-0.000 282 005 914 247 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 247(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
-0.000 282 005 914 247(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:

|-0.000 282 005 914 247| = 0.000 282 005 914 247


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: 0.000 282 005 914 247.

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) 0.000 282 005 914 247 × 2 = 0 + 0.000 564 011 828 494;
  • 2) 0.000 564 011 828 494 × 2 = 0 + 0.001 128 023 656 988;
  • 3) 0.001 128 023 656 988 × 2 = 0 + 0.002 256 047 313 976;
  • 4) 0.002 256 047 313 976 × 2 = 0 + 0.004 512 094 627 952;
  • 5) 0.004 512 094 627 952 × 2 = 0 + 0.009 024 189 255 904;
  • 6) 0.009 024 189 255 904 × 2 = 0 + 0.018 048 378 511 808;
  • 7) 0.018 048 378 511 808 × 2 = 0 + 0.036 096 757 023 616;
  • 8) 0.036 096 757 023 616 × 2 = 0 + 0.072 193 514 047 232;
  • 9) 0.072 193 514 047 232 × 2 = 0 + 0.144 387 028 094 464;
  • 10) 0.144 387 028 094 464 × 2 = 0 + 0.288 774 056 188 928;
  • 11) 0.288 774 056 188 928 × 2 = 0 + 0.577 548 112 377 856;
  • 12) 0.577 548 112 377 856 × 2 = 1 + 0.155 096 224 755 712;
  • 13) 0.155 096 224 755 712 × 2 = 0 + 0.310 192 449 511 424;
  • 14) 0.310 192 449 511 424 × 2 = 0 + 0.620 384 899 022 848;
  • 15) 0.620 384 899 022 848 × 2 = 1 + 0.240 769 798 045 696;
  • 16) 0.240 769 798 045 696 × 2 = 0 + 0.481 539 596 091 392;
  • 17) 0.481 539 596 091 392 × 2 = 0 + 0.963 079 192 182 784;
  • 18) 0.963 079 192 182 784 × 2 = 1 + 0.926 158 384 365 568;
  • 19) 0.926 158 384 365 568 × 2 = 1 + 0.852 316 768 731 136;
  • 20) 0.852 316 768 731 136 × 2 = 1 + 0.704 633 537 462 272;
  • 21) 0.704 633 537 462 272 × 2 = 1 + 0.409 267 074 924 544;
  • 22) 0.409 267 074 924 544 × 2 = 0 + 0.818 534 149 849 088;
  • 23) 0.818 534 149 849 088 × 2 = 1 + 0.637 068 299 698 176;
  • 24) 0.637 068 299 698 176 × 2 = 1 + 0.274 136 599 396 352;
  • 25) 0.274 136 599 396 352 × 2 = 0 + 0.548 273 198 792 704;
  • 26) 0.548 273 198 792 704 × 2 = 1 + 0.096 546 397 585 408;
  • 27) 0.096 546 397 585 408 × 2 = 0 + 0.193 092 795 170 816;
  • 28) 0.193 092 795 170 816 × 2 = 0 + 0.386 185 590 341 632;
  • 29) 0.386 185 590 341 632 × 2 = 0 + 0.772 371 180 683 264;
  • 30) 0.772 371 180 683 264 × 2 = 1 + 0.544 742 361 366 528;
  • 31) 0.544 742 361 366 528 × 2 = 1 + 0.089 484 722 733 056;
  • 32) 0.089 484 722 733 056 × 2 = 0 + 0.178 969 445 466 112;
  • 33) 0.178 969 445 466 112 × 2 = 0 + 0.357 938 890 932 224;
  • 34) 0.357 938 890 932 224 × 2 = 0 + 0.715 877 781 864 448;
  • 35) 0.715 877 781 864 448 × 2 = 1 + 0.431 755 563 728 896;
  • 36) 0.431 755 563 728 896 × 2 = 0 + 0.863 511 127 457 792;
  • 37) 0.863 511 127 457 792 × 2 = 1 + 0.727 022 254 915 584;
  • 38) 0.727 022 254 915 584 × 2 = 1 + 0.454 044 509 831 168;
  • 39) 0.454 044 509 831 168 × 2 = 0 + 0.908 089 019 662 336;
  • 40) 0.908 089 019 662 336 × 2 = 1 + 0.816 178 039 324 672;
  • 41) 0.816 178 039 324 672 × 2 = 1 + 0.632 356 078 649 344;
  • 42) 0.632 356 078 649 344 × 2 = 1 + 0.264 712 157 298 688;
  • 43) 0.264 712 157 298 688 × 2 = 0 + 0.529 424 314 597 376;
  • 44) 0.529 424 314 597 376 × 2 = 1 + 0.058 848 629 194 752;
  • 45) 0.058 848 629 194 752 × 2 = 0 + 0.117 697 258 389 504;
  • 46) 0.117 697 258 389 504 × 2 = 0 + 0.235 394 516 779 008;
  • 47) 0.235 394 516 779 008 × 2 = 0 + 0.470 789 033 558 016;
  • 48) 0.470 789 033 558 016 × 2 = 0 + 0.941 578 067 116 032;
  • 49) 0.941 578 067 116 032 × 2 = 1 + 0.883 156 134 232 064;
  • 50) 0.883 156 134 232 064 × 2 = 1 + 0.766 312 268 464 128;
  • 51) 0.766 312 268 464 128 × 2 = 1 + 0.532 624 536 928 256;
  • 52) 0.532 624 536 928 256 × 2 = 1 + 0.065 249 073 856 512;
  • 53) 0.065 249 073 856 512 × 2 = 0 + 0.130 498 147 713 024;
  • 54) 0.130 498 147 713 024 × 2 = 0 + 0.260 996 295 426 048;
  • 55) 0.260 996 295 426 048 × 2 = 0 + 0.521 992 590 852 096;
  • 56) 0.521 992 590 852 096 × 2 = 1 + 0.043 985 181 704 192;
  • 57) 0.043 985 181 704 192 × 2 = 0 + 0.087 970 363 408 384;
  • 58) 0.087 970 363 408 384 × 2 = 0 + 0.175 940 726 816 768;
  • 59) 0.175 940 726 816 768 × 2 = 0 + 0.351 881 453 633 536;
  • 60) 0.351 881 453 633 536 × 2 = 0 + 0.703 762 907 267 072;
  • 61) 0.703 762 907 267 072 × 2 = 1 + 0.407 525 814 534 144;
  • 62) 0.407 525 814 534 144 × 2 = 0 + 0.815 051 629 068 288;
  • 63) 0.815 051 629 068 288 × 2 = 1 + 0.630 103 258 136 576;
  • 64) 0.630 103 258 136 576 × 2 = 1 + 0.260 206 516 273 152;

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:


0.000 282 005 914 247(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1111 0001 0000 1011(2)

6. Positive number before normalization:

0.000 282 005 914 247(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1111 0001 0000 1011(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:


0.000 282 005 914 247(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1111 0001 0000 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1111 0001 0000 1011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1101 0000 1111 0001 0000 1011(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 1101 1101 0000 1111 0001 0000 1011


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 1101 1101 0000 1111 0001 0000 1011 =

0010 0111 1011 0100 0110 0010 1101 1101 0000 1111 0001 0000 1011


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 1101 1101 0000 1111 0001 0000 1011


Decimal number -0.000 282 005 914 247 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1101 0000 1111 0001 0000 1011

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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