-0.000 282 022 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 022 2(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
-0.000 282 022 2(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:

|-0.000 282 022 2| = 0.000 282 022 2


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: 0.000 282 022 2.

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) 0.000 282 022 2 × 2 = 0 + 0.000 564 044 4;
  • 2) 0.000 564 044 4 × 2 = 0 + 0.001 128 088 8;
  • 3) 0.001 128 088 8 × 2 = 0 + 0.002 256 177 6;
  • 4) 0.002 256 177 6 × 2 = 0 + 0.004 512 355 2;
  • 5) 0.004 512 355 2 × 2 = 0 + 0.009 024 710 4;
  • 6) 0.009 024 710 4 × 2 = 0 + 0.018 049 420 8;
  • 7) 0.018 049 420 8 × 2 = 0 + 0.036 098 841 6;
  • 8) 0.036 098 841 6 × 2 = 0 + 0.072 197 683 2;
  • 9) 0.072 197 683 2 × 2 = 0 + 0.144 395 366 4;
  • 10) 0.144 395 366 4 × 2 = 0 + 0.288 790 732 8;
  • 11) 0.288 790 732 8 × 2 = 0 + 0.577 581 465 6;
  • 12) 0.577 581 465 6 × 2 = 1 + 0.155 162 931 2;
  • 13) 0.155 162 931 2 × 2 = 0 + 0.310 325 862 4;
  • 14) 0.310 325 862 4 × 2 = 0 + 0.620 651 724 8;
  • 15) 0.620 651 724 8 × 2 = 1 + 0.241 303 449 6;
  • 16) 0.241 303 449 6 × 2 = 0 + 0.482 606 899 2;
  • 17) 0.482 606 899 2 × 2 = 0 + 0.965 213 798 4;
  • 18) 0.965 213 798 4 × 2 = 1 + 0.930 427 596 8;
  • 19) 0.930 427 596 8 × 2 = 1 + 0.860 855 193 6;
  • 20) 0.860 855 193 6 × 2 = 1 + 0.721 710 387 2;
  • 21) 0.721 710 387 2 × 2 = 1 + 0.443 420 774 4;
  • 22) 0.443 420 774 4 × 2 = 0 + 0.886 841 548 8;
  • 23) 0.886 841 548 8 × 2 = 1 + 0.773 683 097 6;
  • 24) 0.773 683 097 6 × 2 = 1 + 0.547 366 195 2;
  • 25) 0.547 366 195 2 × 2 = 1 + 0.094 732 390 4;
  • 26) 0.094 732 390 4 × 2 = 0 + 0.189 464 780 8;
  • 27) 0.189 464 780 8 × 2 = 0 + 0.378 929 561 6;
  • 28) 0.378 929 561 6 × 2 = 0 + 0.757 859 123 2;
  • 29) 0.757 859 123 2 × 2 = 1 + 0.515 718 246 4;
  • 30) 0.515 718 246 4 × 2 = 1 + 0.031 436 492 8;
  • 31) 0.031 436 492 8 × 2 = 0 + 0.062 872 985 6;
  • 32) 0.062 872 985 6 × 2 = 0 + 0.125 745 971 2;
  • 33) 0.125 745 971 2 × 2 = 0 + 0.251 491 942 4;
  • 34) 0.251 491 942 4 × 2 = 0 + 0.502 983 884 8;
  • 35) 0.502 983 884 8 × 2 = 1 + 0.005 967 769 6;
  • 36) 0.005 967 769 6 × 2 = 0 + 0.011 935 539 2;
  • 37) 0.011 935 539 2 × 2 = 0 + 0.023 871 078 4;
  • 38) 0.023 871 078 4 × 2 = 0 + 0.047 742 156 8;
  • 39) 0.047 742 156 8 × 2 = 0 + 0.095 484 313 6;
  • 40) 0.095 484 313 6 × 2 = 0 + 0.190 968 627 2;
  • 41) 0.190 968 627 2 × 2 = 0 + 0.381 937 254 4;
  • 42) 0.381 937 254 4 × 2 = 0 + 0.763 874 508 8;
  • 43) 0.763 874 508 8 × 2 = 1 + 0.527 749 017 6;
  • 44) 0.527 749 017 6 × 2 = 1 + 0.055 498 035 2;
  • 45) 0.055 498 035 2 × 2 = 0 + 0.110 996 070 4;
  • 46) 0.110 996 070 4 × 2 = 0 + 0.221 992 140 8;
  • 47) 0.221 992 140 8 × 2 = 0 + 0.443 984 281 6;
  • 48) 0.443 984 281 6 × 2 = 0 + 0.887 968 563 2;
  • 49) 0.887 968 563 2 × 2 = 1 + 0.775 937 126 4;
  • 50) 0.775 937 126 4 × 2 = 1 + 0.551 874 252 8;
  • 51) 0.551 874 252 8 × 2 = 1 + 0.103 748 505 6;
  • 52) 0.103 748 505 6 × 2 = 0 + 0.207 497 011 2;
  • 53) 0.207 497 011 2 × 2 = 0 + 0.414 994 022 4;
  • 54) 0.414 994 022 4 × 2 = 0 + 0.829 988 044 8;
  • 55) 0.829 988 044 8 × 2 = 1 + 0.659 976 089 6;
  • 56) 0.659 976 089 6 × 2 = 1 + 0.319 952 179 2;
  • 57) 0.319 952 179 2 × 2 = 0 + 0.639 904 358 4;
  • 58) 0.639 904 358 4 × 2 = 1 + 0.279 808 716 8;
  • 59) 0.279 808 716 8 × 2 = 0 + 0.559 617 433 6;
  • 60) 0.559 617 433 6 × 2 = 1 + 0.119 234 867 2;
  • 61) 0.119 234 867 2 × 2 = 0 + 0.238 469 734 4;
  • 62) 0.238 469 734 4 × 2 = 0 + 0.476 939 468 8;
  • 63) 0.476 939 468 8 × 2 = 0 + 0.953 878 937 6;
  • 64) 0.953 878 937 6 × 2 = 1 + 0.907 757 875 2;

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:


0.000 282 022 2(10) =

0.0000 0000 0001 0010 0111 1011 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001(2)

6. Positive number before normalization:

0.000 282 022 2(10) =

0.0000 0000 0001 0010 0111 1011 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001(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:


0.000 282 022 2(10) =

0.0000 0000 0001 0010 0111 1011 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001(2) =

0.0000 0000 0001 0010 0111 1011 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001(2) × 20 =

1.0010 0111 1011 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001(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 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001


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 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001 =

0010 0111 1011 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001


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 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001


Decimal number -0.000 282 022 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1000 1100 0010 0000 0011 0000 1110 0011 0101 0001

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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