-0.000 283 13 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 283 13(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
-0.000 283 13(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:

|-0.000 283 13| = 0.000 283 13


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: 0.000 283 13.

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) 0.000 283 13 × 2 = 0 + 0.000 566 26;
  • 2) 0.000 566 26 × 2 = 0 + 0.001 132 52;
  • 3) 0.001 132 52 × 2 = 0 + 0.002 265 04;
  • 4) 0.002 265 04 × 2 = 0 + 0.004 530 08;
  • 5) 0.004 530 08 × 2 = 0 + 0.009 060 16;
  • 6) 0.009 060 16 × 2 = 0 + 0.018 120 32;
  • 7) 0.018 120 32 × 2 = 0 + 0.036 240 64;
  • 8) 0.036 240 64 × 2 = 0 + 0.072 481 28;
  • 9) 0.072 481 28 × 2 = 0 + 0.144 962 56;
  • 10) 0.144 962 56 × 2 = 0 + 0.289 925 12;
  • 11) 0.289 925 12 × 2 = 0 + 0.579 850 24;
  • 12) 0.579 850 24 × 2 = 1 + 0.159 700 48;
  • 13) 0.159 700 48 × 2 = 0 + 0.319 400 96;
  • 14) 0.319 400 96 × 2 = 0 + 0.638 801 92;
  • 15) 0.638 801 92 × 2 = 1 + 0.277 603 84;
  • 16) 0.277 603 84 × 2 = 0 + 0.555 207 68;
  • 17) 0.555 207 68 × 2 = 1 + 0.110 415 36;
  • 18) 0.110 415 36 × 2 = 0 + 0.220 830 72;
  • 19) 0.220 830 72 × 2 = 0 + 0.441 661 44;
  • 20) 0.441 661 44 × 2 = 0 + 0.883 322 88;
  • 21) 0.883 322 88 × 2 = 1 + 0.766 645 76;
  • 22) 0.766 645 76 × 2 = 1 + 0.533 291 52;
  • 23) 0.533 291 52 × 2 = 1 + 0.066 583 04;
  • 24) 0.066 583 04 × 2 = 0 + 0.133 166 08;
  • 25) 0.133 166 08 × 2 = 0 + 0.266 332 16;
  • 26) 0.266 332 16 × 2 = 0 + 0.532 664 32;
  • 27) 0.532 664 32 × 2 = 1 + 0.065 328 64;
  • 28) 0.065 328 64 × 2 = 0 + 0.130 657 28;
  • 29) 0.130 657 28 × 2 = 0 + 0.261 314 56;
  • 30) 0.261 314 56 × 2 = 0 + 0.522 629 12;
  • 31) 0.522 629 12 × 2 = 1 + 0.045 258 24;
  • 32) 0.045 258 24 × 2 = 0 + 0.090 516 48;
  • 33) 0.090 516 48 × 2 = 0 + 0.181 032 96;
  • 34) 0.181 032 96 × 2 = 0 + 0.362 065 92;
  • 35) 0.362 065 92 × 2 = 0 + 0.724 131 84;
  • 36) 0.724 131 84 × 2 = 1 + 0.448 263 68;
  • 37) 0.448 263 68 × 2 = 0 + 0.896 527 36;
  • 38) 0.896 527 36 × 2 = 1 + 0.793 054 72;
  • 39) 0.793 054 72 × 2 = 1 + 0.586 109 44;
  • 40) 0.586 109 44 × 2 = 1 + 0.172 218 88;
  • 41) 0.172 218 88 × 2 = 0 + 0.344 437 76;
  • 42) 0.344 437 76 × 2 = 0 + 0.688 875 52;
  • 43) 0.688 875 52 × 2 = 1 + 0.377 751 04;
  • 44) 0.377 751 04 × 2 = 0 + 0.755 502 08;
  • 45) 0.755 502 08 × 2 = 1 + 0.511 004 16;
  • 46) 0.511 004 16 × 2 = 1 + 0.022 008 32;
  • 47) 0.022 008 32 × 2 = 0 + 0.044 016 64;
  • 48) 0.044 016 64 × 2 = 0 + 0.088 033 28;
  • 49) 0.088 033 28 × 2 = 0 + 0.176 066 56;
  • 50) 0.176 066 56 × 2 = 0 + 0.352 133 12;
  • 51) 0.352 133 12 × 2 = 0 + 0.704 266 24;
  • 52) 0.704 266 24 × 2 = 1 + 0.408 532 48;
  • 53) 0.408 532 48 × 2 = 0 + 0.817 064 96;
  • 54) 0.817 064 96 × 2 = 1 + 0.634 129 92;
  • 55) 0.634 129 92 × 2 = 1 + 0.268 259 84;
  • 56) 0.268 259 84 × 2 = 0 + 0.536 519 68;
  • 57) 0.536 519 68 × 2 = 1 + 0.073 039 36;
  • 58) 0.073 039 36 × 2 = 0 + 0.146 078 72;
  • 59) 0.146 078 72 × 2 = 0 + 0.292 157 44;
  • 60) 0.292 157 44 × 2 = 0 + 0.584 314 88;
  • 61) 0.584 314 88 × 2 = 1 + 0.168 629 76;
  • 62) 0.168 629 76 × 2 = 0 + 0.337 259 52;
  • 63) 0.337 259 52 × 2 = 0 + 0.674 519 04;
  • 64) 0.674 519 04 × 2 = 1 + 0.349 038 08;

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:


0.000 283 13(10) =

0.0000 0000 0001 0010 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001(2)

6. Positive number before normalization:

0.000 283 13(10) =

0.0000 0000 0001 0010 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001(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:


0.000 283 13(10) =

0.0000 0000 0001 0010 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001(2) =

0.0000 0000 0001 0010 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001(2) × 20 =

1.0010 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001(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 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001


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 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001 =

0010 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001


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 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001


Decimal number -0.000 283 13 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 1110 0010 0010 0001 0111 0010 1100 0001 0110 1000 1001

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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