-0.000 282 008 96 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 008 96(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
-0.000 282 008 96(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:

|-0.000 282 008 96| = 0.000 282 008 96


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: 0.000 282 008 96.

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) 0.000 282 008 96 × 2 = 0 + 0.000 564 017 92;
  • 2) 0.000 564 017 92 × 2 = 0 + 0.001 128 035 84;
  • 3) 0.001 128 035 84 × 2 = 0 + 0.002 256 071 68;
  • 4) 0.002 256 071 68 × 2 = 0 + 0.004 512 143 36;
  • 5) 0.004 512 143 36 × 2 = 0 + 0.009 024 286 72;
  • 6) 0.009 024 286 72 × 2 = 0 + 0.018 048 573 44;
  • 7) 0.018 048 573 44 × 2 = 0 + 0.036 097 146 88;
  • 8) 0.036 097 146 88 × 2 = 0 + 0.072 194 293 76;
  • 9) 0.072 194 293 76 × 2 = 0 + 0.144 388 587 52;
  • 10) 0.144 388 587 52 × 2 = 0 + 0.288 777 175 04;
  • 11) 0.288 777 175 04 × 2 = 0 + 0.577 554 350 08;
  • 12) 0.577 554 350 08 × 2 = 1 + 0.155 108 700 16;
  • 13) 0.155 108 700 16 × 2 = 0 + 0.310 217 400 32;
  • 14) 0.310 217 400 32 × 2 = 0 + 0.620 434 800 64;
  • 15) 0.620 434 800 64 × 2 = 1 + 0.240 869 601 28;
  • 16) 0.240 869 601 28 × 2 = 0 + 0.481 739 202 56;
  • 17) 0.481 739 202 56 × 2 = 0 + 0.963 478 405 12;
  • 18) 0.963 478 405 12 × 2 = 1 + 0.926 956 810 24;
  • 19) 0.926 956 810 24 × 2 = 1 + 0.853 913 620 48;
  • 20) 0.853 913 620 48 × 2 = 1 + 0.707 827 240 96;
  • 21) 0.707 827 240 96 × 2 = 1 + 0.415 654 481 92;
  • 22) 0.415 654 481 92 × 2 = 0 + 0.831 308 963 84;
  • 23) 0.831 308 963 84 × 2 = 1 + 0.662 617 927 68;
  • 24) 0.662 617 927 68 × 2 = 1 + 0.325 235 855 36;
  • 25) 0.325 235 855 36 × 2 = 0 + 0.650 471 710 72;
  • 26) 0.650 471 710 72 × 2 = 1 + 0.300 943 421 44;
  • 27) 0.300 943 421 44 × 2 = 0 + 0.601 886 842 88;
  • 28) 0.601 886 842 88 × 2 = 1 + 0.203 773 685 76;
  • 29) 0.203 773 685 76 × 2 = 0 + 0.407 547 371 52;
  • 30) 0.407 547 371 52 × 2 = 0 + 0.815 094 743 04;
  • 31) 0.815 094 743 04 × 2 = 1 + 0.630 189 486 08;
  • 32) 0.630 189 486 08 × 2 = 1 + 0.260 378 972 16;
  • 33) 0.260 378 972 16 × 2 = 0 + 0.520 757 944 32;
  • 34) 0.520 757 944 32 × 2 = 1 + 0.041 515 888 64;
  • 35) 0.041 515 888 64 × 2 = 0 + 0.083 031 777 28;
  • 36) 0.083 031 777 28 × 2 = 0 + 0.166 063 554 56;
  • 37) 0.166 063 554 56 × 2 = 0 + 0.332 127 109 12;
  • 38) 0.332 127 109 12 × 2 = 0 + 0.664 254 218 24;
  • 39) 0.664 254 218 24 × 2 = 1 + 0.328 508 436 48;
  • 40) 0.328 508 436 48 × 2 = 0 + 0.657 016 872 96;
  • 41) 0.657 016 872 96 × 2 = 1 + 0.314 033 745 92;
  • 42) 0.314 033 745 92 × 2 = 0 + 0.628 067 491 84;
  • 43) 0.628 067 491 84 × 2 = 1 + 0.256 134 983 68;
  • 44) 0.256 134 983 68 × 2 = 0 + 0.512 269 967 36;
  • 45) 0.512 269 967 36 × 2 = 1 + 0.024 539 934 72;
  • 46) 0.024 539 934 72 × 2 = 0 + 0.049 079 869 44;
  • 47) 0.049 079 869 44 × 2 = 0 + 0.098 159 738 88;
  • 48) 0.098 159 738 88 × 2 = 0 + 0.196 319 477 76;
  • 49) 0.196 319 477 76 × 2 = 0 + 0.392 638 955 52;
  • 50) 0.392 638 955 52 × 2 = 0 + 0.785 277 911 04;
  • 51) 0.785 277 911 04 × 2 = 1 + 0.570 555 822 08;
  • 52) 0.570 555 822 08 × 2 = 1 + 0.141 111 644 16;
  • 53) 0.141 111 644 16 × 2 = 0 + 0.282 223 288 32;
  • 54) 0.282 223 288 32 × 2 = 0 + 0.564 446 576 64;
  • 55) 0.564 446 576 64 × 2 = 1 + 0.128 893 153 28;
  • 56) 0.128 893 153 28 × 2 = 0 + 0.257 786 306 56;
  • 57) 0.257 786 306 56 × 2 = 0 + 0.515 572 613 12;
  • 58) 0.515 572 613 12 × 2 = 1 + 0.031 145 226 24;
  • 59) 0.031 145 226 24 × 2 = 0 + 0.062 290 452 48;
  • 60) 0.062 290 452 48 × 2 = 0 + 0.124 580 904 96;
  • 61) 0.124 580 904 96 × 2 = 0 + 0.249 161 809 92;
  • 62) 0.249 161 809 92 × 2 = 0 + 0.498 323 619 84;
  • 63) 0.498 323 619 84 × 2 = 0 + 0.996 647 239 68;
  • 64) 0.996 647 239 68 × 2 = 1 + 0.993 294 479 36;

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:


0.000 282 008 96(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001(2)

6. Positive number before normalization:

0.000 282 008 96(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001(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:


0.000 282 008 96(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001(2) =

0.0000 0000 0001 0010 0111 1011 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001(2) × 20 =

1.0010 0111 1011 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001(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 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001


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 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001 =

0010 0111 1011 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001


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 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001


Decimal number -0.000 282 008 96 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0011 0100 0010 1010 1000 0011 0010 0100 0001

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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