-0.000 282 003 61 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 61(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
-0.000 282 003 61(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. Start with the positive version of the number:

|-0.000 282 003 61| = 0.000 282 003 61


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: 0.000 282 003 61.

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) 0.000 282 003 61 × 2 = 0 + 0.000 564 007 22;
  • 2) 0.000 564 007 22 × 2 = 0 + 0.001 128 014 44;
  • 3) 0.001 128 014 44 × 2 = 0 + 0.002 256 028 88;
  • 4) 0.002 256 028 88 × 2 = 0 + 0.004 512 057 76;
  • 5) 0.004 512 057 76 × 2 = 0 + 0.009 024 115 52;
  • 6) 0.009 024 115 52 × 2 = 0 + 0.018 048 231 04;
  • 7) 0.018 048 231 04 × 2 = 0 + 0.036 096 462 08;
  • 8) 0.036 096 462 08 × 2 = 0 + 0.072 192 924 16;
  • 9) 0.072 192 924 16 × 2 = 0 + 0.144 385 848 32;
  • 10) 0.144 385 848 32 × 2 = 0 + 0.288 771 696 64;
  • 11) 0.288 771 696 64 × 2 = 0 + 0.577 543 393 28;
  • 12) 0.577 543 393 28 × 2 = 1 + 0.155 086 786 56;
  • 13) 0.155 086 786 56 × 2 = 0 + 0.310 173 573 12;
  • 14) 0.310 173 573 12 × 2 = 0 + 0.620 347 146 24;
  • 15) 0.620 347 146 24 × 2 = 1 + 0.240 694 292 48;
  • 16) 0.240 694 292 48 × 2 = 0 + 0.481 388 584 96;
  • 17) 0.481 388 584 96 × 2 = 0 + 0.962 777 169 92;
  • 18) 0.962 777 169 92 × 2 = 1 + 0.925 554 339 84;
  • 19) 0.925 554 339 84 × 2 = 1 + 0.851 108 679 68;
  • 20) 0.851 108 679 68 × 2 = 1 + 0.702 217 359 36;
  • 21) 0.702 217 359 36 × 2 = 1 + 0.404 434 718 72;
  • 22) 0.404 434 718 72 × 2 = 0 + 0.808 869 437 44;
  • 23) 0.808 869 437 44 × 2 = 1 + 0.617 738 874 88;
  • 24) 0.617 738 874 88 × 2 = 1 + 0.235 477 749 76;
  • 25) 0.235 477 749 76 × 2 = 0 + 0.470 955 499 52;
  • 26) 0.470 955 499 52 × 2 = 0 + 0.941 910 999 04;
  • 27) 0.941 910 999 04 × 2 = 1 + 0.883 821 998 08;
  • 28) 0.883 821 998 08 × 2 = 1 + 0.767 643 996 16;
  • 29) 0.767 643 996 16 × 2 = 1 + 0.535 287 992 32;
  • 30) 0.535 287 992 32 × 2 = 1 + 0.070 575 984 64;
  • 31) 0.070 575 984 64 × 2 = 0 + 0.141 151 969 28;
  • 32) 0.141 151 969 28 × 2 = 0 + 0.282 303 938 56;
  • 33) 0.282 303 938 56 × 2 = 0 + 0.564 607 877 12;
  • 34) 0.564 607 877 12 × 2 = 1 + 0.129 215 754 24;
  • 35) 0.129 215 754 24 × 2 = 0 + 0.258 431 508 48;
  • 36) 0.258 431 508 48 × 2 = 0 + 0.516 863 016 96;
  • 37) 0.516 863 016 96 × 2 = 1 + 0.033 726 033 92;
  • 38) 0.033 726 033 92 × 2 = 0 + 0.067 452 067 84;
  • 39) 0.067 452 067 84 × 2 = 0 + 0.134 904 135 68;
  • 40) 0.134 904 135 68 × 2 = 0 + 0.269 808 271 36;
  • 41) 0.269 808 271 36 × 2 = 0 + 0.539 616 542 72;
  • 42) 0.539 616 542 72 × 2 = 1 + 0.079 233 085 44;
  • 43) 0.079 233 085 44 × 2 = 0 + 0.158 466 170 88;
  • 44) 0.158 466 170 88 × 2 = 0 + 0.316 932 341 76;
  • 45) 0.316 932 341 76 × 2 = 0 + 0.633 864 683 52;
  • 46) 0.633 864 683 52 × 2 = 1 + 0.267 729 367 04;
  • 47) 0.267 729 367 04 × 2 = 0 + 0.535 458 734 08;
  • 48) 0.535 458 734 08 × 2 = 1 + 0.070 917 468 16;
  • 49) 0.070 917 468 16 × 2 = 0 + 0.141 834 936 32;
  • 50) 0.141 834 936 32 × 2 = 0 + 0.283 669 872 64;
  • 51) 0.283 669 872 64 × 2 = 0 + 0.567 339 745 28;
  • 52) 0.567 339 745 28 × 2 = 1 + 0.134 679 490 56;
  • 53) 0.134 679 490 56 × 2 = 0 + 0.269 358 981 12;
  • 54) 0.269 358 981 12 × 2 = 0 + 0.538 717 962 24;
  • 55) 0.538 717 962 24 × 2 = 1 + 0.077 435 924 48;
  • 56) 0.077 435 924 48 × 2 = 0 + 0.154 871 848 96;
  • 57) 0.154 871 848 96 × 2 = 0 + 0.309 743 697 92;
  • 58) 0.309 743 697 92 × 2 = 0 + 0.619 487 395 84;
  • 59) 0.619 487 395 84 × 2 = 1 + 0.238 974 791 68;
  • 60) 0.238 974 791 68 × 2 = 0 + 0.477 949 583 36;
  • 61) 0.477 949 583 36 × 2 = 0 + 0.955 899 166 72;
  • 62) 0.955 899 166 72 × 2 = 1 + 0.911 798 333 44;
  • 63) 0.911 798 333 44 × 2 = 1 + 0.823 596 666 88;
  • 64) 0.823 596 666 88 × 2 = 1 + 0.647 193 333 76;

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).


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:


0.000 282 003 61(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111(2)

6. Positive number before normalization:

0.000 282 003 61(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 003 61(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111(2) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111(2) × 20 =

1.0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111 =

0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111


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 0011

Mantissa (52 bits) =
0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 0010 0111


Decimal number -0.000 282 003 61 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1100 0100 1000 0100 0101 0001 0010 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