-0.000 282 002 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 002 93(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
-0.000 282 002 93(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:

|-0.000 282 002 93| = 0.000 282 002 93


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: 0.000 282 002 93.

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) 0.000 282 002 93 × 2 = 0 + 0.000 564 005 86;
  • 2) 0.000 564 005 86 × 2 = 0 + 0.001 128 011 72;
  • 3) 0.001 128 011 72 × 2 = 0 + 0.002 256 023 44;
  • 4) 0.002 256 023 44 × 2 = 0 + 0.004 512 046 88;
  • 5) 0.004 512 046 88 × 2 = 0 + 0.009 024 093 76;
  • 6) 0.009 024 093 76 × 2 = 0 + 0.018 048 187 52;
  • 7) 0.018 048 187 52 × 2 = 0 + 0.036 096 375 04;
  • 8) 0.036 096 375 04 × 2 = 0 + 0.072 192 750 08;
  • 9) 0.072 192 750 08 × 2 = 0 + 0.144 385 500 16;
  • 10) 0.144 385 500 16 × 2 = 0 + 0.288 771 000 32;
  • 11) 0.288 771 000 32 × 2 = 0 + 0.577 542 000 64;
  • 12) 0.577 542 000 64 × 2 = 1 + 0.155 084 001 28;
  • 13) 0.155 084 001 28 × 2 = 0 + 0.310 168 002 56;
  • 14) 0.310 168 002 56 × 2 = 0 + 0.620 336 005 12;
  • 15) 0.620 336 005 12 × 2 = 1 + 0.240 672 010 24;
  • 16) 0.240 672 010 24 × 2 = 0 + 0.481 344 020 48;
  • 17) 0.481 344 020 48 × 2 = 0 + 0.962 688 040 96;
  • 18) 0.962 688 040 96 × 2 = 1 + 0.925 376 081 92;
  • 19) 0.925 376 081 92 × 2 = 1 + 0.850 752 163 84;
  • 20) 0.850 752 163 84 × 2 = 1 + 0.701 504 327 68;
  • 21) 0.701 504 327 68 × 2 = 1 + 0.403 008 655 36;
  • 22) 0.403 008 655 36 × 2 = 0 + 0.806 017 310 72;
  • 23) 0.806 017 310 72 × 2 = 1 + 0.612 034 621 44;
  • 24) 0.612 034 621 44 × 2 = 1 + 0.224 069 242 88;
  • 25) 0.224 069 242 88 × 2 = 0 + 0.448 138 485 76;
  • 26) 0.448 138 485 76 × 2 = 0 + 0.896 276 971 52;
  • 27) 0.896 276 971 52 × 2 = 1 + 0.792 553 943 04;
  • 28) 0.792 553 943 04 × 2 = 1 + 0.585 107 886 08;
  • 29) 0.585 107 886 08 × 2 = 1 + 0.170 215 772 16;
  • 30) 0.170 215 772 16 × 2 = 0 + 0.340 431 544 32;
  • 31) 0.340 431 544 32 × 2 = 0 + 0.680 863 088 64;
  • 32) 0.680 863 088 64 × 2 = 1 + 0.361 726 177 28;
  • 33) 0.361 726 177 28 × 2 = 0 + 0.723 452 354 56;
  • 34) 0.723 452 354 56 × 2 = 1 + 0.446 904 709 12;
  • 35) 0.446 904 709 12 × 2 = 0 + 0.893 809 418 24;
  • 36) 0.893 809 418 24 × 2 = 1 + 0.787 618 836 48;
  • 37) 0.787 618 836 48 × 2 = 1 + 0.575 237 672 96;
  • 38) 0.575 237 672 96 × 2 = 1 + 0.150 475 345 92;
  • 39) 0.150 475 345 92 × 2 = 0 + 0.300 950 691 84;
  • 40) 0.300 950 691 84 × 2 = 0 + 0.601 901 383 68;
  • 41) 0.601 901 383 68 × 2 = 1 + 0.203 802 767 36;
  • 42) 0.203 802 767 36 × 2 = 0 + 0.407 605 534 72;
  • 43) 0.407 605 534 72 × 2 = 0 + 0.815 211 069 44;
  • 44) 0.815 211 069 44 × 2 = 1 + 0.630 422 138 88;
  • 45) 0.630 422 138 88 × 2 = 1 + 0.260 844 277 76;
  • 46) 0.260 844 277 76 × 2 = 0 + 0.521 688 555 52;
  • 47) 0.521 688 555 52 × 2 = 1 + 0.043 377 111 04;
  • 48) 0.043 377 111 04 × 2 = 0 + 0.086 754 222 08;
  • 49) 0.086 754 222 08 × 2 = 0 + 0.173 508 444 16;
  • 50) 0.173 508 444 16 × 2 = 0 + 0.347 016 888 32;
  • 51) 0.347 016 888 32 × 2 = 0 + 0.694 033 776 64;
  • 52) 0.694 033 776 64 × 2 = 1 + 0.388 067 553 28;
  • 53) 0.388 067 553 28 × 2 = 0 + 0.776 135 106 56;
  • 54) 0.776 135 106 56 × 2 = 1 + 0.552 270 213 12;
  • 55) 0.552 270 213 12 × 2 = 1 + 0.104 540 426 24;
  • 56) 0.104 540 426 24 × 2 = 0 + 0.209 080 852 48;
  • 57) 0.209 080 852 48 × 2 = 0 + 0.418 161 704 96;
  • 58) 0.418 161 704 96 × 2 = 0 + 0.836 323 409 92;
  • 59) 0.836 323 409 92 × 2 = 1 + 0.672 646 819 84;
  • 60) 0.672 646 819 84 × 2 = 1 + 0.345 293 639 68;
  • 61) 0.345 293 639 68 × 2 = 0 + 0.690 587 279 36;
  • 62) 0.690 587 279 36 × 2 = 1 + 0.381 174 558 72;
  • 63) 0.381 174 558 72 × 2 = 0 + 0.762 349 117 44;
  • 64) 0.762 349 117 44 × 2 = 1 + 0.524 698 234 88;

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:


0.000 282 002 93(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 1100 1001 1010 0001 0110 0011 0101(2)

6. Positive number before normalization:

0.000 282 002 93(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 1100 1001 1010 0001 0110 0011 0101(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:


0.000 282 002 93(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 1100 1001 1010 0001 0110 0011 0101(2) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 1100 1001 1010 0001 0110 0011 0101(2) × 20 =

1.0010 0111 1011 0011 1001 0101 1100 1001 1010 0001 0110 0011 0101(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 1001 0101 1100 1001 1010 0001 0110 0011 0101


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 1001 0101 1100 1001 1010 0001 0110 0011 0101 =

0010 0111 1011 0011 1001 0101 1100 1001 1010 0001 0110 0011 0101


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 1001 0101 1100 1001 1010 0001 0110 0011 0101


Decimal number -0.000 282 002 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1001 0101 1100 1001 1010 0001 0110 0011 0101

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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