0.000 025 656 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 025 656 41(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 41(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 41.

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 41 × 2 = 0 + 0.000 051 312 82;
  • 2) 0.000 051 312 82 × 2 = 0 + 0.000 102 625 64;
  • 3) 0.000 102 625 64 × 2 = 0 + 0.000 205 251 28;
  • 4) 0.000 205 251 28 × 2 = 0 + 0.000 410 502 56;
  • 5) 0.000 410 502 56 × 2 = 0 + 0.000 821 005 12;
  • 6) 0.000 821 005 12 × 2 = 0 + 0.001 642 010 24;
  • 7) 0.001 642 010 24 × 2 = 0 + 0.003 284 020 48;
  • 8) 0.003 284 020 48 × 2 = 0 + 0.006 568 040 96;
  • 9) 0.006 568 040 96 × 2 = 0 + 0.013 136 081 92;
  • 10) 0.013 136 081 92 × 2 = 0 + 0.026 272 163 84;
  • 11) 0.026 272 163 84 × 2 = 0 + 0.052 544 327 68;
  • 12) 0.052 544 327 68 × 2 = 0 + 0.105 088 655 36;
  • 13) 0.105 088 655 36 × 2 = 0 + 0.210 177 310 72;
  • 14) 0.210 177 310 72 × 2 = 0 + 0.420 354 621 44;
  • 15) 0.420 354 621 44 × 2 = 0 + 0.840 709 242 88;
  • 16) 0.840 709 242 88 × 2 = 1 + 0.681 418 485 76;
  • 17) 0.681 418 485 76 × 2 = 1 + 0.362 836 971 52;
  • 18) 0.362 836 971 52 × 2 = 0 + 0.725 673 943 04;
  • 19) 0.725 673 943 04 × 2 = 1 + 0.451 347 886 08;
  • 20) 0.451 347 886 08 × 2 = 0 + 0.902 695 772 16;
  • 21) 0.902 695 772 16 × 2 = 1 + 0.805 391 544 32;
  • 22) 0.805 391 544 32 × 2 = 1 + 0.610 783 088 64;
  • 23) 0.610 783 088 64 × 2 = 1 + 0.221 566 177 28;
  • 24) 0.221 566 177 28 × 2 = 0 + 0.443 132 354 56;
  • 25) 0.443 132 354 56 × 2 = 0 + 0.886 264 709 12;
  • 26) 0.886 264 709 12 × 2 = 1 + 0.772 529 418 24;
  • 27) 0.772 529 418 24 × 2 = 1 + 0.545 058 836 48;
  • 28) 0.545 058 836 48 × 2 = 1 + 0.090 117 672 96;
  • 29) 0.090 117 672 96 × 2 = 0 + 0.180 235 345 92;
  • 30) 0.180 235 345 92 × 2 = 0 + 0.360 470 691 84;
  • 31) 0.360 470 691 84 × 2 = 0 + 0.720 941 383 68;
  • 32) 0.720 941 383 68 × 2 = 1 + 0.441 882 767 36;
  • 33) 0.441 882 767 36 × 2 = 0 + 0.883 765 534 72;
  • 34) 0.883 765 534 72 × 2 = 1 + 0.767 531 069 44;
  • 35) 0.767 531 069 44 × 2 = 1 + 0.535 062 138 88;
  • 36) 0.535 062 138 88 × 2 = 1 + 0.070 124 277 76;
  • 37) 0.070 124 277 76 × 2 = 0 + 0.140 248 555 52;
  • 38) 0.140 248 555 52 × 2 = 0 + 0.280 497 111 04;
  • 39) 0.280 497 111 04 × 2 = 0 + 0.560 994 222 08;
  • 40) 0.560 994 222 08 × 2 = 1 + 0.121 988 444 16;
  • 41) 0.121 988 444 16 × 2 = 0 + 0.243 976 888 32;
  • 42) 0.243 976 888 32 × 2 = 0 + 0.487 953 776 64;
  • 43) 0.487 953 776 64 × 2 = 0 + 0.975 907 553 28;
  • 44) 0.975 907 553 28 × 2 = 1 + 0.951 815 106 56;
  • 45) 0.951 815 106 56 × 2 = 1 + 0.903 630 213 12;
  • 46) 0.903 630 213 12 × 2 = 1 + 0.807 260 426 24;
  • 47) 0.807 260 426 24 × 2 = 1 + 0.614 520 852 48;
  • 48) 0.614 520 852 48 × 2 = 1 + 0.229 041 704 96;
  • 49) 0.229 041 704 96 × 2 = 0 + 0.458 083 409 92;
  • 50) 0.458 083 409 92 × 2 = 0 + 0.916 166 819 84;
  • 51) 0.916 166 819 84 × 2 = 1 + 0.832 333 639 68;
  • 52) 0.832 333 639 68 × 2 = 1 + 0.664 667 279 36;
  • 53) 0.664 667 279 36 × 2 = 1 + 0.329 334 558 72;
  • 54) 0.329 334 558 72 × 2 = 0 + 0.658 669 117 44;
  • 55) 0.658 669 117 44 × 2 = 1 + 0.317 338 234 88;
  • 56) 0.317 338 234 88 × 2 = 0 + 0.634 676 469 76;
  • 57) 0.634 676 469 76 × 2 = 1 + 0.269 352 939 52;
  • 58) 0.269 352 939 52 × 2 = 0 + 0.538 705 879 04;
  • 59) 0.538 705 879 04 × 2 = 1 + 0.077 411 758 08;
  • 60) 0.077 411 758 08 × 2 = 0 + 0.154 823 516 16;
  • 61) 0.154 823 516 16 × 2 = 0 + 0.309 647 032 32;
  • 62) 0.309 647 032 32 × 2 = 0 + 0.619 294 064 64;
  • 63) 0.619 294 064 64 × 2 = 1 + 0.238 588 129 28;
  • 64) 0.238 588 129 28 × 2 = 0 + 0.477 176 258 56;
  • 65) 0.477 176 258 56 × 2 = 0 + 0.954 352 517 12;
  • 66) 0.954 352 517 12 × 2 = 1 + 0.908 705 034 24;
  • 67) 0.908 705 034 24 × 2 = 1 + 0.817 410 068 48;
  • 68) 0.817 410 068 48 × 2 = 1 + 0.634 820 136 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 41(10) =

0.0000 0000 0000 0001 1010 1110 0111 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111(2)

5. Positive number before normalization:

0.000 025 656 41(10) =

0.0000 0000 0000 0001 1010 1110 0111 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111(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 41(10) =

0.0000 0000 0000 0001 1010 1110 0111 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111(2) =

0.0000 0000 0000 0001 1010 1110 0111 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111(2) × 20 =

1.1010 1110 0111 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111(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 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111


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 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111 =

1010 1110 0111 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111


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 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111


Decimal number 0.000 025 656 41 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1111 - 1010 1110 0111 0001 0111 0001 0001 1111 0011 1010 1010 0010 0111

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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