-0.000 282 010 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 010 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
-0.000 282 010 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:

|-0.000 282 010 71| = 0.000 282 010 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: 0.000 282 010 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) 0.000 282 010 71 × 2 = 0 + 0.000 564 021 42;
  • 2) 0.000 564 021 42 × 2 = 0 + 0.001 128 042 84;
  • 3) 0.001 128 042 84 × 2 = 0 + 0.002 256 085 68;
  • 4) 0.002 256 085 68 × 2 = 0 + 0.004 512 171 36;
  • 5) 0.004 512 171 36 × 2 = 0 + 0.009 024 342 72;
  • 6) 0.009 024 342 72 × 2 = 0 + 0.018 048 685 44;
  • 7) 0.018 048 685 44 × 2 = 0 + 0.036 097 370 88;
  • 8) 0.036 097 370 88 × 2 = 0 + 0.072 194 741 76;
  • 9) 0.072 194 741 76 × 2 = 0 + 0.144 389 483 52;
  • 10) 0.144 389 483 52 × 2 = 0 + 0.288 778 967 04;
  • 11) 0.288 778 967 04 × 2 = 0 + 0.577 557 934 08;
  • 12) 0.577 557 934 08 × 2 = 1 + 0.155 115 868 16;
  • 13) 0.155 115 868 16 × 2 = 0 + 0.310 231 736 32;
  • 14) 0.310 231 736 32 × 2 = 0 + 0.620 463 472 64;
  • 15) 0.620 463 472 64 × 2 = 1 + 0.240 926 945 28;
  • 16) 0.240 926 945 28 × 2 = 0 + 0.481 853 890 56;
  • 17) 0.481 853 890 56 × 2 = 0 + 0.963 707 781 12;
  • 18) 0.963 707 781 12 × 2 = 1 + 0.927 415 562 24;
  • 19) 0.927 415 562 24 × 2 = 1 + 0.854 831 124 48;
  • 20) 0.854 831 124 48 × 2 = 1 + 0.709 662 248 96;
  • 21) 0.709 662 248 96 × 2 = 1 + 0.419 324 497 92;
  • 22) 0.419 324 497 92 × 2 = 0 + 0.838 648 995 84;
  • 23) 0.838 648 995 84 × 2 = 1 + 0.677 297 991 68;
  • 24) 0.677 297 991 68 × 2 = 1 + 0.354 595 983 36;
  • 25) 0.354 595 983 36 × 2 = 0 + 0.709 191 966 72;
  • 26) 0.709 191 966 72 × 2 = 1 + 0.418 383 933 44;
  • 27) 0.418 383 933 44 × 2 = 0 + 0.836 767 866 88;
  • 28) 0.836 767 866 88 × 2 = 1 + 0.673 535 733 76;
  • 29) 0.673 535 733 76 × 2 = 1 + 0.347 071 467 52;
  • 30) 0.347 071 467 52 × 2 = 0 + 0.694 142 935 04;
  • 31) 0.694 142 935 04 × 2 = 1 + 0.388 285 870 08;
  • 32) 0.388 285 870 08 × 2 = 0 + 0.776 571 740 16;
  • 33) 0.776 571 740 16 × 2 = 1 + 0.553 143 480 32;
  • 34) 0.553 143 480 32 × 2 = 1 + 0.106 286 960 64;
  • 35) 0.106 286 960 64 × 2 = 0 + 0.212 573 921 28;
  • 36) 0.212 573 921 28 × 2 = 0 + 0.425 147 842 56;
  • 37) 0.425 147 842 56 × 2 = 0 + 0.850 295 685 12;
  • 38) 0.850 295 685 12 × 2 = 1 + 0.700 591 370 24;
  • 39) 0.700 591 370 24 × 2 = 1 + 0.401 182 740 48;
  • 40) 0.401 182 740 48 × 2 = 0 + 0.802 365 480 96;
  • 41) 0.802 365 480 96 × 2 = 1 + 0.604 730 961 92;
  • 42) 0.604 730 961 92 × 2 = 1 + 0.209 461 923 84;
  • 43) 0.209 461 923 84 × 2 = 0 + 0.418 923 847 68;
  • 44) 0.418 923 847 68 × 2 = 0 + 0.837 847 695 36;
  • 45) 0.837 847 695 36 × 2 = 1 + 0.675 695 390 72;
  • 46) 0.675 695 390 72 × 2 = 1 + 0.351 390 781 44;
  • 47) 0.351 390 781 44 × 2 = 0 + 0.702 781 562 88;
  • 48) 0.702 781 562 88 × 2 = 1 + 0.405 563 125 76;
  • 49) 0.405 563 125 76 × 2 = 0 + 0.811 126 251 52;
  • 50) 0.811 126 251 52 × 2 = 1 + 0.622 252 503 04;
  • 51) 0.622 252 503 04 × 2 = 1 + 0.244 505 006 08;
  • 52) 0.244 505 006 08 × 2 = 0 + 0.489 010 012 16;
  • 53) 0.489 010 012 16 × 2 = 0 + 0.978 020 024 32;
  • 54) 0.978 020 024 32 × 2 = 1 + 0.956 040 048 64;
  • 55) 0.956 040 048 64 × 2 = 1 + 0.912 080 097 28;
  • 56) 0.912 080 097 28 × 2 = 1 + 0.824 160 194 56;
  • 57) 0.824 160 194 56 × 2 = 1 + 0.648 320 389 12;
  • 58) 0.648 320 389 12 × 2 = 1 + 0.296 640 778 24;
  • 59) 0.296 640 778 24 × 2 = 0 + 0.593 281 556 48;
  • 60) 0.593 281 556 48 × 2 = 1 + 0.186 563 112 96;
  • 61) 0.186 563 112 96 × 2 = 0 + 0.373 126 225 92;
  • 62) 0.373 126 225 92 × 2 = 0 + 0.746 252 451 84;
  • 63) 0.746 252 451 84 × 2 = 1 + 0.492 504 903 68;
  • 64) 0.492 504 903 68 × 2 = 0 + 0.985 009 807 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:


0.000 282 010 71(10) =

0.0000 0000 0001 0010 0111 1011 0101 1010 1100 0110 1100 1101 0110 0111 1101 0010(2)

6. Positive number before normalization:

0.000 282 010 71(10) =

0.0000 0000 0001 0010 0111 1011 0101 1010 1100 0110 1100 1101 0110 0111 1101 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:


0.000 282 010 71(10) =

0.0000 0000 0001 0010 0111 1011 0101 1010 1100 0110 1100 1101 0110 0111 1101 0010(2) =

0.0000 0000 0001 0010 0111 1011 0101 1010 1100 0110 1100 1101 0110 0111 1101 0010(2) × 20 =

1.0010 0111 1011 0101 1010 1100 0110 1100 1101 0110 0111 1101 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 0101 1010 1100 0110 1100 1101 0110 0111 1101 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 0101 1010 1100 0110 1100 1101 0110 0111 1101 0010 =

0010 0111 1011 0101 1010 1100 0110 1100 1101 0110 0111 1101 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 0101 1010 1100 0110 1100 1101 0110 0111 1101 0010


Decimal number -0.000 282 010 71 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1010 1100 0110 1100 1101 0110 0111 1101 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