-0.000 280 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 280 12(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
-0.000 280 12(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:

|-0.000 280 12| = 0.000 280 12


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: 0.000 280 12.

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) 0.000 280 12 × 2 = 0 + 0.000 560 24;
  • 2) 0.000 560 24 × 2 = 0 + 0.001 120 48;
  • 3) 0.001 120 48 × 2 = 0 + 0.002 240 96;
  • 4) 0.002 240 96 × 2 = 0 + 0.004 481 92;
  • 5) 0.004 481 92 × 2 = 0 + 0.008 963 84;
  • 6) 0.008 963 84 × 2 = 0 + 0.017 927 68;
  • 7) 0.017 927 68 × 2 = 0 + 0.035 855 36;
  • 8) 0.035 855 36 × 2 = 0 + 0.071 710 72;
  • 9) 0.071 710 72 × 2 = 0 + 0.143 421 44;
  • 10) 0.143 421 44 × 2 = 0 + 0.286 842 88;
  • 11) 0.286 842 88 × 2 = 0 + 0.573 685 76;
  • 12) 0.573 685 76 × 2 = 1 + 0.147 371 52;
  • 13) 0.147 371 52 × 2 = 0 + 0.294 743 04;
  • 14) 0.294 743 04 × 2 = 0 + 0.589 486 08;
  • 15) 0.589 486 08 × 2 = 1 + 0.178 972 16;
  • 16) 0.178 972 16 × 2 = 0 + 0.357 944 32;
  • 17) 0.357 944 32 × 2 = 0 + 0.715 888 64;
  • 18) 0.715 888 64 × 2 = 1 + 0.431 777 28;
  • 19) 0.431 777 28 × 2 = 0 + 0.863 554 56;
  • 20) 0.863 554 56 × 2 = 1 + 0.727 109 12;
  • 21) 0.727 109 12 × 2 = 1 + 0.454 218 24;
  • 22) 0.454 218 24 × 2 = 0 + 0.908 436 48;
  • 23) 0.908 436 48 × 2 = 1 + 0.816 872 96;
  • 24) 0.816 872 96 × 2 = 1 + 0.633 745 92;
  • 25) 0.633 745 92 × 2 = 1 + 0.267 491 84;
  • 26) 0.267 491 84 × 2 = 0 + 0.534 983 68;
  • 27) 0.534 983 68 × 2 = 1 + 0.069 967 36;
  • 28) 0.069 967 36 × 2 = 0 + 0.139 934 72;
  • 29) 0.139 934 72 × 2 = 0 + 0.279 869 44;
  • 30) 0.279 869 44 × 2 = 0 + 0.559 738 88;
  • 31) 0.559 738 88 × 2 = 1 + 0.119 477 76;
  • 32) 0.119 477 76 × 2 = 0 + 0.238 955 52;
  • 33) 0.238 955 52 × 2 = 0 + 0.477 911 04;
  • 34) 0.477 911 04 × 2 = 0 + 0.955 822 08;
  • 35) 0.955 822 08 × 2 = 1 + 0.911 644 16;
  • 36) 0.911 644 16 × 2 = 1 + 0.823 288 32;
  • 37) 0.823 288 32 × 2 = 1 + 0.646 576 64;
  • 38) 0.646 576 64 × 2 = 1 + 0.293 153 28;
  • 39) 0.293 153 28 × 2 = 0 + 0.586 306 56;
  • 40) 0.586 306 56 × 2 = 1 + 0.172 613 12;
  • 41) 0.172 613 12 × 2 = 0 + 0.345 226 24;
  • 42) 0.345 226 24 × 2 = 0 + 0.690 452 48;
  • 43) 0.690 452 48 × 2 = 1 + 0.380 904 96;
  • 44) 0.380 904 96 × 2 = 0 + 0.761 809 92;
  • 45) 0.761 809 92 × 2 = 1 + 0.523 619 84;
  • 46) 0.523 619 84 × 2 = 1 + 0.047 239 68;
  • 47) 0.047 239 68 × 2 = 0 + 0.094 479 36;
  • 48) 0.094 479 36 × 2 = 0 + 0.188 958 72;
  • 49) 0.188 958 72 × 2 = 0 + 0.377 917 44;
  • 50) 0.377 917 44 × 2 = 0 + 0.755 834 88;
  • 51) 0.755 834 88 × 2 = 1 + 0.511 669 76;
  • 52) 0.511 669 76 × 2 = 1 + 0.023 339 52;
  • 53) 0.023 339 52 × 2 = 0 + 0.046 679 04;
  • 54) 0.046 679 04 × 2 = 0 + 0.093 358 08;
  • 55) 0.093 358 08 × 2 = 0 + 0.186 716 16;
  • 56) 0.186 716 16 × 2 = 0 + 0.373 432 32;
  • 57) 0.373 432 32 × 2 = 0 + 0.746 864 64;
  • 58) 0.746 864 64 × 2 = 1 + 0.493 729 28;
  • 59) 0.493 729 28 × 2 = 0 + 0.987 458 56;
  • 60) 0.987 458 56 × 2 = 1 + 0.974 917 12;
  • 61) 0.974 917 12 × 2 = 1 + 0.949 834 24;
  • 62) 0.949 834 24 × 2 = 1 + 0.899 668 48;
  • 63) 0.899 668 48 × 2 = 1 + 0.799 336 96;
  • 64) 0.799 336 96 × 2 = 1 + 0.598 673 92;

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:


0.000 280 12(10) =

0.0000 0000 0001 0010 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 1111(2)

6. Positive number before normalization:

0.000 280 12(10) =

0.0000 0000 0001 0010 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 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:


0.000 280 12(10) =

0.0000 0000 0001 0010 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 1111(2) =

0.0000 0000 0001 0010 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 1111(2) × 20 =

1.0010 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 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 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 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 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 1111 =

0010 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 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 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 1111


Decimal number -0.000 280 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0101 1011 1010 0010 0011 1101 0010 1100 0011 0000 0101 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