64;
  • 12) 0.052 546 928 64 × 2 = 0 + 0.105 093 857 28;
  • 13) 0.105 093 857 28 × 2 = 0 + 0.210 187 714 56;
  • 14) 0.210 187 714 56 × 2 = 0 + 0.420 375 429 12;
  • 15) 0.420 375 429 12 × 2 = 0 + 0.840 750 858 24;
  • 16) 0.840 750 858 24 × 2 = 1 + 0.681 501 716 48;
  • 17) 0.681 501 716 48 × 2 = 1 + 0.363 003 432 96;
  • 18) 0.363 003 432 96 × 2 = 0 + 0.726 006 865 92;
  • 19) 0.726 006 865 92 × 2 = 1 + 0.452 013 731 84;
  • 20) 0.452 013 731 84 × 2 = 0 + 0.904 027 463 68;
  • 21) 0.904 027 463 68 × 2 = 1 + 0.808 054 927 36;
  • 22) 0.808 054 927 36 × 2 = 1 + 0.616 109 854 72;
  • 23) 0.616 109 854 72 × 2 = 1 + 0.232 219 709 44;
  • 24) 0.232 219 709 44 × 2 = 0 + 0.464 439 418 88;
  • 25) 0.464 439 418 88 × 2 = 0 + 0.928 878 837 76;
  • 26) 0.928 878 837 76 × 2 = 1 + 0.857 757 675 52;
  • 27) 0.857 757 675 52 × 2 = 1 + 0.715 515 351 04;
  • 28) 0.715 515 351 04 × 2 = 1 + 0.431 030 702 08;
  • 29) 0.431 030 702 08 × 2 = 0 + 0.862 061 404 16;
  • 30) 0.862 061 404 16 × 2 = 1 + 0.724 122 808 32;
  • 31) 0.724 122 808 32 × 2 = 1 + 0.448 245 616 64;
  • 32) 0.448 245 616 64 × 2 = 0 + 0.896 491 233 28;
  • 33) 0.896 491 233 28 × 2 = 1 + 0.792 982 466 56;
  • 34) 0.792 982 466 56 × 2 = 1 + 0.585 964 933 12;
  • 35) 0.585 964 933 12 × 2 = 1 + 0.171 929 866 24;
  • 36) 0.171 929 866 24 × 2 = 0 + 0.343 859 732 48;
  • 37) 0.343 859 732 48 × 2 = 0 + 0.687 719 464 96;
  • 38) 0.687 719 464 96 × 2 = 1 + 0.375 438 929 92;
  • 39) 0.375 438 929 92 × 2 = 0 + 0.750 877 859 84;
  • 40) 0.750 877 859 84 × 2 = 1 + 0.501 755 719 68;
  • 41) 0.501 755 719 68 × 2 = 1 + 0.003 511 439 36;
  • 42) 0.003 511 439 36 × 2 = 0 + 0.007 022 878 72;
  • 43) 0.007 022 878 72 × 2 = 0 + 0.014 045 757 44;
  • 44) 0.014 045 757 44 × 2 = 0 + 0.028 091 514 88;
  • 45) 0.028 091 514 88 × 2 = 0 + 0.056 183 029 76;
  • 46) 0.056 183 029 76 × 2 = 0 + 0.112 366 059 52;
  • 47) 0.112 366 059 52 × 2 = 0 + 0.224 732 119 04;
  • 48) 0.224 732 119 04 × 2 = 0 + 0.449 464 238 08;
  • 49) 0.449 464 238 08 × 2 = 0 + 0.898 928 476 16;
  • 50) 0.898 928 476 16 × 2 = 1 + 0.797 856 952 32;
  • 51) 0.797 856 952 32 × 2 = 1 + 0.595 713 904 64;
  • 52) 0.595 713 904 64 × 2 = 1 + 0.191 427 809 28;
  • 53) 0.191 427 809 28 × 2 = 0 + 0.382 855 618 56;
  • 54) 0.382 855 618 56 × 2 = 0 + 0.765 711 237 12;
  • 55) 0.765 711 237 12 × 2 = 1 + 0.531 422 474 24;
  • 56) 0.531 422 474 24 × 2 = 1 + 0.062 844 948 48;
  • 57) 0.062 844 948 48 × 2 = 0 + 0.125 689 896 96;
  • 58) 0.125 689 896 96 × 2 = 0 + 0.251 379 793 92;
  • 59) 0.251 379 793 92 × 2 = 0 + 0.502 759 587 84;
  • 60) 0.502 759 587 84 × 2 = 1 + 0.005 519 175 68;
  • 61) 0.005 519 175 68 × 2 = 0 + 0.011 038 351 36;
  • 62) 0.011 038 351 36 × 2 = 0 + 0.022 076 702 72;
  • 63) 0.022 076 702 72 × 2 = 0 + 0.044 153 405 44;
  • 64) 0.044 153 405 44 × 2 = 0 + 0.088 306 810 88;
  • 65) 0.088 306 810 88 × 2 = 0 + 0.176 613 621 76;
  • 66) 0.176 613 621 76 × 2 = 0 + 0.353 227 243 52;
  • 67) 0.353 227 243 52 × 2 = 0 + 0.706 454 487 04;
  • 68) 0.706 454 487 04 × 2 = 1 + 0.412 908 974 08;

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:


0.000 025 657 68(10) =

0.0000 0000 0000 0001 1010 1110 0111 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001(2)

5. Positive number before normalization:

0.000 025 657 68(10) =

0.0000 0000 0000 0001 1010 1110 0111 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001(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:


0.000 025 657 68(10) =

0.0000 0000 0000 0001 1010 1110 0111 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001(2) =

0.0000 0000 0000 0001 1010 1110 0111 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001(2) × 20 =

1.1010 1110 0111 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001(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 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001


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 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001 =

1010 1110 0111 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001


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 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001


Decimal number 0.000 025 657 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1111 - 1010 1110 0111 0110 1110 0101 1000 0000 0111 0011 0001 0000 0001

» Convert decimal 0.000 025 656 91 to 64 bit double precision IEEE 754 binary floating point representation standard

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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