-0.000 282 367 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 367(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
-0.000 282 367(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:

|-0.000 282 367| = 0.000 282 367


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: 0.000 282 367.

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) 0.000 282 367 × 2 = 0 + 0.000 564 734;
  • 2) 0.000 564 734 × 2 = 0 + 0.001 129 468;
  • 3) 0.001 129 468 × 2 = 0 + 0.002 258 936;
  • 4) 0.002 258 936 × 2 = 0 + 0.004 517 872;
  • 5) 0.004 517 872 × 2 = 0 + 0.009 035 744;
  • 6) 0.009 035 744 × 2 = 0 + 0.018 071 488;
  • 7) 0.018 071 488 × 2 = 0 + 0.036 142 976;
  • 8) 0.036 142 976 × 2 = 0 + 0.072 285 952;
  • 9) 0.072 285 952 × 2 = 0 + 0.144 571 904;
  • 10) 0.144 571 904 × 2 = 0 + 0.289 143 808;
  • 11) 0.289 143 808 × 2 = 0 + 0.578 287 616;
  • 12) 0.578 287 616 × 2 = 1 + 0.156 575 232;
  • 13) 0.156 575 232 × 2 = 0 + 0.313 150 464;
  • 14) 0.313 150 464 × 2 = 0 + 0.626 300 928;
  • 15) 0.626 300 928 × 2 = 1 + 0.252 601 856;
  • 16) 0.252 601 856 × 2 = 0 + 0.505 203 712;
  • 17) 0.505 203 712 × 2 = 1 + 0.010 407 424;
  • 18) 0.010 407 424 × 2 = 0 + 0.020 814 848;
  • 19) 0.020 814 848 × 2 = 0 + 0.041 629 696;
  • 20) 0.041 629 696 × 2 = 0 + 0.083 259 392;
  • 21) 0.083 259 392 × 2 = 0 + 0.166 518 784;
  • 22) 0.166 518 784 × 2 = 0 + 0.333 037 568;
  • 23) 0.333 037 568 × 2 = 0 + 0.666 075 136;
  • 24) 0.666 075 136 × 2 = 1 + 0.332 150 272;
  • 25) 0.332 150 272 × 2 = 0 + 0.664 300 544;
  • 26) 0.664 300 544 × 2 = 1 + 0.328 601 088;
  • 27) 0.328 601 088 × 2 = 0 + 0.657 202 176;
  • 28) 0.657 202 176 × 2 = 1 + 0.314 404 352;
  • 29) 0.314 404 352 × 2 = 0 + 0.628 808 704;
  • 30) 0.628 808 704 × 2 = 1 + 0.257 617 408;
  • 31) 0.257 617 408 × 2 = 0 + 0.515 234 816;
  • 32) 0.515 234 816 × 2 = 1 + 0.030 469 632;
  • 33) 0.030 469 632 × 2 = 0 + 0.060 939 264;
  • 34) 0.060 939 264 × 2 = 0 + 0.121 878 528;
  • 35) 0.121 878 528 × 2 = 0 + 0.243 757 056;
  • 36) 0.243 757 056 × 2 = 0 + 0.487 514 112;
  • 37) 0.487 514 112 × 2 = 0 + 0.975 028 224;
  • 38) 0.975 028 224 × 2 = 1 + 0.950 056 448;
  • 39) 0.950 056 448 × 2 = 1 + 0.900 112 896;
  • 40) 0.900 112 896 × 2 = 1 + 0.800 225 792;
  • 41) 0.800 225 792 × 2 = 1 + 0.600 451 584;
  • 42) 0.600 451 584 × 2 = 1 + 0.200 903 168;
  • 43) 0.200 903 168 × 2 = 0 + 0.401 806 336;
  • 44) 0.401 806 336 × 2 = 0 + 0.803 612 672;
  • 45) 0.803 612 672 × 2 = 1 + 0.607 225 344;
  • 46) 0.607 225 344 × 2 = 1 + 0.214 450 688;
  • 47) 0.214 450 688 × 2 = 0 + 0.428 901 376;
  • 48) 0.428 901 376 × 2 = 0 + 0.857 802 752;
  • 49) 0.857 802 752 × 2 = 1 + 0.715 605 504;
  • 50) 0.715 605 504 × 2 = 1 + 0.431 211 008;
  • 51) 0.431 211 008 × 2 = 0 + 0.862 422 016;
  • 52) 0.862 422 016 × 2 = 1 + 0.724 844 032;
  • 53) 0.724 844 032 × 2 = 1 + 0.449 688 064;
  • 54) 0.449 688 064 × 2 = 0 + 0.899 376 128;
  • 55) 0.899 376 128 × 2 = 1 + 0.798 752 256;
  • 56) 0.798 752 256 × 2 = 1 + 0.597 504 512;
  • 57) 0.597 504 512 × 2 = 1 + 0.195 009 024;
  • 58) 0.195 009 024 × 2 = 0 + 0.390 018 048;
  • 59) 0.390 018 048 × 2 = 0 + 0.780 036 096;
  • 60) 0.780 036 096 × 2 = 1 + 0.560 072 192;
  • 61) 0.560 072 192 × 2 = 1 + 0.120 144 384;
  • 62) 0.120 144 384 × 2 = 0 + 0.240 288 768;
  • 63) 0.240 288 768 × 2 = 0 + 0.480 577 536;
  • 64) 0.480 577 536 × 2 = 0 + 0.961 155 072;

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:


0.000 282 367(10) =

0.0000 0000 0001 0010 1000 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000(2)

6. Positive number before normalization:

0.000 282 367(10) =

0.0000 0000 0001 0010 1000 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000(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:


0.000 282 367(10) =

0.0000 0000 0001 0010 1000 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000(2) =

0.0000 0000 0001 0010 1000 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000(2) × 20 =

1.0010 1000 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000(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 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000


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 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000 =

0010 1000 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000


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 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000


Decimal number -0.000 282 367 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0001 0101 0101 0000 0111 1100 1100 1101 1011 1001 1000

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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