-0.000 282 005 904 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 904(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
-0.000 282 005 904(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:

|-0.000 282 005 904| = 0.000 282 005 904


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: 0.000 282 005 904.

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) 0.000 282 005 904 × 2 = 0 + 0.000 564 011 808;
  • 2) 0.000 564 011 808 × 2 = 0 + 0.001 128 023 616;
  • 3) 0.001 128 023 616 × 2 = 0 + 0.002 256 047 232;
  • 4) 0.002 256 047 232 × 2 = 0 + 0.004 512 094 464;
  • 5) 0.004 512 094 464 × 2 = 0 + 0.009 024 188 928;
  • 6) 0.009 024 188 928 × 2 = 0 + 0.018 048 377 856;
  • 7) 0.018 048 377 856 × 2 = 0 + 0.036 096 755 712;
  • 8) 0.036 096 755 712 × 2 = 0 + 0.072 193 511 424;
  • 9) 0.072 193 511 424 × 2 = 0 + 0.144 387 022 848;
  • 10) 0.144 387 022 848 × 2 = 0 + 0.288 774 045 696;
  • 11) 0.288 774 045 696 × 2 = 0 + 0.577 548 091 392;
  • 12) 0.577 548 091 392 × 2 = 1 + 0.155 096 182 784;
  • 13) 0.155 096 182 784 × 2 = 0 + 0.310 192 365 568;
  • 14) 0.310 192 365 568 × 2 = 0 + 0.620 384 731 136;
  • 15) 0.620 384 731 136 × 2 = 1 + 0.240 769 462 272;
  • 16) 0.240 769 462 272 × 2 = 0 + 0.481 538 924 544;
  • 17) 0.481 538 924 544 × 2 = 0 + 0.963 077 849 088;
  • 18) 0.963 077 849 088 × 2 = 1 + 0.926 155 698 176;
  • 19) 0.926 155 698 176 × 2 = 1 + 0.852 311 396 352;
  • 20) 0.852 311 396 352 × 2 = 1 + 0.704 622 792 704;
  • 21) 0.704 622 792 704 × 2 = 1 + 0.409 245 585 408;
  • 22) 0.409 245 585 408 × 2 = 0 + 0.818 491 170 816;
  • 23) 0.818 491 170 816 × 2 = 1 + 0.636 982 341 632;
  • 24) 0.636 982 341 632 × 2 = 1 + 0.273 964 683 264;
  • 25) 0.273 964 683 264 × 2 = 0 + 0.547 929 366 528;
  • 26) 0.547 929 366 528 × 2 = 1 + 0.095 858 733 056;
  • 27) 0.095 858 733 056 × 2 = 0 + 0.191 717 466 112;
  • 28) 0.191 717 466 112 × 2 = 0 + 0.383 434 932 224;
  • 29) 0.383 434 932 224 × 2 = 0 + 0.766 869 864 448;
  • 30) 0.766 869 864 448 × 2 = 1 + 0.533 739 728 896;
  • 31) 0.533 739 728 896 × 2 = 1 + 0.067 479 457 792;
  • 32) 0.067 479 457 792 × 2 = 0 + 0.134 958 915 584;
  • 33) 0.134 958 915 584 × 2 = 0 + 0.269 917 831 168;
  • 34) 0.269 917 831 168 × 2 = 0 + 0.539 835 662 336;
  • 35) 0.539 835 662 336 × 2 = 1 + 0.079 671 324 672;
  • 36) 0.079 671 324 672 × 2 = 0 + 0.159 342 649 344;
  • 37) 0.159 342 649 344 × 2 = 0 + 0.318 685 298 688;
  • 38) 0.318 685 298 688 × 2 = 0 + 0.637 370 597 376;
  • 39) 0.637 370 597 376 × 2 = 1 + 0.274 741 194 752;
  • 40) 0.274 741 194 752 × 2 = 0 + 0.549 482 389 504;
  • 41) 0.549 482 389 504 × 2 = 1 + 0.098 964 779 008;
  • 42) 0.098 964 779 008 × 2 = 0 + 0.197 929 558 016;
  • 43) 0.197 929 558 016 × 2 = 0 + 0.395 859 116 032;
  • 44) 0.395 859 116 032 × 2 = 0 + 0.791 718 232 064;
  • 45) 0.791 718 232 064 × 2 = 1 + 0.583 436 464 128;
  • 46) 0.583 436 464 128 × 2 = 1 + 0.166 872 928 256;
  • 47) 0.166 872 928 256 × 2 = 0 + 0.333 745 856 512;
  • 48) 0.333 745 856 512 × 2 = 0 + 0.667 491 713 024;
  • 49) 0.667 491 713 024 × 2 = 1 + 0.334 983 426 048;
  • 50) 0.334 983 426 048 × 2 = 0 + 0.669 966 852 096;
  • 51) 0.669 966 852 096 × 2 = 1 + 0.339 933 704 192;
  • 52) 0.339 933 704 192 × 2 = 0 + 0.679 867 408 384;
  • 53) 0.679 867 408 384 × 2 = 1 + 0.359 734 816 768;
  • 54) 0.359 734 816 768 × 2 = 0 + 0.719 469 633 536;
  • 55) 0.719 469 633 536 × 2 = 1 + 0.438 939 267 072;
  • 56) 0.438 939 267 072 × 2 = 0 + 0.877 878 534 144;
  • 57) 0.877 878 534 144 × 2 = 1 + 0.755 757 068 288;
  • 58) 0.755 757 068 288 × 2 = 1 + 0.511 514 136 576;
  • 59) 0.511 514 136 576 × 2 = 1 + 0.023 028 273 152;
  • 60) 0.023 028 273 152 × 2 = 0 + 0.046 056 546 304;
  • 61) 0.046 056 546 304 × 2 = 0 + 0.092 113 092 608;
  • 62) 0.092 113 092 608 × 2 = 0 + 0.184 226 185 216;
  • 63) 0.184 226 185 216 × 2 = 0 + 0.368 452 370 432;
  • 64) 0.368 452 370 432 × 2 = 0 + 0.736 904 740 864;

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:


0.000 282 005 904(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0010 1000 1100 1010 1010 1110 0000(2)

6. Positive number before normalization:

0.000 282 005 904(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0010 1000 1100 1010 1010 1110 0000(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:


0.000 282 005 904(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0010 1000 1100 1010 1010 1110 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0010 1000 1100 1010 1010 1110 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0010 1000 1100 1010 1010 1110 0000(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 0110 0010 0010 1000 1100 1010 1010 1110 0000


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 0110 0010 0010 1000 1100 1010 1010 1110 0000 =

0010 0111 1011 0100 0110 0010 0010 1000 1100 1010 1010 1110 0000


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 0110 0010 0010 1000 1100 1010 1010 1110 0000


Decimal number -0.000 282 005 904 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0010 1000 1100 1010 1010 1110 0000

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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