-0.000 282 004 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 06(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
-0.000 282 004 06(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:

|-0.000 282 004 06| = 0.000 282 004 06


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: 0.000 282 004 06.

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) 0.000 282 004 06 × 2 = 0 + 0.000 564 008 12;
  • 2) 0.000 564 008 12 × 2 = 0 + 0.001 128 016 24;
  • 3) 0.001 128 016 24 × 2 = 0 + 0.002 256 032 48;
  • 4) 0.002 256 032 48 × 2 = 0 + 0.004 512 064 96;
  • 5) 0.004 512 064 96 × 2 = 0 + 0.009 024 129 92;
  • 6) 0.009 024 129 92 × 2 = 0 + 0.018 048 259 84;
  • 7) 0.018 048 259 84 × 2 = 0 + 0.036 096 519 68;
  • 8) 0.036 096 519 68 × 2 = 0 + 0.072 193 039 36;
  • 9) 0.072 193 039 36 × 2 = 0 + 0.144 386 078 72;
  • 10) 0.144 386 078 72 × 2 = 0 + 0.288 772 157 44;
  • 11) 0.288 772 157 44 × 2 = 0 + 0.577 544 314 88;
  • 12) 0.577 544 314 88 × 2 = 1 + 0.155 088 629 76;
  • 13) 0.155 088 629 76 × 2 = 0 + 0.310 177 259 52;
  • 14) 0.310 177 259 52 × 2 = 0 + 0.620 354 519 04;
  • 15) 0.620 354 519 04 × 2 = 1 + 0.240 709 038 08;
  • 16) 0.240 709 038 08 × 2 = 0 + 0.481 418 076 16;
  • 17) 0.481 418 076 16 × 2 = 0 + 0.962 836 152 32;
  • 18) 0.962 836 152 32 × 2 = 1 + 0.925 672 304 64;
  • 19) 0.925 672 304 64 × 2 = 1 + 0.851 344 609 28;
  • 20) 0.851 344 609 28 × 2 = 1 + 0.702 689 218 56;
  • 21) 0.702 689 218 56 × 2 = 1 + 0.405 378 437 12;
  • 22) 0.405 378 437 12 × 2 = 0 + 0.810 756 874 24;
  • 23) 0.810 756 874 24 × 2 = 1 + 0.621 513 748 48;
  • 24) 0.621 513 748 48 × 2 = 1 + 0.243 027 496 96;
  • 25) 0.243 027 496 96 × 2 = 0 + 0.486 054 993 92;
  • 26) 0.486 054 993 92 × 2 = 0 + 0.972 109 987 84;
  • 27) 0.972 109 987 84 × 2 = 1 + 0.944 219 975 68;
  • 28) 0.944 219 975 68 × 2 = 1 + 0.888 439 951 36;
  • 29) 0.888 439 951 36 × 2 = 1 + 0.776 879 902 72;
  • 30) 0.776 879 902 72 × 2 = 1 + 0.553 759 805 44;
  • 31) 0.553 759 805 44 × 2 = 1 + 0.107 519 610 88;
  • 32) 0.107 519 610 88 × 2 = 0 + 0.215 039 221 76;
  • 33) 0.215 039 221 76 × 2 = 0 + 0.430 078 443 52;
  • 34) 0.430 078 443 52 × 2 = 0 + 0.860 156 887 04;
  • 35) 0.860 156 887 04 × 2 = 1 + 0.720 313 774 08;
  • 36) 0.720 313 774 08 × 2 = 1 + 0.440 627 548 16;
  • 37) 0.440 627 548 16 × 2 = 0 + 0.881 255 096 32;
  • 38) 0.881 255 096 32 × 2 = 1 + 0.762 510 192 64;
  • 39) 0.762 510 192 64 × 2 = 1 + 0.525 020 385 28;
  • 40) 0.525 020 385 28 × 2 = 1 + 0.050 040 770 56;
  • 41) 0.050 040 770 56 × 2 = 0 + 0.100 081 541 12;
  • 42) 0.100 081 541 12 × 2 = 0 + 0.200 163 082 24;
  • 43) 0.200 163 082 24 × 2 = 0 + 0.400 326 164 48;
  • 44) 0.400 326 164 48 × 2 = 0 + 0.800 652 328 96;
  • 45) 0.800 652 328 96 × 2 = 1 + 0.601 304 657 92;
  • 46) 0.601 304 657 92 × 2 = 1 + 0.202 609 315 84;
  • 47) 0.202 609 315 84 × 2 = 0 + 0.405 218 631 68;
  • 48) 0.405 218 631 68 × 2 = 0 + 0.810 437 263 36;
  • 49) 0.810 437 263 36 × 2 = 1 + 0.620 874 526 72;
  • 50) 0.620 874 526 72 × 2 = 1 + 0.241 749 053 44;
  • 51) 0.241 749 053 44 × 2 = 0 + 0.483 498 106 88;
  • 52) 0.483 498 106 88 × 2 = 0 + 0.966 996 213 76;
  • 53) 0.966 996 213 76 × 2 = 1 + 0.933 992 427 52;
  • 54) 0.933 992 427 52 × 2 = 1 + 0.867 984 855 04;
  • 55) 0.867 984 855 04 × 2 = 1 + 0.735 969 710 08;
  • 56) 0.735 969 710 08 × 2 = 1 + 0.471 939 420 16;
  • 57) 0.471 939 420 16 × 2 = 0 + 0.943 878 840 32;
  • 58) 0.943 878 840 32 × 2 = 1 + 0.887 757 680 64;
  • 59) 0.887 757 680 64 × 2 = 1 + 0.775 515 361 28;
  • 60) 0.775 515 361 28 × 2 = 1 + 0.551 030 722 56;
  • 61) 0.551 030 722 56 × 2 = 1 + 0.102 061 445 12;
  • 62) 0.102 061 445 12 × 2 = 0 + 0.204 122 890 24;
  • 63) 0.204 122 890 24 × 2 = 0 + 0.408 245 780 48;
  • 64) 0.408 245 780 48 × 2 = 0 + 0.816 491 560 96;

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:


0.000 282 004 06(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0011 0111 0000 1100 1100 1111 0111 1000(2)

6. Positive number before normalization:

0.000 282 004 06(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0011 0111 0000 1100 1100 1111 0111 1000(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:


0.000 282 004 06(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0011 0111 0000 1100 1100 1111 0111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0011 0111 0000 1100 1100 1111 0111 1000(2) × 20 =

1.0010 0111 1011 0011 1110 0011 0111 0000 1100 1100 1111 0111 1000(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 1110 0011 0111 0000 1100 1100 1111 0111 1000


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 1110 0011 0111 0000 1100 1100 1111 0111 1000 =

0010 0111 1011 0011 1110 0011 0111 0000 1100 1100 1111 0111 1000


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 1110 0011 0111 0000 1100 1100 1111 0111 1000


Decimal number -0.000 282 004 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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