-0.000 282 006 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 71(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
-0.000 282 006 71(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:

|-0.000 282 006 71| = 0.000 282 006 71


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: 0.000 282 006 71.

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) 0.000 282 006 71 × 2 = 0 + 0.000 564 013 42;
  • 2) 0.000 564 013 42 × 2 = 0 + 0.001 128 026 84;
  • 3) 0.001 128 026 84 × 2 = 0 + 0.002 256 053 68;
  • 4) 0.002 256 053 68 × 2 = 0 + 0.004 512 107 36;
  • 5) 0.004 512 107 36 × 2 = 0 + 0.009 024 214 72;
  • 6) 0.009 024 214 72 × 2 = 0 + 0.018 048 429 44;
  • 7) 0.018 048 429 44 × 2 = 0 + 0.036 096 858 88;
  • 8) 0.036 096 858 88 × 2 = 0 + 0.072 193 717 76;
  • 9) 0.072 193 717 76 × 2 = 0 + 0.144 387 435 52;
  • 10) 0.144 387 435 52 × 2 = 0 + 0.288 774 871 04;
  • 11) 0.288 774 871 04 × 2 = 0 + 0.577 549 742 08;
  • 12) 0.577 549 742 08 × 2 = 1 + 0.155 099 484 16;
  • 13) 0.155 099 484 16 × 2 = 0 + 0.310 198 968 32;
  • 14) 0.310 198 968 32 × 2 = 0 + 0.620 397 936 64;
  • 15) 0.620 397 936 64 × 2 = 1 + 0.240 795 873 28;
  • 16) 0.240 795 873 28 × 2 = 0 + 0.481 591 746 56;
  • 17) 0.481 591 746 56 × 2 = 0 + 0.963 183 493 12;
  • 18) 0.963 183 493 12 × 2 = 1 + 0.926 366 986 24;
  • 19) 0.926 366 986 24 × 2 = 1 + 0.852 733 972 48;
  • 20) 0.852 733 972 48 × 2 = 1 + 0.705 467 944 96;
  • 21) 0.705 467 944 96 × 2 = 1 + 0.410 935 889 92;
  • 22) 0.410 935 889 92 × 2 = 0 + 0.821 871 779 84;
  • 23) 0.821 871 779 84 × 2 = 1 + 0.643 743 559 68;
  • 24) 0.643 743 559 68 × 2 = 1 + 0.287 487 119 36;
  • 25) 0.287 487 119 36 × 2 = 0 + 0.574 974 238 72;
  • 26) 0.574 974 238 72 × 2 = 1 + 0.149 948 477 44;
  • 27) 0.149 948 477 44 × 2 = 0 + 0.299 896 954 88;
  • 28) 0.299 896 954 88 × 2 = 0 + 0.599 793 909 76;
  • 29) 0.599 793 909 76 × 2 = 1 + 0.199 587 819 52;
  • 30) 0.199 587 819 52 × 2 = 0 + 0.399 175 639 04;
  • 31) 0.399 175 639 04 × 2 = 0 + 0.798 351 278 08;
  • 32) 0.798 351 278 08 × 2 = 1 + 0.596 702 556 16;
  • 33) 0.596 702 556 16 × 2 = 1 + 0.193 405 112 32;
  • 34) 0.193 405 112 32 × 2 = 0 + 0.386 810 224 64;
  • 35) 0.386 810 224 64 × 2 = 0 + 0.773 620 449 28;
  • 36) 0.773 620 449 28 × 2 = 1 + 0.547 240 898 56;
  • 37) 0.547 240 898 56 × 2 = 1 + 0.094 481 797 12;
  • 38) 0.094 481 797 12 × 2 = 0 + 0.188 963 594 24;
  • 39) 0.188 963 594 24 × 2 = 0 + 0.377 927 188 48;
  • 40) 0.377 927 188 48 × 2 = 0 + 0.755 854 376 96;
  • 41) 0.755 854 376 96 × 2 = 1 + 0.511 708 753 92;
  • 42) 0.511 708 753 92 × 2 = 1 + 0.023 417 507 84;
  • 43) 0.023 417 507 84 × 2 = 0 + 0.046 835 015 68;
  • 44) 0.046 835 015 68 × 2 = 0 + 0.093 670 031 36;
  • 45) 0.093 670 031 36 × 2 = 0 + 0.187 340 062 72;
  • 46) 0.187 340 062 72 × 2 = 0 + 0.374 680 125 44;
  • 47) 0.374 680 125 44 × 2 = 0 + 0.749 360 250 88;
  • 48) 0.749 360 250 88 × 2 = 1 + 0.498 720 501 76;
  • 49) 0.498 720 501 76 × 2 = 0 + 0.997 441 003 52;
  • 50) 0.997 441 003 52 × 2 = 1 + 0.994 882 007 04;
  • 51) 0.994 882 007 04 × 2 = 1 + 0.989 764 014 08;
  • 52) 0.989 764 014 08 × 2 = 1 + 0.979 528 028 16;
  • 53) 0.979 528 028 16 × 2 = 1 + 0.959 056 056 32;
  • 54) 0.959 056 056 32 × 2 = 1 + 0.918 112 112 64;
  • 55) 0.918 112 112 64 × 2 = 1 + 0.836 224 225 28;
  • 56) 0.836 224 225 28 × 2 = 1 + 0.672 448 450 56;
  • 57) 0.672 448 450 56 × 2 = 1 + 0.344 896 901 12;
  • 58) 0.344 896 901 12 × 2 = 0 + 0.689 793 802 24;
  • 59) 0.689 793 802 24 × 2 = 1 + 0.379 587 604 48;
  • 60) 0.379 587 604 48 × 2 = 0 + 0.759 175 208 96;
  • 61) 0.759 175 208 96 × 2 = 1 + 0.518 350 417 92;
  • 62) 0.518 350 417 92 × 2 = 1 + 0.036 700 835 84;
  • 63) 0.036 700 835 84 × 2 = 0 + 0.073 401 671 68;
  • 64) 0.073 401 671 68 × 2 = 0 + 0.146 803 343 36;

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:


0.000 282 006 71(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1001 1000 1100 0001 0111 1111 1010 1100(2)

6. Positive number before normalization:

0.000 282 006 71(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1001 1000 1100 0001 0111 1111 1010 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:


0.000 282 006 71(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1001 1000 1100 0001 0111 1111 1010 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1001 1000 1100 0001 0111 1111 1010 1100(2) × 20 =

1.0010 0111 1011 0100 1001 1001 1000 1100 0001 0111 1111 1010 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 1001 1001 1000 1100 0001 0111 1111 1010 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 1001 1001 1000 1100 0001 0111 1111 1010 1100 =

0010 0111 1011 0100 1001 1001 1000 1100 0001 0111 1111 1010 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 1001 1001 1000 1100 0001 0111 1111 1010 1100


Decimal number -0.000 282 006 71 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1001 1001 1000 1100 0001 0111 1111 1010 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