-0.000 282 027 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 027 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
-0.000 282 027 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:

|-0.000 282 027 2| = 0.000 282 027 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: 0.000 282 027 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) 0.000 282 027 2 × 2 = 0 + 0.000 564 054 4;
  • 2) 0.000 564 054 4 × 2 = 0 + 0.001 128 108 8;
  • 3) 0.001 128 108 8 × 2 = 0 + 0.002 256 217 6;
  • 4) 0.002 256 217 6 × 2 = 0 + 0.004 512 435 2;
  • 5) 0.004 512 435 2 × 2 = 0 + 0.009 024 870 4;
  • 6) 0.009 024 870 4 × 2 = 0 + 0.018 049 740 8;
  • 7) 0.018 049 740 8 × 2 = 0 + 0.036 099 481 6;
  • 8) 0.036 099 481 6 × 2 = 0 + 0.072 198 963 2;
  • 9) 0.072 198 963 2 × 2 = 0 + 0.144 397 926 4;
  • 10) 0.144 397 926 4 × 2 = 0 + 0.288 795 852 8;
  • 11) 0.288 795 852 8 × 2 = 0 + 0.577 591 705 6;
  • 12) 0.577 591 705 6 × 2 = 1 + 0.155 183 411 2;
  • 13) 0.155 183 411 2 × 2 = 0 + 0.310 366 822 4;
  • 14) 0.310 366 822 4 × 2 = 0 + 0.620 733 644 8;
  • 15) 0.620 733 644 8 × 2 = 1 + 0.241 467 289 6;
  • 16) 0.241 467 289 6 × 2 = 0 + 0.482 934 579 2;
  • 17) 0.482 934 579 2 × 2 = 0 + 0.965 869 158 4;
  • 18) 0.965 869 158 4 × 2 = 1 + 0.931 738 316 8;
  • 19) 0.931 738 316 8 × 2 = 1 + 0.863 476 633 6;
  • 20) 0.863 476 633 6 × 2 = 1 + 0.726 953 267 2;
  • 21) 0.726 953 267 2 × 2 = 1 + 0.453 906 534 4;
  • 22) 0.453 906 534 4 × 2 = 0 + 0.907 813 068 8;
  • 23) 0.907 813 068 8 × 2 = 1 + 0.815 626 137 6;
  • 24) 0.815 626 137 6 × 2 = 1 + 0.631 252 275 2;
  • 25) 0.631 252 275 2 × 2 = 1 + 0.262 504 550 4;
  • 26) 0.262 504 550 4 × 2 = 0 + 0.525 009 100 8;
  • 27) 0.525 009 100 8 × 2 = 1 + 0.050 018 201 6;
  • 28) 0.050 018 201 6 × 2 = 0 + 0.100 036 403 2;
  • 29) 0.100 036 403 2 × 2 = 0 + 0.200 072 806 4;
  • 30) 0.200 072 806 4 × 2 = 0 + 0.400 145 612 8;
  • 31) 0.400 145 612 8 × 2 = 0 + 0.800 291 225 6;
  • 32) 0.800 291 225 6 × 2 = 1 + 0.600 582 451 2;
  • 33) 0.600 582 451 2 × 2 = 1 + 0.201 164 902 4;
  • 34) 0.201 164 902 4 × 2 = 0 + 0.402 329 804 8;
  • 35) 0.402 329 804 8 × 2 = 0 + 0.804 659 609 6;
  • 36) 0.804 659 609 6 × 2 = 1 + 0.609 319 219 2;
  • 37) 0.609 319 219 2 × 2 = 1 + 0.218 638 438 4;
  • 38) 0.218 638 438 4 × 2 = 0 + 0.437 276 876 8;
  • 39) 0.437 276 876 8 × 2 = 0 + 0.874 553 753 6;
  • 40) 0.874 553 753 6 × 2 = 1 + 0.749 107 507 2;
  • 41) 0.749 107 507 2 × 2 = 1 + 0.498 215 014 4;
  • 42) 0.498 215 014 4 × 2 = 0 + 0.996 430 028 8;
  • 43) 0.996 430 028 8 × 2 = 1 + 0.992 860 057 6;
  • 44) 0.992 860 057 6 × 2 = 1 + 0.985 720 115 2;
  • 45) 0.985 720 115 2 × 2 = 1 + 0.971 440 230 4;
  • 46) 0.971 440 230 4 × 2 = 1 + 0.942 880 460 8;
  • 47) 0.942 880 460 8 × 2 = 1 + 0.885 760 921 6;
  • 48) 0.885 760 921 6 × 2 = 1 + 0.771 521 843 2;
  • 49) 0.771 521 843 2 × 2 = 1 + 0.543 043 686 4;
  • 50) 0.543 043 686 4 × 2 = 1 + 0.086 087 372 8;
  • 51) 0.086 087 372 8 × 2 = 0 + 0.172 174 745 6;
  • 52) 0.172 174 745 6 × 2 = 0 + 0.344 349 491 2;
  • 53) 0.344 349 491 2 × 2 = 0 + 0.688 698 982 4;
  • 54) 0.688 698 982 4 × 2 = 1 + 0.377 397 964 8;
  • 55) 0.377 397 964 8 × 2 = 0 + 0.754 795 929 6;
  • 56) 0.754 795 929 6 × 2 = 1 + 0.509 591 859 2;
  • 57) 0.509 591 859 2 × 2 = 1 + 0.019 183 718 4;
  • 58) 0.019 183 718 4 × 2 = 0 + 0.038 367 436 8;
  • 59) 0.038 367 436 8 × 2 = 0 + 0.076 734 873 6;
  • 60) 0.076 734 873 6 × 2 = 0 + 0.153 469 747 2;
  • 61) 0.153 469 747 2 × 2 = 0 + 0.306 939 494 4;
  • 62) 0.306 939 494 4 × 2 = 0 + 0.613 878 988 8;
  • 63) 0.613 878 988 8 × 2 = 1 + 0.227 757 977 6;
  • 64) 0.227 757 977 6 × 2 = 0 + 0.455 515 955 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:


0.000 282 027 2(10) =

0.0000 0000 0001 0010 0111 1011 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010(2)

6. Positive number before normalization:

0.000 282 027 2(10) =

0.0000 0000 0001 0010 0111 1011 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010(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:


0.000 282 027 2(10) =

0.0000 0000 0001 0010 0111 1011 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010(2) =

0.0000 0000 0001 0010 0111 1011 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010(2) × 20 =

1.0010 0111 1011 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010(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 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010


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 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010 =

0010 0111 1011 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010


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 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010


Decimal number -0.000 282 027 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1010 0001 1001 1001 1011 1111 1100 0101 1000 0010

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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