0.000 000 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 67(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
0.000 000 67(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: 0.000 000 67.

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) 0.000 000 67 × 2 = 0 + 0.000 001 34;
  • 2) 0.000 001 34 × 2 = 0 + 0.000 002 68;
  • 3) 0.000 002 68 × 2 = 0 + 0.000 005 36;
  • 4) 0.000 005 36 × 2 = 0 + 0.000 010 72;
  • 5) 0.000 010 72 × 2 = 0 + 0.000 021 44;
  • 6) 0.000 021 44 × 2 = 0 + 0.000 042 88;
  • 7) 0.000 042 88 × 2 = 0 + 0.000 085 76;
  • 8) 0.000 085 76 × 2 = 0 + 0.000 171 52;
  • 9) 0.000 171 52 × 2 = 0 + 0.000 343 04;
  • 10) 0.000 343 04 × 2 = 0 + 0.000 686 08;
  • 11) 0.000 686 08 × 2 = 0 + 0.001 372 16;
  • 12) 0.001 372 16 × 2 = 0 + 0.002 744 32;
  • 13) 0.002 744 32 × 2 = 0 + 0.005 488 64;
  • 14) 0.005 488 64 × 2 = 0 + 0.010 977 28;
  • 15) 0.010 977 28 × 2 = 0 + 0.021 954 56;
  • 16) 0.021 954 56 × 2 = 0 + 0.043 909 12;
  • 17) 0.043 909 12 × 2 = 0 + 0.087 818 24;
  • 18) 0.087 818 24 × 2 = 0 + 0.175 636 48;
  • 19) 0.175 636 48 × 2 = 0 + 0.351 272 96;
  • 20) 0.351 272 96 × 2 = 0 + 0.702 545 92;
  • 21) 0.702 545 92 × 2 = 1 + 0.405 091 84;
  • 22) 0.405 091 84 × 2 = 0 + 0.810 183 68;
  • 23) 0.810 183 68 × 2 = 1 + 0.620 367 36;
  • 24) 0.620 367 36 × 2 = 1 + 0.240 734 72;
  • 25) 0.240 734 72 × 2 = 0 + 0.481 469 44;
  • 26) 0.481 469 44 × 2 = 0 + 0.962 938 88;
  • 27) 0.962 938 88 × 2 = 1 + 0.925 877 76;
  • 28) 0.925 877 76 × 2 = 1 + 0.851 755 52;
  • 29) 0.851 755 52 × 2 = 1 + 0.703 511 04;
  • 30) 0.703 511 04 × 2 = 1 + 0.407 022 08;
  • 31) 0.407 022 08 × 2 = 0 + 0.814 044 16;
  • 32) 0.814 044 16 × 2 = 1 + 0.628 088 32;
  • 33) 0.628 088 32 × 2 = 1 + 0.256 176 64;
  • 34) 0.256 176 64 × 2 = 0 + 0.512 353 28;
  • 35) 0.512 353 28 × 2 = 1 + 0.024 706 56;
  • 36) 0.024 706 56 × 2 = 0 + 0.049 413 12;
  • 37) 0.049 413 12 × 2 = 0 + 0.098 826 24;
  • 38) 0.098 826 24 × 2 = 0 + 0.197 652 48;
  • 39) 0.197 652 48 × 2 = 0 + 0.395 304 96;
  • 40) 0.395 304 96 × 2 = 0 + 0.790 609 92;
  • 41) 0.790 609 92 × 2 = 1 + 0.581 219 84;
  • 42) 0.581 219 84 × 2 = 1 + 0.162 439 68;
  • 43) 0.162 439 68 × 2 = 0 + 0.324 879 36;
  • 44) 0.324 879 36 × 2 = 0 + 0.649 758 72;
  • 45) 0.649 758 72 × 2 = 1 + 0.299 517 44;
  • 46) 0.299 517 44 × 2 = 0 + 0.599 034 88;
  • 47) 0.599 034 88 × 2 = 1 + 0.198 069 76;
  • 48) 0.198 069 76 × 2 = 0 + 0.396 139 52;
  • 49) 0.396 139 52 × 2 = 0 + 0.792 279 04;
  • 50) 0.792 279 04 × 2 = 1 + 0.584 558 08;
  • 51) 0.584 558 08 × 2 = 1 + 0.169 116 16;
  • 52) 0.169 116 16 × 2 = 0 + 0.338 232 32;
  • 53) 0.338 232 32 × 2 = 0 + 0.676 464 64;
  • 54) 0.676 464 64 × 2 = 1 + 0.352 929 28;
  • 55) 0.352 929 28 × 2 = 0 + 0.705 858 56;
  • 56) 0.705 858 56 × 2 = 1 + 0.411 717 12;
  • 57) 0.411 717 12 × 2 = 0 + 0.823 434 24;
  • 58) 0.823 434 24 × 2 = 1 + 0.646 868 48;
  • 59) 0.646 868 48 × 2 = 1 + 0.293 736 96;
  • 60) 0.293 736 96 × 2 = 0 + 0.587 473 92;
  • 61) 0.587 473 92 × 2 = 1 + 0.174 947 84;
  • 62) 0.174 947 84 × 2 = 0 + 0.349 895 68;
  • 63) 0.349 895 68 × 2 = 0 + 0.699 791 36;
  • 64) 0.699 791 36 × 2 = 1 + 0.399 582 72;
  • 65) 0.399 582 72 × 2 = 0 + 0.799 165 44;
  • 66) 0.799 165 44 × 2 = 1 + 0.598 330 88;
  • 67) 0.598 330 88 × 2 = 1 + 0.196 661 76;
  • 68) 0.196 661 76 × 2 = 0 + 0.393 323 52;
  • 69) 0.393 323 52 × 2 = 0 + 0.786 647 04;
  • 70) 0.786 647 04 × 2 = 1 + 0.573 294 08;
  • 71) 0.573 294 08 × 2 = 1 + 0.146 588 16;
  • 72) 0.146 588 16 × 2 = 0 + 0.293 176 32;
  • 73) 0.293 176 32 × 2 = 0 + 0.586 352 64;

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:


0.000 000 67(10) =

0.0000 0000 0000 0000 0000 1011 0011 1101 1010 0000 1100 1010 0110 0101 0110 1001 0110 0110 0(2)

5. Positive number before normalization:

0.000 000 67(10) =

0.0000 0000 0000 0000 0000 1011 0011 1101 1010 0000 1100 1010 0110 0101 0110 1001 0110 0110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 67(10) =

0.0000 0000 0000 0000 0000 1011 0011 1101 1010 0000 1100 1010 0110 0101 0110 1001 0110 0110 0(2) =

0.0000 0000 0000 0000 0000 1011 0011 1101 1010 0000 1100 1010 0110 0101 0110 1001 0110 0110 0(2) × 20 =

1.0110 0111 1011 0100 0001 1001 0100 1100 1010 1101 0010 1100 1100(2) × 2-21


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): -21

Mantissa (not normalized):
1.0110 0111 1011 0100 0001 1001 0100 1100 1010 1101 0010 1100 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(11-1) - 1 =

(-21 + 1 023)(10) =

1 002(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;
  • 1 002 ÷ 2 = 501 + 0;
  • 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) =

1002(10) =

011 1110 1010(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. 0110 0111 1011 0100 0001 1001 0100 1100 1010 1101 0010 1100 1100 =

0110 0111 1011 0100 0001 1001 0100 1100 1010 1101 0010 1100 1100


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 1010

Mantissa (52 bits) =
0110 0111 1011 0100 0001 1001 0100 1100 1010 1101 0010 1100 1100


Decimal number 0.000 000 67 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0110 0111 1011 0100 0001 1001 0100 1100 1010 1101 0010 1100 1100

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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