-0.000 282 81 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 81(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 81(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 81| = 0.000 282 81


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 81.

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 81 × 2 = 0 + 0.000 565 62;
  • 2) 0.000 565 62 × 2 = 0 + 0.001 131 24;
  • 3) 0.001 131 24 × 2 = 0 + 0.002 262 48;
  • 4) 0.002 262 48 × 2 = 0 + 0.004 524 96;
  • 5) 0.004 524 96 × 2 = 0 + 0.009 049 92;
  • 6) 0.009 049 92 × 2 = 0 + 0.018 099 84;
  • 7) 0.018 099 84 × 2 = 0 + 0.036 199 68;
  • 8) 0.036 199 68 × 2 = 0 + 0.072 399 36;
  • 9) 0.072 399 36 × 2 = 0 + 0.144 798 72;
  • 10) 0.144 798 72 × 2 = 0 + 0.289 597 44;
  • 11) 0.289 597 44 × 2 = 0 + 0.579 194 88;
  • 12) 0.579 194 88 × 2 = 1 + 0.158 389 76;
  • 13) 0.158 389 76 × 2 = 0 + 0.316 779 52;
  • 14) 0.316 779 52 × 2 = 0 + 0.633 559 04;
  • 15) 0.633 559 04 × 2 = 1 + 0.267 118 08;
  • 16) 0.267 118 08 × 2 = 0 + 0.534 236 16;
  • 17) 0.534 236 16 × 2 = 1 + 0.068 472 32;
  • 18) 0.068 472 32 × 2 = 0 + 0.136 944 64;
  • 19) 0.136 944 64 × 2 = 0 + 0.273 889 28;
  • 20) 0.273 889 28 × 2 = 0 + 0.547 778 56;
  • 21) 0.547 778 56 × 2 = 1 + 0.095 557 12;
  • 22) 0.095 557 12 × 2 = 0 + 0.191 114 24;
  • 23) 0.191 114 24 × 2 = 0 + 0.382 228 48;
  • 24) 0.382 228 48 × 2 = 0 + 0.764 456 96;
  • 25) 0.764 456 96 × 2 = 1 + 0.528 913 92;
  • 26) 0.528 913 92 × 2 = 1 + 0.057 827 84;
  • 27) 0.057 827 84 × 2 = 0 + 0.115 655 68;
  • 28) 0.115 655 68 × 2 = 0 + 0.231 311 36;
  • 29) 0.231 311 36 × 2 = 0 + 0.462 622 72;
  • 30) 0.462 622 72 × 2 = 0 + 0.925 245 44;
  • 31) 0.925 245 44 × 2 = 1 + 0.850 490 88;
  • 32) 0.850 490 88 × 2 = 1 + 0.700 981 76;
  • 33) 0.700 981 76 × 2 = 1 + 0.401 963 52;
  • 34) 0.401 963 52 × 2 = 0 + 0.803 927 04;
  • 35) 0.803 927 04 × 2 = 1 + 0.607 854 08;
  • 36) 0.607 854 08 × 2 = 1 + 0.215 708 16;
  • 37) 0.215 708 16 × 2 = 0 + 0.431 416 32;
  • 38) 0.431 416 32 × 2 = 0 + 0.862 832 64;
  • 39) 0.862 832 64 × 2 = 1 + 0.725 665 28;
  • 40) 0.725 665 28 × 2 = 1 + 0.451 330 56;
  • 41) 0.451 330 56 × 2 = 0 + 0.902 661 12;
  • 42) 0.902 661 12 × 2 = 1 + 0.805 322 24;
  • 43) 0.805 322 24 × 2 = 1 + 0.610 644 48;
  • 44) 0.610 644 48 × 2 = 1 + 0.221 288 96;
  • 45) 0.221 288 96 × 2 = 0 + 0.442 577 92;
  • 46) 0.442 577 92 × 2 = 0 + 0.885 155 84;
  • 47) 0.885 155 84 × 2 = 1 + 0.770 311 68;
  • 48) 0.770 311 68 × 2 = 1 + 0.540 623 36;
  • 49) 0.540 623 36 × 2 = 1 + 0.081 246 72;
  • 50) 0.081 246 72 × 2 = 0 + 0.162 493 44;
  • 51) 0.162 493 44 × 2 = 0 + 0.324 986 88;
  • 52) 0.324 986 88 × 2 = 0 + 0.649 973 76;
  • 53) 0.649 973 76 × 2 = 1 + 0.299 947 52;
  • 54) 0.299 947 52 × 2 = 0 + 0.599 895 04;
  • 55) 0.599 895 04 × 2 = 1 + 0.199 790 08;
  • 56) 0.199 790 08 × 2 = 0 + 0.399 580 16;
  • 57) 0.399 580 16 × 2 = 0 + 0.799 160 32;
  • 58) 0.799 160 32 × 2 = 1 + 0.598 320 64;
  • 59) 0.598 320 64 × 2 = 1 + 0.196 641 28;
  • 60) 0.196 641 28 × 2 = 0 + 0.393 282 56;
  • 61) 0.393 282 56 × 2 = 0 + 0.786 565 12;
  • 62) 0.786 565 12 × 2 = 1 + 0.573 130 24;
  • 63) 0.573 130 24 × 2 = 1 + 0.146 260 48;
  • 64) 0.146 260 48 × 2 = 0 + 0.292 520 96;

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 81(10) =

0.0000 0000 0001 0010 1000 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110(2)

6. Positive number before normalization:

0.000 282 81(10) =

0.0000 0000 0001 0010 1000 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110(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 81(10) =

0.0000 0000 0001 0010 1000 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110(2) =

0.0000 0000 0001 0010 1000 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110(2) × 20 =

1.0010 1000 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110(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 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110


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 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110 =

0010 1000 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110


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 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110


Decimal number -0.000 282 81 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 1000 1100 0011 1011 0011 0111 0011 1000 1010 0110 0110

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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