64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -0.154 999 999 999 94 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -0.154 999 999 999 94(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.154 999 999 999 94| = 0.154 999 999 999 94

2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.154 999 999 999 94.

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) 0.154 999 999 999 94 × 2 = 0 + 0.309 999 999 999 88;
  • 2) 0.309 999 999 999 88 × 2 = 0 + 0.619 999 999 999 76;
  • 3) 0.619 999 999 999 76 × 2 = 1 + 0.239 999 999 999 52;
  • 4) 0.239 999 999 999 52 × 2 = 0 + 0.479 999 999 999 04;
  • 5) 0.479 999 999 999 04 × 2 = 0 + 0.959 999 999 998 08;
  • 6) 0.959 999 999 998 08 × 2 = 1 + 0.919 999 999 996 16;
  • 7) 0.919 999 999 996 16 × 2 = 1 + 0.839 999 999 992 32;
  • 8) 0.839 999 999 992 32 × 2 = 1 + 0.679 999 999 984 64;
  • 9) 0.679 999 999 984 64 × 2 = 1 + 0.359 999 999 969 28;
  • 10) 0.359 999 999 969 28 × 2 = 0 + 0.719 999 999 938 56;
  • 11) 0.719 999 999 938 56 × 2 = 1 + 0.439 999 999 877 12;
  • 12) 0.439 999 999 877 12 × 2 = 0 + 0.879 999 999 754 24;
  • 13) 0.879 999 999 754 24 × 2 = 1 + 0.759 999 999 508 48;
  • 14) 0.759 999 999 508 48 × 2 = 1 + 0.519 999 999 016 96;
  • 15) 0.519 999 999 016 96 × 2 = 1 + 0.039 999 998 033 92;
  • 16) 0.039 999 998 033 92 × 2 = 0 + 0.079 999 996 067 84;
  • 17) 0.079 999 996 067 84 × 2 = 0 + 0.159 999 992 135 68;
  • 18) 0.159 999 992 135 68 × 2 = 0 + 0.319 999 984 271 36;
  • 19) 0.319 999 984 271 36 × 2 = 0 + 0.639 999 968 542 72;
  • 20) 0.639 999 968 542 72 × 2 = 1 + 0.279 999 937 085 44;
  • 21) 0.279 999 937 085 44 × 2 = 0 + 0.559 999 874 170 88;
  • 22) 0.559 999 874 170 88 × 2 = 1 + 0.119 999 748 341 76;
  • 23) 0.119 999 748 341 76 × 2 = 0 + 0.239 999 496 683 52;
  • 24) 0.239 999 496 683 52 × 2 = 0 + 0.479 998 993 367 04;
  • 25) 0.479 998 993 367 04 × 2 = 0 + 0.959 997 986 734 08;
  • 26) 0.959 997 986 734 08 × 2 = 1 + 0.919 995 973 468 16;
  • 27) 0.919 995 973 468 16 × 2 = 1 + 0.839 991 946 936 32;
  • 28) 0.839 991 946 936 32 × 2 = 1 + 0.679 983 893 872 64;
  • 29) 0.679 983 893 872 64 × 2 = 1 + 0.359 967 787 745 28;
  • 30) 0.359 967 787 745 28 × 2 = 0 + 0.719 935 575 490 56;
  • 31) 0.719 935 575 490 56 × 2 = 1 + 0.439 871 150 981 12;
  • 32) 0.439 871 150 981 12 × 2 = 0 + 0.879 742 301 962 24;
  • 33) 0.879 742 301 962 24 × 2 = 1 + 0.759 484 603 924 48;
  • 34) 0.759 484 603 924 48 × 2 = 1 + 0.518 969 207 848 96;
  • 35) 0.518 969 207 848 96 × 2 = 1 + 0.037 938 415 697 92;
  • 36) 0.037 938 415 697 92 × 2 = 0 + 0.075 876 831 395 84;
  • 37) 0.075 876 831 395 84 × 2 = 0 + 0.151 753 662 791 68;
  • 38) 0.151 753 662 791 68 × 2 = 0 + 0.303 507 325 583 36;
  • 39) 0.303 507 325 583 36 × 2 = 0 + 0.607 014 651 166 72;
  • 40) 0.607 014 651 166 72 × 2 = 1 + 0.214 029 302 333 44;
  • 41) 0.214 029 302 333 44 × 2 = 0 + 0.428 058 604 666 88;
  • 42) 0.428 058 604 666 88 × 2 = 0 + 0.856 117 209 333 76;
  • 43) 0.856 117 209 333 76 × 2 = 1 + 0.712 234 418 667 52;
  • 44) 0.712 234 418 667 52 × 2 = 1 + 0.424 468 837 335 04;
  • 45) 0.424 468 837 335 04 × 2 = 0 + 0.848 937 674 670 08;
  • 46) 0.848 937 674 670 08 × 2 = 1 + 0.697 875 349 340 16;
  • 47) 0.697 875 349 340 16 × 2 = 1 + 0.395 750 698 680 32;
  • 48) 0.395 750 698 680 32 × 2 = 0 + 0.791 501 397 360 64;
  • 49) 0.791 501 397 360 64 × 2 = 1 + 0.583 002 794 721 28;
  • 50) 0.583 002 794 721 28 × 2 = 1 + 0.166 005 589 442 56;
  • 51) 0.166 005 589 442 56 × 2 = 0 + 0.332 011 178 885 12;
  • 52) 0.332 011 178 885 12 × 2 = 0 + 0.664 022 357 770 24;
  • 53) 0.664 022 357 770 24 × 2 = 1 + 0.328 044 715 540 48;
  • 54) 0.328 044 715 540 48 × 2 = 0 + 0.656 089 431 080 96;
  • 55) 0.656 089 431 080 96 × 2 = 1 + 0.312 178 862 161 92;

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...)


5. 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:


0.154 999 999 999 94(10) =


0.0010 0111 1010 1110 0001 0100 0111 1010 1110 0001 0011 0110 1100 101(2)


6. Positive number before normalization:

0.154 999 999 999 94(10) =


0.0010 0111 1010 1110 0001 0100 0111 1010 1110 0001 0011 0110 1100 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.154 999 999 999 94(10) =


0.0010 0111 1010 1110 0001 0100 0111 1010 1110 0001 0011 0110 1100 101(2) =


0.0010 0111 1010 1110 0001 0100 0111 1010 1110 0001 0011 0110 1100 101(2) × 20 =


1.0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 1011 0110 0101(2) × 2-3


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): -3


Mantissa (not normalized):
1.0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 1011 0110 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


-3 + 2(11-1) - 1 =


(-3 + 1 023)(10) =


1 020(10)


10. 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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1020(10) =


011 1111 1100(2)


12. 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. 0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 1011 0110 0101 =


0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 1011 0110 0101


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
011 1111 1100


Mantissa (52 bits) =
0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 1011 0110 0101


The base ten decimal number -0.154 999 999 999 94 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 1100 - 0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 1011 0110 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