-0.000 282 005 913 82 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 82(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
-0.000 282 005 913 82(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:

|-0.000 282 005 913 82| = 0.000 282 005 913 82


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: 0.000 282 005 913 82.

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) 0.000 282 005 913 82 × 2 = 0 + 0.000 564 011 827 64;
  • 2) 0.000 564 011 827 64 × 2 = 0 + 0.001 128 023 655 28;
  • 3) 0.001 128 023 655 28 × 2 = 0 + 0.002 256 047 310 56;
  • 4) 0.002 256 047 310 56 × 2 = 0 + 0.004 512 094 621 12;
  • 5) 0.004 512 094 621 12 × 2 = 0 + 0.009 024 189 242 24;
  • 6) 0.009 024 189 242 24 × 2 = 0 + 0.018 048 378 484 48;
  • 7) 0.018 048 378 484 48 × 2 = 0 + 0.036 096 756 968 96;
  • 8) 0.036 096 756 968 96 × 2 = 0 + 0.072 193 513 937 92;
  • 9) 0.072 193 513 937 92 × 2 = 0 + 0.144 387 027 875 84;
  • 10) 0.144 387 027 875 84 × 2 = 0 + 0.288 774 055 751 68;
  • 11) 0.288 774 055 751 68 × 2 = 0 + 0.577 548 111 503 36;
  • 12) 0.577 548 111 503 36 × 2 = 1 + 0.155 096 223 006 72;
  • 13) 0.155 096 223 006 72 × 2 = 0 + 0.310 192 446 013 44;
  • 14) 0.310 192 446 013 44 × 2 = 0 + 0.620 384 892 026 88;
  • 15) 0.620 384 892 026 88 × 2 = 1 + 0.240 769 784 053 76;
  • 16) 0.240 769 784 053 76 × 2 = 0 + 0.481 539 568 107 52;
  • 17) 0.481 539 568 107 52 × 2 = 0 + 0.963 079 136 215 04;
  • 18) 0.963 079 136 215 04 × 2 = 1 + 0.926 158 272 430 08;
  • 19) 0.926 158 272 430 08 × 2 = 1 + 0.852 316 544 860 16;
  • 20) 0.852 316 544 860 16 × 2 = 1 + 0.704 633 089 720 32;
  • 21) 0.704 633 089 720 32 × 2 = 1 + 0.409 266 179 440 64;
  • 22) 0.409 266 179 440 64 × 2 = 0 + 0.818 532 358 881 28;
  • 23) 0.818 532 358 881 28 × 2 = 1 + 0.637 064 717 762 56;
  • 24) 0.637 064 717 762 56 × 2 = 1 + 0.274 129 435 525 12;
  • 25) 0.274 129 435 525 12 × 2 = 0 + 0.548 258 871 050 24;
  • 26) 0.548 258 871 050 24 × 2 = 1 + 0.096 517 742 100 48;
  • 27) 0.096 517 742 100 48 × 2 = 0 + 0.193 035 484 200 96;
  • 28) 0.193 035 484 200 96 × 2 = 0 + 0.386 070 968 401 92;
  • 29) 0.386 070 968 401 92 × 2 = 0 + 0.772 141 936 803 84;
  • 30) 0.772 141 936 803 84 × 2 = 1 + 0.544 283 873 607 68;
  • 31) 0.544 283 873 607 68 × 2 = 1 + 0.088 567 747 215 36;
  • 32) 0.088 567 747 215 36 × 2 = 0 + 0.177 135 494 430 72;
  • 33) 0.177 135 494 430 72 × 2 = 0 + 0.354 270 988 861 44;
  • 34) 0.354 270 988 861 44 × 2 = 0 + 0.708 541 977 722 88;
  • 35) 0.708 541 977 722 88 × 2 = 1 + 0.417 083 955 445 76;
  • 36) 0.417 083 955 445 76 × 2 = 0 + 0.834 167 910 891 52;
  • 37) 0.834 167 910 891 52 × 2 = 1 + 0.668 335 821 783 04;
  • 38) 0.668 335 821 783 04 × 2 = 1 + 0.336 671 643 566 08;
  • 39) 0.336 671 643 566 08 × 2 = 0 + 0.673 343 287 132 16;
  • 40) 0.673 343 287 132 16 × 2 = 1 + 0.346 686 574 264 32;
  • 41) 0.346 686 574 264 32 × 2 = 0 + 0.693 373 148 528 64;
  • 42) 0.693 373 148 528 64 × 2 = 1 + 0.386 746 297 057 28;
  • 43) 0.386 746 297 057 28 × 2 = 0 + 0.773 492 594 114 56;
  • 44) 0.773 492 594 114 56 × 2 = 1 + 0.546 985 188 229 12;
  • 45) 0.546 985 188 229 12 × 2 = 1 + 0.093 970 376 458 24;
  • 46) 0.093 970 376 458 24 × 2 = 0 + 0.187 940 752 916 48;
  • 47) 0.187 940 752 916 48 × 2 = 0 + 0.375 881 505 832 96;
  • 48) 0.375 881 505 832 96 × 2 = 0 + 0.751 763 011 665 92;
  • 49) 0.751 763 011 665 92 × 2 = 1 + 0.503 526 023 331 84;
  • 50) 0.503 526 023 331 84 × 2 = 1 + 0.007 052 046 663 68;
  • 51) 0.007 052 046 663 68 × 2 = 0 + 0.014 104 093 327 36;
  • 52) 0.014 104 093 327 36 × 2 = 0 + 0.028 208 186 654 72;
  • 53) 0.028 208 186 654 72 × 2 = 0 + 0.056 416 373 309 44;
  • 54) 0.056 416 373 309 44 × 2 = 0 + 0.112 832 746 618 88;
  • 55) 0.112 832 746 618 88 × 2 = 0 + 0.225 665 493 237 76;
  • 56) 0.225 665 493 237 76 × 2 = 0 + 0.451 330 986 475 52;
  • 57) 0.451 330 986 475 52 × 2 = 0 + 0.902 661 972 951 04;
  • 58) 0.902 661 972 951 04 × 2 = 1 + 0.805 323 945 902 08;
  • 59) 0.805 323 945 902 08 × 2 = 1 + 0.610 647 891 804 16;
  • 60) 0.610 647 891 804 16 × 2 = 1 + 0.221 295 783 608 32;
  • 61) 0.221 295 783 608 32 × 2 = 0 + 0.442 591 567 216 64;
  • 62) 0.442 591 567 216 64 × 2 = 0 + 0.885 183 134 433 28;
  • 63) 0.885 183 134 433 28 × 2 = 1 + 0.770 366 268 866 56;
  • 64) 0.770 366 268 866 56 × 2 = 1 + 0.540 732 537 733 12;

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:


0.000 282 005 913 82(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0101 1000 1100 0000 0111 0011(2)

6. Positive number before normalization:

0.000 282 005 913 82(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0101 1000 1100 0000 0111 0011(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:


0.000 282 005 913 82(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0101 1000 1100 0000 0111 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0101 1000 1100 0000 0111 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 0101 1000 1100 0000 0111 0011(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 0110 0010 1101 0101 1000 1100 0000 0111 0011


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 0110 0010 1101 0101 1000 1100 0000 0111 0011 =

0010 0111 1011 0100 0110 0010 1101 0101 1000 1100 0000 0111 0011


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 0110 0010 1101 0101 1000 1100 0000 0111 0011


Decimal number -0.000 282 005 913 82 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 0101 1000 1100 0000 0111 0011

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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