-0.000 282 88 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 88(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 88(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 88| = 0.000 282 88


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 88.

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 88 × 2 = 0 + 0.000 565 76;
  • 2) 0.000 565 76 × 2 = 0 + 0.001 131 52;
  • 3) 0.001 131 52 × 2 = 0 + 0.002 263 04;
  • 4) 0.002 263 04 × 2 = 0 + 0.004 526 08;
  • 5) 0.004 526 08 × 2 = 0 + 0.009 052 16;
  • 6) 0.009 052 16 × 2 = 0 + 0.018 104 32;
  • 7) 0.018 104 32 × 2 = 0 + 0.036 208 64;
  • 8) 0.036 208 64 × 2 = 0 + 0.072 417 28;
  • 9) 0.072 417 28 × 2 = 0 + 0.144 834 56;
  • 10) 0.144 834 56 × 2 = 0 + 0.289 669 12;
  • 11) 0.289 669 12 × 2 = 0 + 0.579 338 24;
  • 12) 0.579 338 24 × 2 = 1 + 0.158 676 48;
  • 13) 0.158 676 48 × 2 = 0 + 0.317 352 96;
  • 14) 0.317 352 96 × 2 = 0 + 0.634 705 92;
  • 15) 0.634 705 92 × 2 = 1 + 0.269 411 84;
  • 16) 0.269 411 84 × 2 = 0 + 0.538 823 68;
  • 17) 0.538 823 68 × 2 = 1 + 0.077 647 36;
  • 18) 0.077 647 36 × 2 = 0 + 0.155 294 72;
  • 19) 0.155 294 72 × 2 = 0 + 0.310 589 44;
  • 20) 0.310 589 44 × 2 = 0 + 0.621 178 88;
  • 21) 0.621 178 88 × 2 = 1 + 0.242 357 76;
  • 22) 0.242 357 76 × 2 = 0 + 0.484 715 52;
  • 23) 0.484 715 52 × 2 = 0 + 0.969 431 04;
  • 24) 0.969 431 04 × 2 = 1 + 0.938 862 08;
  • 25) 0.938 862 08 × 2 = 1 + 0.877 724 16;
  • 26) 0.877 724 16 × 2 = 1 + 0.755 448 32;
  • 27) 0.755 448 32 × 2 = 1 + 0.510 896 64;
  • 28) 0.510 896 64 × 2 = 1 + 0.021 793 28;
  • 29) 0.021 793 28 × 2 = 0 + 0.043 586 56;
  • 30) 0.043 586 56 × 2 = 0 + 0.087 173 12;
  • 31) 0.087 173 12 × 2 = 0 + 0.174 346 24;
  • 32) 0.174 346 24 × 2 = 0 + 0.348 692 48;
  • 33) 0.348 692 48 × 2 = 0 + 0.697 384 96;
  • 34) 0.697 384 96 × 2 = 1 + 0.394 769 92;
  • 35) 0.394 769 92 × 2 = 0 + 0.789 539 84;
  • 36) 0.789 539 84 × 2 = 1 + 0.579 079 68;
  • 37) 0.579 079 68 × 2 = 1 + 0.158 159 36;
  • 38) 0.158 159 36 × 2 = 0 + 0.316 318 72;
  • 39) 0.316 318 72 × 2 = 0 + 0.632 637 44;
  • 40) 0.632 637 44 × 2 = 1 + 0.265 274 88;
  • 41) 0.265 274 88 × 2 = 0 + 0.530 549 76;
  • 42) 0.530 549 76 × 2 = 1 + 0.061 099 52;
  • 43) 0.061 099 52 × 2 = 0 + 0.122 199 04;
  • 44) 0.122 199 04 × 2 = 0 + 0.244 398 08;
  • 45) 0.244 398 08 × 2 = 0 + 0.488 796 16;
  • 46) 0.488 796 16 × 2 = 0 + 0.977 592 32;
  • 47) 0.977 592 32 × 2 = 1 + 0.955 184 64;
  • 48) 0.955 184 64 × 2 = 1 + 0.910 369 28;
  • 49) 0.910 369 28 × 2 = 1 + 0.820 738 56;
  • 50) 0.820 738 56 × 2 = 1 + 0.641 477 12;
  • 51) 0.641 477 12 × 2 = 1 + 0.282 954 24;
  • 52) 0.282 954 24 × 2 = 0 + 0.565 908 48;
  • 53) 0.565 908 48 × 2 = 1 + 0.131 816 96;
  • 54) 0.131 816 96 × 2 = 0 + 0.263 633 92;
  • 55) 0.263 633 92 × 2 = 0 + 0.527 267 84;
  • 56) 0.527 267 84 × 2 = 1 + 0.054 535 68;
  • 57) 0.054 535 68 × 2 = 0 + 0.109 071 36;
  • 58) 0.109 071 36 × 2 = 0 + 0.218 142 72;
  • 59) 0.218 142 72 × 2 = 0 + 0.436 285 44;
  • 60) 0.436 285 44 × 2 = 0 + 0.872 570 88;
  • 61) 0.872 570 88 × 2 = 1 + 0.745 141 76;
  • 62) 0.745 141 76 × 2 = 1 + 0.490 283 52;
  • 63) 0.490 283 52 × 2 = 0 + 0.980 567 04;
  • 64) 0.980 567 04 × 2 = 1 + 0.961 134 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:


0.000 282 88(10) =

0.0000 0000 0001 0010 1000 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101(2)

6. Positive number before normalization:

0.000 282 88(10) =

0.0000 0000 0001 0010 1000 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101(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 88(10) =

0.0000 0000 0001 0010 1000 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101(2) =

0.0000 0000 0001 0010 1000 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101(2) × 20 =

1.0010 1000 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101(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 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101


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 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101 =

0010 1000 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101


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 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101


Decimal number -0.000 282 88 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 1001 1111 0000 0101 1001 0100 0011 1110 1001 0000 1101

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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