0.000 001 506 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 506 37(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
0.000 001 506 37(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: 0.000 001 506 37.

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) 0.000 001 506 37 × 2 = 0 + 0.000 003 012 74;
  • 2) 0.000 003 012 74 × 2 = 0 + 0.000 006 025 48;
  • 3) 0.000 006 025 48 × 2 = 0 + 0.000 012 050 96;
  • 4) 0.000 012 050 96 × 2 = 0 + 0.000 024 101 92;
  • 5) 0.000 024 101 92 × 2 = 0 + 0.000 048 203 84;
  • 6) 0.000 048 203 84 × 2 = 0 + 0.000 096 407 68;
  • 7) 0.000 096 407 68 × 2 = 0 + 0.000 192 815 36;
  • 8) 0.000 192 815 36 × 2 = 0 + 0.000 385 630 72;
  • 9) 0.000 385 630 72 × 2 = 0 + 0.000 771 261 44;
  • 10) 0.000 771 261 44 × 2 = 0 + 0.001 542 522 88;
  • 11) 0.001 542 522 88 × 2 = 0 + 0.003 085 045 76;
  • 12) 0.003 085 045 76 × 2 = 0 + 0.006 170 091 52;
  • 13) 0.006 170 091 52 × 2 = 0 + 0.012 340 183 04;
  • 14) 0.012 340 183 04 × 2 = 0 + 0.024 680 366 08;
  • 15) 0.024 680 366 08 × 2 = 0 + 0.049 360 732 16;
  • 16) 0.049 360 732 16 × 2 = 0 + 0.098 721 464 32;
  • 17) 0.098 721 464 32 × 2 = 0 + 0.197 442 928 64;
  • 18) 0.197 442 928 64 × 2 = 0 + 0.394 885 857 28;
  • 19) 0.394 885 857 28 × 2 = 0 + 0.789 771 714 56;
  • 20) 0.789 771 714 56 × 2 = 1 + 0.579 543 429 12;
  • 21) 0.579 543 429 12 × 2 = 1 + 0.159 086 858 24;
  • 22) 0.159 086 858 24 × 2 = 0 + 0.318 173 716 48;
  • 23) 0.318 173 716 48 × 2 = 0 + 0.636 347 432 96;
  • 24) 0.636 347 432 96 × 2 = 1 + 0.272 694 865 92;
  • 25) 0.272 694 865 92 × 2 = 0 + 0.545 389 731 84;
  • 26) 0.545 389 731 84 × 2 = 1 + 0.090 779 463 68;
  • 27) 0.090 779 463 68 × 2 = 0 + 0.181 558 927 36;
  • 28) 0.181 558 927 36 × 2 = 0 + 0.363 117 854 72;
  • 29) 0.363 117 854 72 × 2 = 0 + 0.726 235 709 44;
  • 30) 0.726 235 709 44 × 2 = 1 + 0.452 471 418 88;
  • 31) 0.452 471 418 88 × 2 = 0 + 0.904 942 837 76;
  • 32) 0.904 942 837 76 × 2 = 1 + 0.809 885 675 52;
  • 33) 0.809 885 675 52 × 2 = 1 + 0.619 771 351 04;
  • 34) 0.619 771 351 04 × 2 = 1 + 0.239 542 702 08;
  • 35) 0.239 542 702 08 × 2 = 0 + 0.479 085 404 16;
  • 36) 0.479 085 404 16 × 2 = 0 + 0.958 170 808 32;
  • 37) 0.958 170 808 32 × 2 = 1 + 0.916 341 616 64;
  • 38) 0.916 341 616 64 × 2 = 1 + 0.832 683 233 28;
  • 39) 0.832 683 233 28 × 2 = 1 + 0.665 366 466 56;
  • 40) 0.665 366 466 56 × 2 = 1 + 0.330 732 933 12;
  • 41) 0.330 732 933 12 × 2 = 0 + 0.661 465 866 24;
  • 42) 0.661 465 866 24 × 2 = 1 + 0.322 931 732 48;
  • 43) 0.322 931 732 48 × 2 = 0 + 0.645 863 464 96;
  • 44) 0.645 863 464 96 × 2 = 1 + 0.291 726 929 92;
  • 45) 0.291 726 929 92 × 2 = 0 + 0.583 453 859 84;
  • 46) 0.583 453 859 84 × 2 = 1 + 0.166 907 719 68;
  • 47) 0.166 907 719 68 × 2 = 0 + 0.333 815 439 36;
  • 48) 0.333 815 439 36 × 2 = 0 + 0.667 630 878 72;
  • 49) 0.667 630 878 72 × 2 = 1 + 0.335 261 757 44;
  • 50) 0.335 261 757 44 × 2 = 0 + 0.670 523 514 88;
  • 51) 0.670 523 514 88 × 2 = 1 + 0.341 047 029 76;
  • 52) 0.341 047 029 76 × 2 = 0 + 0.682 094 059 52;
  • 53) 0.682 094 059 52 × 2 = 1 + 0.364 188 119 04;
  • 54) 0.364 188 119 04 × 2 = 0 + 0.728 376 238 08;
  • 55) 0.728 376 238 08 × 2 = 1 + 0.456 752 476 16;
  • 56) 0.456 752 476 16 × 2 = 0 + 0.913 504 952 32;
  • 57) 0.913 504 952 32 × 2 = 1 + 0.827 009 904 64;
  • 58) 0.827 009 904 64 × 2 = 1 + 0.654 019 809 28;
  • 59) 0.654 019 809 28 × 2 = 1 + 0.308 039 618 56;
  • 60) 0.308 039 618 56 × 2 = 0 + 0.616 079 237 12;
  • 61) 0.616 079 237 12 × 2 = 1 + 0.232 158 474 24;
  • 62) 0.232 158 474 24 × 2 = 0 + 0.464 316 948 48;
  • 63) 0.464 316 948 48 × 2 = 0 + 0.928 633 896 96;
  • 64) 0.928 633 896 96 × 2 = 1 + 0.857 267 793 92;
  • 65) 0.857 267 793 92 × 2 = 1 + 0.714 535 587 84;
  • 66) 0.714 535 587 84 × 2 = 1 + 0.429 071 175 68;
  • 67) 0.429 071 175 68 × 2 = 0 + 0.858 142 351 36;
  • 68) 0.858 142 351 36 × 2 = 1 + 0.716 284 702 72;
  • 69) 0.716 284 702 72 × 2 = 1 + 0.432 569 405 44;
  • 70) 0.432 569 405 44 × 2 = 0 + 0.865 138 810 88;
  • 71) 0.865 138 810 88 × 2 = 1 + 0.730 277 621 76;
  • 72) 0.730 277 621 76 × 2 = 1 + 0.460 555 243 52;

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:


0.000 001 506 37(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011(2)

5. Positive number before normalization:

0.000 001 506 37(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 506 37(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011(2) =

0.0000 0000 0000 0000 0001 1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011(2) × 20 =

1.1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011(2) × 2-20


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): -20

Mantissa (not normalized):
1.1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 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) =

1003(10) =

011 1110 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. 1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011 =

1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011


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 1011

Mantissa (52 bits) =
1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011


Decimal number 0.000 001 506 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1011 - 1001 0100 0101 1100 1111 0101 0100 1010 1010 1110 1001 1101 1011

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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