-0.000 282 004 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 11(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
-0.000 282 004 11(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:

|-0.000 282 004 11| = 0.000 282 004 11


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: 0.000 282 004 11.

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) 0.000 282 004 11 × 2 = 0 + 0.000 564 008 22;
  • 2) 0.000 564 008 22 × 2 = 0 + 0.001 128 016 44;
  • 3) 0.001 128 016 44 × 2 = 0 + 0.002 256 032 88;
  • 4) 0.002 256 032 88 × 2 = 0 + 0.004 512 065 76;
  • 5) 0.004 512 065 76 × 2 = 0 + 0.009 024 131 52;
  • 6) 0.009 024 131 52 × 2 = 0 + 0.018 048 263 04;
  • 7) 0.018 048 263 04 × 2 = 0 + 0.036 096 526 08;
  • 8) 0.036 096 526 08 × 2 = 0 + 0.072 193 052 16;
  • 9) 0.072 193 052 16 × 2 = 0 + 0.144 386 104 32;
  • 10) 0.144 386 104 32 × 2 = 0 + 0.288 772 208 64;
  • 11) 0.288 772 208 64 × 2 = 0 + 0.577 544 417 28;
  • 12) 0.577 544 417 28 × 2 = 1 + 0.155 088 834 56;
  • 13) 0.155 088 834 56 × 2 = 0 + 0.310 177 669 12;
  • 14) 0.310 177 669 12 × 2 = 0 + 0.620 355 338 24;
  • 15) 0.620 355 338 24 × 2 = 1 + 0.240 710 676 48;
  • 16) 0.240 710 676 48 × 2 = 0 + 0.481 421 352 96;
  • 17) 0.481 421 352 96 × 2 = 0 + 0.962 842 705 92;
  • 18) 0.962 842 705 92 × 2 = 1 + 0.925 685 411 84;
  • 19) 0.925 685 411 84 × 2 = 1 + 0.851 370 823 68;
  • 20) 0.851 370 823 68 × 2 = 1 + 0.702 741 647 36;
  • 21) 0.702 741 647 36 × 2 = 1 + 0.405 483 294 72;
  • 22) 0.405 483 294 72 × 2 = 0 + 0.810 966 589 44;
  • 23) 0.810 966 589 44 × 2 = 1 + 0.621 933 178 88;
  • 24) 0.621 933 178 88 × 2 = 1 + 0.243 866 357 76;
  • 25) 0.243 866 357 76 × 2 = 0 + 0.487 732 715 52;
  • 26) 0.487 732 715 52 × 2 = 0 + 0.975 465 431 04;
  • 27) 0.975 465 431 04 × 2 = 1 + 0.950 930 862 08;
  • 28) 0.950 930 862 08 × 2 = 1 + 0.901 861 724 16;
  • 29) 0.901 861 724 16 × 2 = 1 + 0.803 723 448 32;
  • 30) 0.803 723 448 32 × 2 = 1 + 0.607 446 896 64;
  • 31) 0.607 446 896 64 × 2 = 1 + 0.214 893 793 28;
  • 32) 0.214 893 793 28 × 2 = 0 + 0.429 787 586 56;
  • 33) 0.429 787 586 56 × 2 = 0 + 0.859 575 173 12;
  • 34) 0.859 575 173 12 × 2 = 1 + 0.719 150 346 24;
  • 35) 0.719 150 346 24 × 2 = 1 + 0.438 300 692 48;
  • 36) 0.438 300 692 48 × 2 = 0 + 0.876 601 384 96;
  • 37) 0.876 601 384 96 × 2 = 1 + 0.753 202 769 92;
  • 38) 0.753 202 769 92 × 2 = 1 + 0.506 405 539 84;
  • 39) 0.506 405 539 84 × 2 = 1 + 0.012 811 079 68;
  • 40) 0.012 811 079 68 × 2 = 0 + 0.025 622 159 36;
  • 41) 0.025 622 159 36 × 2 = 0 + 0.051 244 318 72;
  • 42) 0.051 244 318 72 × 2 = 0 + 0.102 488 637 44;
  • 43) 0.102 488 637 44 × 2 = 0 + 0.204 977 274 88;
  • 44) 0.204 977 274 88 × 2 = 0 + 0.409 954 549 76;
  • 45) 0.409 954 549 76 × 2 = 0 + 0.819 909 099 52;
  • 46) 0.819 909 099 52 × 2 = 1 + 0.639 818 199 04;
  • 47) 0.639 818 199 04 × 2 = 1 + 0.279 636 398 08;
  • 48) 0.279 636 398 08 × 2 = 0 + 0.559 272 796 16;
  • 49) 0.559 272 796 16 × 2 = 1 + 0.118 545 592 32;
  • 50) 0.118 545 592 32 × 2 = 0 + 0.237 091 184 64;
  • 51) 0.237 091 184 64 × 2 = 0 + 0.474 182 369 28;
  • 52) 0.474 182 369 28 × 2 = 0 + 0.948 364 738 56;
  • 53) 0.948 364 738 56 × 2 = 1 + 0.896 729 477 12;
  • 54) 0.896 729 477 12 × 2 = 1 + 0.793 458 954 24;
  • 55) 0.793 458 954 24 × 2 = 1 + 0.586 917 908 48;
  • 56) 0.586 917 908 48 × 2 = 1 + 0.173 835 816 96;
  • 57) 0.173 835 816 96 × 2 = 0 + 0.347 671 633 92;
  • 58) 0.347 671 633 92 × 2 = 0 + 0.695 343 267 84;
  • 59) 0.695 343 267 84 × 2 = 1 + 0.390 686 535 68;
  • 60) 0.390 686 535 68 × 2 = 0 + 0.781 373 071 36;
  • 61) 0.781 373 071 36 × 2 = 1 + 0.562 746 142 72;
  • 62) 0.562 746 142 72 × 2 = 1 + 0.125 492 285 44;
  • 63) 0.125 492 285 44 × 2 = 0 + 0.250 984 570 88;
  • 64) 0.250 984 570 88 × 2 = 0 + 0.501 969 141 76;

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:


0.000 282 004 11(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100(2)

6. Positive number before normalization:

0.000 282 004 11(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100(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:


0.000 282 004 11(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100(2) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100(2) × 20 =

1.0010 0111 1011 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100(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 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100


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 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100 =

0010 0111 1011 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100


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 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100


Decimal number -0.000 282 004 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1110 0110 1110 0000 0110 1000 1111 0010 1100

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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