-0.000 282 004 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 57(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 57(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 57| = 0.000 282 004 57


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 57.

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 57 × 2 = 0 + 0.000 564 009 14;
  • 2) 0.000 564 009 14 × 2 = 0 + 0.001 128 018 28;
  • 3) 0.001 128 018 28 × 2 = 0 + 0.002 256 036 56;
  • 4) 0.002 256 036 56 × 2 = 0 + 0.004 512 073 12;
  • 5) 0.004 512 073 12 × 2 = 0 + 0.009 024 146 24;
  • 6) 0.009 024 146 24 × 2 = 0 + 0.018 048 292 48;
  • 7) 0.018 048 292 48 × 2 = 0 + 0.036 096 584 96;
  • 8) 0.036 096 584 96 × 2 = 0 + 0.072 193 169 92;
  • 9) 0.072 193 169 92 × 2 = 0 + 0.144 386 339 84;
  • 10) 0.144 386 339 84 × 2 = 0 + 0.288 772 679 68;
  • 11) 0.288 772 679 68 × 2 = 0 + 0.577 545 359 36;
  • 12) 0.577 545 359 36 × 2 = 1 + 0.155 090 718 72;
  • 13) 0.155 090 718 72 × 2 = 0 + 0.310 181 437 44;
  • 14) 0.310 181 437 44 × 2 = 0 + 0.620 362 874 88;
  • 15) 0.620 362 874 88 × 2 = 1 + 0.240 725 749 76;
  • 16) 0.240 725 749 76 × 2 = 0 + 0.481 451 499 52;
  • 17) 0.481 451 499 52 × 2 = 0 + 0.962 902 999 04;
  • 18) 0.962 902 999 04 × 2 = 1 + 0.925 805 998 08;
  • 19) 0.925 805 998 08 × 2 = 1 + 0.851 611 996 16;
  • 20) 0.851 611 996 16 × 2 = 1 + 0.703 223 992 32;
  • 21) 0.703 223 992 32 × 2 = 1 + 0.406 447 984 64;
  • 22) 0.406 447 984 64 × 2 = 0 + 0.812 895 969 28;
  • 23) 0.812 895 969 28 × 2 = 1 + 0.625 791 938 56;
  • 24) 0.625 791 938 56 × 2 = 1 + 0.251 583 877 12;
  • 25) 0.251 583 877 12 × 2 = 0 + 0.503 167 754 24;
  • 26) 0.503 167 754 24 × 2 = 1 + 0.006 335 508 48;
  • 27) 0.006 335 508 48 × 2 = 0 + 0.012 671 016 96;
  • 28) 0.012 671 016 96 × 2 = 0 + 0.025 342 033 92;
  • 29) 0.025 342 033 92 × 2 = 0 + 0.050 684 067 84;
  • 30) 0.050 684 067 84 × 2 = 0 + 0.101 368 135 68;
  • 31) 0.101 368 135 68 × 2 = 0 + 0.202 736 271 36;
  • 32) 0.202 736 271 36 × 2 = 0 + 0.405 472 542 72;
  • 33) 0.405 472 542 72 × 2 = 0 + 0.810 945 085 44;
  • 34) 0.810 945 085 44 × 2 = 1 + 0.621 890 170 88;
  • 35) 0.621 890 170 88 × 2 = 1 + 0.243 780 341 76;
  • 36) 0.243 780 341 76 × 2 = 0 + 0.487 560 683 52;
  • 37) 0.487 560 683 52 × 2 = 0 + 0.975 121 367 04;
  • 38) 0.975 121 367 04 × 2 = 1 + 0.950 242 734 08;
  • 39) 0.950 242 734 08 × 2 = 1 + 0.900 485 468 16;
  • 40) 0.900 485 468 16 × 2 = 1 + 0.800 970 936 32;
  • 41) 0.800 970 936 32 × 2 = 1 + 0.601 941 872 64;
  • 42) 0.601 941 872 64 × 2 = 1 + 0.203 883 745 28;
  • 43) 0.203 883 745 28 × 2 = 0 + 0.407 767 490 56;
  • 44) 0.407 767 490 56 × 2 = 0 + 0.815 534 981 12;
  • 45) 0.815 534 981 12 × 2 = 1 + 0.631 069 962 24;
  • 46) 0.631 069 962 24 × 2 = 1 + 0.262 139 924 48;
  • 47) 0.262 139 924 48 × 2 = 0 + 0.524 279 848 96;
  • 48) 0.524 279 848 96 × 2 = 1 + 0.048 559 697 92;
  • 49) 0.048 559 697 92 × 2 = 0 + 0.097 119 395 84;
  • 50) 0.097 119 395 84 × 2 = 0 + 0.194 238 791 68;
  • 51) 0.194 238 791 68 × 2 = 0 + 0.388 477 583 36;
  • 52) 0.388 477 583 36 × 2 = 0 + 0.776 955 166 72;
  • 53) 0.776 955 166 72 × 2 = 1 + 0.553 910 333 44;
  • 54) 0.553 910 333 44 × 2 = 1 + 0.107 820 666 88;
  • 55) 0.107 820 666 88 × 2 = 0 + 0.215 641 333 76;
  • 56) 0.215 641 333 76 × 2 = 0 + 0.431 282 667 52;
  • 57) 0.431 282 667 52 × 2 = 0 + 0.862 565 335 04;
  • 58) 0.862 565 335 04 × 2 = 1 + 0.725 130 670 08;
  • 59) 0.725 130 670 08 × 2 = 1 + 0.450 261 340 16;
  • 60) 0.450 261 340 16 × 2 = 0 + 0.900 522 680 32;
  • 61) 0.900 522 680 32 × 2 = 1 + 0.801 045 360 64;
  • 62) 0.801 045 360 64 × 2 = 1 + 0.602 090 721 28;
  • 63) 0.602 090 721 28 × 2 = 1 + 0.204 181 442 56;
  • 64) 0.204 181 442 56 × 2 = 0 + 0.408 362 885 12;

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 57(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110(2)

6. Positive number before normalization:

0.000 282 004 57(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110(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 57(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110(2) × 20 =

1.0010 0111 1011 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110(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 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110


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 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110 =

0010 0111 1011 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110


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 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110


Decimal number -0.000 282 004 57 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0000 0110 0111 1100 1101 0000 1100 0110 1110

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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