-0.000 282 003 78 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 78(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 78(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 78| = 0.000 282 003 78


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 78.

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 78 × 2 = 0 + 0.000 564 007 56;
  • 2) 0.000 564 007 56 × 2 = 0 + 0.001 128 015 12;
  • 3) 0.001 128 015 12 × 2 = 0 + 0.002 256 030 24;
  • 4) 0.002 256 030 24 × 2 = 0 + 0.004 512 060 48;
  • 5) 0.004 512 060 48 × 2 = 0 + 0.009 024 120 96;
  • 6) 0.009 024 120 96 × 2 = 0 + 0.018 048 241 92;
  • 7) 0.018 048 241 92 × 2 = 0 + 0.036 096 483 84;
  • 8) 0.036 096 483 84 × 2 = 0 + 0.072 192 967 68;
  • 9) 0.072 192 967 68 × 2 = 0 + 0.144 385 935 36;
  • 10) 0.144 385 935 36 × 2 = 0 + 0.288 771 870 72;
  • 11) 0.288 771 870 72 × 2 = 0 + 0.577 543 741 44;
  • 12) 0.577 543 741 44 × 2 = 1 + 0.155 087 482 88;
  • 13) 0.155 087 482 88 × 2 = 0 + 0.310 174 965 76;
  • 14) 0.310 174 965 76 × 2 = 0 + 0.620 349 931 52;
  • 15) 0.620 349 931 52 × 2 = 1 + 0.240 699 863 04;
  • 16) 0.240 699 863 04 × 2 = 0 + 0.481 399 726 08;
  • 17) 0.481 399 726 08 × 2 = 0 + 0.962 799 452 16;
  • 18) 0.962 799 452 16 × 2 = 1 + 0.925 598 904 32;
  • 19) 0.925 598 904 32 × 2 = 1 + 0.851 197 808 64;
  • 20) 0.851 197 808 64 × 2 = 1 + 0.702 395 617 28;
  • 21) 0.702 395 617 28 × 2 = 1 + 0.404 791 234 56;
  • 22) 0.404 791 234 56 × 2 = 0 + 0.809 582 469 12;
  • 23) 0.809 582 469 12 × 2 = 1 + 0.619 164 938 24;
  • 24) 0.619 164 938 24 × 2 = 1 + 0.238 329 876 48;
  • 25) 0.238 329 876 48 × 2 = 0 + 0.476 659 752 96;
  • 26) 0.476 659 752 96 × 2 = 0 + 0.953 319 505 92;
  • 27) 0.953 319 505 92 × 2 = 1 + 0.906 639 011 84;
  • 28) 0.906 639 011 84 × 2 = 1 + 0.813 278 023 68;
  • 29) 0.813 278 023 68 × 2 = 1 + 0.626 556 047 36;
  • 30) 0.626 556 047 36 × 2 = 1 + 0.253 112 094 72;
  • 31) 0.253 112 094 72 × 2 = 0 + 0.506 224 189 44;
  • 32) 0.506 224 189 44 × 2 = 1 + 0.012 448 378 88;
  • 33) 0.012 448 378 88 × 2 = 0 + 0.024 896 757 76;
  • 34) 0.024 896 757 76 × 2 = 0 + 0.049 793 515 52;
  • 35) 0.049 793 515 52 × 2 = 0 + 0.099 587 031 04;
  • 36) 0.099 587 031 04 × 2 = 0 + 0.199 174 062 08;
  • 37) 0.199 174 062 08 × 2 = 0 + 0.398 348 124 16;
  • 38) 0.398 348 124 16 × 2 = 0 + 0.796 696 248 32;
  • 39) 0.796 696 248 32 × 2 = 1 + 0.593 392 496 64;
  • 40) 0.593 392 496 64 × 2 = 1 + 0.186 784 993 28;
  • 41) 0.186 784 993 28 × 2 = 0 + 0.373 569 986 56;
  • 42) 0.373 569 986 56 × 2 = 0 + 0.747 139 973 12;
  • 43) 0.747 139 973 12 × 2 = 1 + 0.494 279 946 24;
  • 44) 0.494 279 946 24 × 2 = 0 + 0.988 559 892 48;
  • 45) 0.988 559 892 48 × 2 = 1 + 0.977 119 784 96;
  • 46) 0.977 119 784 96 × 2 = 1 + 0.954 239 569 92;
  • 47) 0.954 239 569 92 × 2 = 1 + 0.908 479 139 84;
  • 48) 0.908 479 139 84 × 2 = 1 + 0.816 958 279 68;
  • 49) 0.816 958 279 68 × 2 = 1 + 0.633 916 559 36;
  • 50) 0.633 916 559 36 × 2 = 1 + 0.267 833 118 72;
  • 51) 0.267 833 118 72 × 2 = 0 + 0.535 666 237 44;
  • 52) 0.535 666 237 44 × 2 = 1 + 0.071 332 474 88;
  • 53) 0.071 332 474 88 × 2 = 0 + 0.142 664 949 76;
  • 54) 0.142 664 949 76 × 2 = 0 + 0.285 329 899 52;
  • 55) 0.285 329 899 52 × 2 = 0 + 0.570 659 799 04;
  • 56) 0.570 659 799 04 × 2 = 1 + 0.141 319 598 08;
  • 57) 0.141 319 598 08 × 2 = 0 + 0.282 639 196 16;
  • 58) 0.282 639 196 16 × 2 = 0 + 0.565 278 392 32;
  • 59) 0.565 278 392 32 × 2 = 1 + 0.130 556 784 64;
  • 60) 0.130 556 784 64 × 2 = 0 + 0.261 113 569 28;
  • 61) 0.261 113 569 28 × 2 = 0 + 0.522 227 138 56;
  • 62) 0.522 227 138 56 × 2 = 1 + 0.044 454 277 12;
  • 63) 0.044 454 277 12 × 2 = 0 + 0.088 908 554 24;
  • 64) 0.088 908 554 24 × 2 = 0 + 0.177 817 108 48;

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 78(10) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0000 0011 0010 1111 1101 0001 0010 0100(2)

6. Positive number before normalization:

0.000 282 003 78(10) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0000 0011 0010 1111 1101 0001 0010 0100(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 78(10) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0000 0011 0010 1111 1101 0001 0010 0100(2) =

0.0000 0000 0001 0010 0111 1011 0011 1101 0000 0011 0010 1111 1101 0001 0010 0100(2) × 20 =

1.0010 0111 1011 0011 1101 0000 0011 0010 1111 1101 0001 0010 0100(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 1101 0000 0011 0010 1111 1101 0001 0010 0100


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 1101 0000 0011 0010 1111 1101 0001 0010 0100 =

0010 0111 1011 0011 1101 0000 0011 0010 1111 1101 0001 0010 0100


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 1101 0000 0011 0010 1111 1101 0001 0010 0100


Decimal number -0.000 282 003 78 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1101 0000 0011 0010 1111 1101 0001 0010 0100

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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