-0.000 282 005 735 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 735(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
-0.000 282 005 735(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:

|-0.000 282 005 735| = 0.000 282 005 735


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: 0.000 282 005 735.

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) 0.000 282 005 735 × 2 = 0 + 0.000 564 011 47;
  • 2) 0.000 564 011 47 × 2 = 0 + 0.001 128 022 94;
  • 3) 0.001 128 022 94 × 2 = 0 + 0.002 256 045 88;
  • 4) 0.002 256 045 88 × 2 = 0 + 0.004 512 091 76;
  • 5) 0.004 512 091 76 × 2 = 0 + 0.009 024 183 52;
  • 6) 0.009 024 183 52 × 2 = 0 + 0.018 048 367 04;
  • 7) 0.018 048 367 04 × 2 = 0 + 0.036 096 734 08;
  • 8) 0.036 096 734 08 × 2 = 0 + 0.072 193 468 16;
  • 9) 0.072 193 468 16 × 2 = 0 + 0.144 386 936 32;
  • 10) 0.144 386 936 32 × 2 = 0 + 0.288 773 872 64;
  • 11) 0.288 773 872 64 × 2 = 0 + 0.577 547 745 28;
  • 12) 0.577 547 745 28 × 2 = 1 + 0.155 095 490 56;
  • 13) 0.155 095 490 56 × 2 = 0 + 0.310 190 981 12;
  • 14) 0.310 190 981 12 × 2 = 0 + 0.620 381 962 24;
  • 15) 0.620 381 962 24 × 2 = 1 + 0.240 763 924 48;
  • 16) 0.240 763 924 48 × 2 = 0 + 0.481 527 848 96;
  • 17) 0.481 527 848 96 × 2 = 0 + 0.963 055 697 92;
  • 18) 0.963 055 697 92 × 2 = 1 + 0.926 111 395 84;
  • 19) 0.926 111 395 84 × 2 = 1 + 0.852 222 791 68;
  • 20) 0.852 222 791 68 × 2 = 1 + 0.704 445 583 36;
  • 21) 0.704 445 583 36 × 2 = 1 + 0.408 891 166 72;
  • 22) 0.408 891 166 72 × 2 = 0 + 0.817 782 333 44;
  • 23) 0.817 782 333 44 × 2 = 1 + 0.635 564 666 88;
  • 24) 0.635 564 666 88 × 2 = 1 + 0.271 129 333 76;
  • 25) 0.271 129 333 76 × 2 = 0 + 0.542 258 667 52;
  • 26) 0.542 258 667 52 × 2 = 1 + 0.084 517 335 04;
  • 27) 0.084 517 335 04 × 2 = 0 + 0.169 034 670 08;
  • 28) 0.169 034 670 08 × 2 = 0 + 0.338 069 340 16;
  • 29) 0.338 069 340 16 × 2 = 0 + 0.676 138 680 32;
  • 30) 0.676 138 680 32 × 2 = 1 + 0.352 277 360 64;
  • 31) 0.352 277 360 64 × 2 = 0 + 0.704 554 721 28;
  • 32) 0.704 554 721 28 × 2 = 1 + 0.409 109 442 56;
  • 33) 0.409 109 442 56 × 2 = 0 + 0.818 218 885 12;
  • 34) 0.818 218 885 12 × 2 = 1 + 0.636 437 770 24;
  • 35) 0.636 437 770 24 × 2 = 1 + 0.272 875 540 48;
  • 36) 0.272 875 540 48 × 2 = 0 + 0.545 751 080 96;
  • 37) 0.545 751 080 96 × 2 = 1 + 0.091 502 161 92;
  • 38) 0.091 502 161 92 × 2 = 0 + 0.183 004 323 84;
  • 39) 0.183 004 323 84 × 2 = 0 + 0.366 008 647 68;
  • 40) 0.366 008 647 68 × 2 = 0 + 0.732 017 295 36;
  • 41) 0.732 017 295 36 × 2 = 1 + 0.464 034 590 72;
  • 42) 0.464 034 590 72 × 2 = 0 + 0.928 069 181 44;
  • 43) 0.928 069 181 44 × 2 = 1 + 0.856 138 362 88;
  • 44) 0.856 138 362 88 × 2 = 1 + 0.712 276 725 76;
  • 45) 0.712 276 725 76 × 2 = 1 + 0.424 553 451 52;
  • 46) 0.424 553 451 52 × 2 = 0 + 0.849 106 903 04;
  • 47) 0.849 106 903 04 × 2 = 1 + 0.698 213 806 08;
  • 48) 0.698 213 806 08 × 2 = 1 + 0.396 427 612 16;
  • 49) 0.396 427 612 16 × 2 = 0 + 0.792 855 224 32;
  • 50) 0.792 855 224 32 × 2 = 1 + 0.585 710 448 64;
  • 51) 0.585 710 448 64 × 2 = 1 + 0.171 420 897 28;
  • 52) 0.171 420 897 28 × 2 = 0 + 0.342 841 794 56;
  • 53) 0.342 841 794 56 × 2 = 0 + 0.685 683 589 12;
  • 54) 0.685 683 589 12 × 2 = 1 + 0.371 367 178 24;
  • 55) 0.371 367 178 24 × 2 = 0 + 0.742 734 356 48;
  • 56) 0.742 734 356 48 × 2 = 1 + 0.485 468 712 96;
  • 57) 0.485 468 712 96 × 2 = 0 + 0.970 937 425 92;
  • 58) 0.970 937 425 92 × 2 = 1 + 0.941 874 851 84;
  • 59) 0.941 874 851 84 × 2 = 1 + 0.883 749 703 68;
  • 60) 0.883 749 703 68 × 2 = 1 + 0.767 499 407 36;
  • 61) 0.767 499 407 36 × 2 = 1 + 0.534 998 814 72;
  • 62) 0.534 998 814 72 × 2 = 1 + 0.069 997 629 44;
  • 63) 0.069 997 629 44 × 2 = 0 + 0.139 995 258 88;
  • 64) 0.139 995 258 88 × 2 = 0 + 0.279 990 517 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:


0.000 282 005 735(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100(2)

6. Positive number before normalization:

0.000 282 005 735(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100(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:


0.000 282 005 735(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100(2) × 20 =

1.0010 0111 1011 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100(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 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100


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 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100 =

0010 0111 1011 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100


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 0100 0101 0110 1000 1011 1011 0110 0101 0111 1100


Decimal number -0.000 282 005 735 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0110 1000 1011 1011 0110 0101 0111 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