-0.000 282 013 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 013 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
-0.000 282 013 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:

|-0.000 282 013 12| = 0.000 282 013 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: 0.000 282 013 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) 0.000 282 013 12 × 2 = 0 + 0.000 564 026 24;
  • 2) 0.000 564 026 24 × 2 = 0 + 0.001 128 052 48;
  • 3) 0.001 128 052 48 × 2 = 0 + 0.002 256 104 96;
  • 4) 0.002 256 104 96 × 2 = 0 + 0.004 512 209 92;
  • 5) 0.004 512 209 92 × 2 = 0 + 0.009 024 419 84;
  • 6) 0.009 024 419 84 × 2 = 0 + 0.018 048 839 68;
  • 7) 0.018 048 839 68 × 2 = 0 + 0.036 097 679 36;
  • 8) 0.036 097 679 36 × 2 = 0 + 0.072 195 358 72;
  • 9) 0.072 195 358 72 × 2 = 0 + 0.144 390 717 44;
  • 10) 0.144 390 717 44 × 2 = 0 + 0.288 781 434 88;
  • 11) 0.288 781 434 88 × 2 = 0 + 0.577 562 869 76;
  • 12) 0.577 562 869 76 × 2 = 1 + 0.155 125 739 52;
  • 13) 0.155 125 739 52 × 2 = 0 + 0.310 251 479 04;
  • 14) 0.310 251 479 04 × 2 = 0 + 0.620 502 958 08;
  • 15) 0.620 502 958 08 × 2 = 1 + 0.241 005 916 16;
  • 16) 0.241 005 916 16 × 2 = 0 + 0.482 011 832 32;
  • 17) 0.482 011 832 32 × 2 = 0 + 0.964 023 664 64;
  • 18) 0.964 023 664 64 × 2 = 1 + 0.928 047 329 28;
  • 19) 0.928 047 329 28 × 2 = 1 + 0.856 094 658 56;
  • 20) 0.856 094 658 56 × 2 = 1 + 0.712 189 317 12;
  • 21) 0.712 189 317 12 × 2 = 1 + 0.424 378 634 24;
  • 22) 0.424 378 634 24 × 2 = 0 + 0.848 757 268 48;
  • 23) 0.848 757 268 48 × 2 = 1 + 0.697 514 536 96;
  • 24) 0.697 514 536 96 × 2 = 1 + 0.395 029 073 92;
  • 25) 0.395 029 073 92 × 2 = 0 + 0.790 058 147 84;
  • 26) 0.790 058 147 84 × 2 = 1 + 0.580 116 295 68;
  • 27) 0.580 116 295 68 × 2 = 1 + 0.160 232 591 36;
  • 28) 0.160 232 591 36 × 2 = 0 + 0.320 465 182 72;
  • 29) 0.320 465 182 72 × 2 = 0 + 0.640 930 365 44;
  • 30) 0.640 930 365 44 × 2 = 1 + 0.281 860 730 88;
  • 31) 0.281 860 730 88 × 2 = 0 + 0.563 721 461 76;
  • 32) 0.563 721 461 76 × 2 = 1 + 0.127 442 923 52;
  • 33) 0.127 442 923 52 × 2 = 0 + 0.254 885 847 04;
  • 34) 0.254 885 847 04 × 2 = 0 + 0.509 771 694 08;
  • 35) 0.509 771 694 08 × 2 = 1 + 0.019 543 388 16;
  • 36) 0.019 543 388 16 × 2 = 0 + 0.039 086 776 32;
  • 37) 0.039 086 776 32 × 2 = 0 + 0.078 173 552 64;
  • 38) 0.078 173 552 64 × 2 = 0 + 0.156 347 105 28;
  • 39) 0.156 347 105 28 × 2 = 0 + 0.312 694 210 56;
  • 40) 0.312 694 210 56 × 2 = 0 + 0.625 388 421 12;
  • 41) 0.625 388 421 12 × 2 = 1 + 0.250 776 842 24;
  • 42) 0.250 776 842 24 × 2 = 0 + 0.501 553 684 48;
  • 43) 0.501 553 684 48 × 2 = 1 + 0.003 107 368 96;
  • 44) 0.003 107 368 96 × 2 = 0 + 0.006 214 737 92;
  • 45) 0.006 214 737 92 × 2 = 0 + 0.012 429 475 84;
  • 46) 0.012 429 475 84 × 2 = 0 + 0.024 858 951 68;
  • 47) 0.024 858 951 68 × 2 = 0 + 0.049 717 903 36;
  • 48) 0.049 717 903 36 × 2 = 0 + 0.099 435 806 72;
  • 49) 0.099 435 806 72 × 2 = 0 + 0.198 871 613 44;
  • 50) 0.198 871 613 44 × 2 = 0 + 0.397 743 226 88;
  • 51) 0.397 743 226 88 × 2 = 0 + 0.795 486 453 76;
  • 52) 0.795 486 453 76 × 2 = 1 + 0.590 972 907 52;
  • 53) 0.590 972 907 52 × 2 = 1 + 0.181 945 815 04;
  • 54) 0.181 945 815 04 × 2 = 0 + 0.363 891 630 08;
  • 55) 0.363 891 630 08 × 2 = 0 + 0.727 783 260 16;
  • 56) 0.727 783 260 16 × 2 = 1 + 0.455 566 520 32;
  • 57) 0.455 566 520 32 × 2 = 0 + 0.911 133 040 64;
  • 58) 0.911 133 040 64 × 2 = 1 + 0.822 266 081 28;
  • 59) 0.822 266 081 28 × 2 = 1 + 0.644 532 162 56;
  • 60) 0.644 532 162 56 × 2 = 1 + 0.289 064 325 12;
  • 61) 0.289 064 325 12 × 2 = 0 + 0.578 128 650 24;
  • 62) 0.578 128 650 24 × 2 = 1 + 0.156 257 300 48;
  • 63) 0.156 257 300 48 × 2 = 0 + 0.312 514 600 96;
  • 64) 0.312 514 600 96 × 2 = 0 + 0.625 029 201 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:


0.000 282 013 12(10) =

0.0000 0000 0001 0010 0111 1011 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100(2)

6. Positive number before normalization:

0.000 282 013 12(10) =

0.0000 0000 0001 0010 0111 1011 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100(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:


0.000 282 013 12(10) =

0.0000 0000 0001 0010 0111 1011 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100(2) =

0.0000 0000 0001 0010 0111 1011 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100(2) × 20 =

1.0010 0111 1011 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100(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 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100


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 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100 =

0010 0111 1011 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100


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 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100


Decimal number -0.000 282 013 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0110 0101 0010 0000 1010 0000 0001 1001 0111 0100

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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