-0.000 282 003 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 72(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
-0.000 282 003 72(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:

|-0.000 282 003 72| = 0.000 282 003 72


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: 0.000 282 003 72.

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) 0.000 282 003 72 × 2 = 0 + 0.000 564 007 44;
  • 2) 0.000 564 007 44 × 2 = 0 + 0.001 128 014 88;
  • 3) 0.001 128 014 88 × 2 = 0 + 0.002 256 029 76;
  • 4) 0.002 256 029 76 × 2 = 0 + 0.004 512 059 52;
  • 5) 0.004 512 059 52 × 2 = 0 + 0.009 024 119 04;
  • 6) 0.009 024 119 04 × 2 = 0 + 0.018 048 238 08;
  • 7) 0.018 048 238 08 × 2 = 0 + 0.036 096 476 16;
  • 8) 0.036 096 476 16 × 2 = 0 + 0.072 192 952 32;
  • 9) 0.072 192 952 32 × 2 = 0 + 0.144 385 904 64;
  • 10) 0.144 385 904 64 × 2 = 0 + 0.288 771 809 28;
  • 11) 0.288 771 809 28 × 2 = 0 + 0.577 543 618 56;
  • 12) 0.577 543 618 56 × 2 = 1 + 0.155 087 237 12;
  • 13) 0.155 087 237 12 × 2 = 0 + 0.310 174 474 24;
  • 14) 0.310 174 474 24 × 2 = 0 + 0.620 348 948 48;
  • 15) 0.620 348 948 48 × 2 = 1 + 0.240 697 896 96;
  • 16) 0.240 697 896 96 × 2 = 0 + 0.481 395 793 92;
  • 17) 0.481 395 793 92 × 2 = 0 + 0.962 791 587 84;
  • 18) 0.962 791 587 84 × 2 = 1 + 0.925 583 175 68;
  • 19) 0.925 583 175 68 × 2 = 1 + 0.851 166 351 36;
  • 20) 0.851 166 351 36 × 2 = 1 + 0.702 332 702 72;
  • 21) 0.702 332 702 72 × 2 = 1 + 0.404 665 405 44;
  • 22) 0.404 665 405 44 × 2 = 0 + 0.809 330 810 88;
  • 23) 0.809 330 810 88 × 2 = 1 + 0.618 661 621 76;
  • 24) 0.618 661 621 76 × 2 = 1 + 0.237 323 243 52;
  • 25) 0.237 323 243 52 × 2 = 0 + 0.474 646 487 04;
  • 26) 0.474 646 487 04 × 2 = 0 + 0.949 292 974 08;
  • 27) 0.949 292 974 08 × 2 = 1 + 0.898 585 948 16;
  • 28) 0.898 585 948 16 × 2 = 1 + 0.797 171 896 32;
  • 29) 0.797 171 896 32 × 2 = 1 + 0.594 343 792 64;
  • 30) 0.594 343 792 64 × 2 = 1 + 0.188 687 585 28;
  • 31) 0.188 687 585 28 × 2 = 0 + 0.377 375 170 56;
  • 32) 0.377 375 170 56 × 2 = 0 + 0.754 750 341 12;
  • 33) 0.754 750 341 12 × 2 = 1 + 0.509 500 682 24;
  • 34) 0.509 500 682 24 × 2 = 1 + 0.019 001 364 48;
  • 35) 0.019 001 364 48 × 2 = 0 + 0.038 002 728 96;
  • 36) 0.038 002 728 96 × 2 = 0 + 0.076 005 457 92;
  • 37) 0.076 005 457 92 × 2 = 0 + 0.152 010 915 84;
  • 38) 0.152 010 915 84 × 2 = 0 + 0.304 021 831 68;
  • 39) 0.304 021 831 68 × 2 = 0 + 0.608 043 663 36;
  • 40) 0.608 043 663 36 × 2 = 1 + 0.216 087 326 72;
  • 41) 0.216 087 326 72 × 2 = 0 + 0.432 174 653 44;
  • 42) 0.432 174 653 44 × 2 = 0 + 0.864 349 306 88;
  • 43) 0.864 349 306 88 × 2 = 1 + 0.728 698 613 76;
  • 44) 0.728 698 613 76 × 2 = 1 + 0.457 397 227 52;
  • 45) 0.457 397 227 52 × 2 = 0 + 0.914 794 455 04;
  • 46) 0.914 794 455 04 × 2 = 1 + 0.829 588 910 08;
  • 47) 0.829 588 910 08 × 2 = 1 + 0.659 177 820 16;
  • 48) 0.659 177 820 16 × 2 = 1 + 0.318 355 640 32;
  • 49) 0.318 355 640 32 × 2 = 0 + 0.636 711 280 64;
  • 50) 0.636 711 280 64 × 2 = 1 + 0.273 422 561 28;
  • 51) 0.273 422 561 28 × 2 = 0 + 0.546 845 122 56;
  • 52) 0.546 845 122 56 × 2 = 1 + 0.093 690 245 12;
  • 53) 0.093 690 245 12 × 2 = 0 + 0.187 380 490 24;
  • 54) 0.187 380 490 24 × 2 = 0 + 0.374 760 980 48;
  • 55) 0.374 760 980 48 × 2 = 0 + 0.749 521 960 96;
  • 56) 0.749 521 960 96 × 2 = 1 + 0.499 043 921 92;
  • 57) 0.499 043 921 92 × 2 = 0 + 0.998 087 843 84;
  • 58) 0.998 087 843 84 × 2 = 1 + 0.996 175 687 68;
  • 59) 0.996 175 687 68 × 2 = 1 + 0.992 351 375 36;
  • 60) 0.992 351 375 36 × 2 = 1 + 0.984 702 750 72;
  • 61) 0.984 702 750 72 × 2 = 1 + 0.969 405 501 44;
  • 62) 0.969 405 501 44 × 2 = 1 + 0.938 811 002 88;
  • 63) 0.938 811 002 88 × 2 = 1 + 0.877 622 005 76;
  • 64) 0.877 622 005 76 × 2 = 1 + 0.755 244 011 52;

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:


0.000 282 003 72(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 1100 0001 0011 0111 0101 0001 0111 1111(2)

6. Positive number before normalization:

0.000 282 003 72(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 1100 0001 0011 0111 0101 0001 0111 1111(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:


0.000 282 003 72(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 1100 0001 0011 0111 0101 0001 0111 1111(2) =

0.0000 0000 0001 0010 0111 1011 0011 1100 1100 0001 0011 0111 0101 0001 0111 1111(2) × 20 =

1.0010 0111 1011 0011 1100 1100 0001 0011 0111 0101 0001 0111 1111(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 1100 1100 0001 0011 0111 0101 0001 0111 1111


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 1100 1100 0001 0011 0111 0101 0001 0111 1111 =

0010 0111 1011 0011 1100 1100 0001 0011 0111 0101 0001 0111 1111


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 1100 1100 0001 0011 0111 0101 0001 0111 1111


Decimal number -0.000 282 003 72 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1100 1100 0001 0011 0111 0101 0001 0111 1111

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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