-0.000 282 005 914 479 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 479(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
-0.000 282 005 914 479(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:

|-0.000 282 005 914 479| = 0.000 282 005 914 479


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: 0.000 282 005 914 479.

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) 0.000 282 005 914 479 × 2 = 0 + 0.000 564 011 828 958;
  • 2) 0.000 564 011 828 958 × 2 = 0 + 0.001 128 023 657 916;
  • 3) 0.001 128 023 657 916 × 2 = 0 + 0.002 256 047 315 832;
  • 4) 0.002 256 047 315 832 × 2 = 0 + 0.004 512 094 631 664;
  • 5) 0.004 512 094 631 664 × 2 = 0 + 0.009 024 189 263 328;
  • 6) 0.009 024 189 263 328 × 2 = 0 + 0.018 048 378 526 656;
  • 7) 0.018 048 378 526 656 × 2 = 0 + 0.036 096 757 053 312;
  • 8) 0.036 096 757 053 312 × 2 = 0 + 0.072 193 514 106 624;
  • 9) 0.072 193 514 106 624 × 2 = 0 + 0.144 387 028 213 248;
  • 10) 0.144 387 028 213 248 × 2 = 0 + 0.288 774 056 426 496;
  • 11) 0.288 774 056 426 496 × 2 = 0 + 0.577 548 112 852 992;
  • 12) 0.577 548 112 852 992 × 2 = 1 + 0.155 096 225 705 984;
  • 13) 0.155 096 225 705 984 × 2 = 0 + 0.310 192 451 411 968;
  • 14) 0.310 192 451 411 968 × 2 = 0 + 0.620 384 902 823 936;
  • 15) 0.620 384 902 823 936 × 2 = 1 + 0.240 769 805 647 872;
  • 16) 0.240 769 805 647 872 × 2 = 0 + 0.481 539 611 295 744;
  • 17) 0.481 539 611 295 744 × 2 = 0 + 0.963 079 222 591 488;
  • 18) 0.963 079 222 591 488 × 2 = 1 + 0.926 158 445 182 976;
  • 19) 0.926 158 445 182 976 × 2 = 1 + 0.852 316 890 365 952;
  • 20) 0.852 316 890 365 952 × 2 = 1 + 0.704 633 780 731 904;
  • 21) 0.704 633 780 731 904 × 2 = 1 + 0.409 267 561 463 808;
  • 22) 0.409 267 561 463 808 × 2 = 0 + 0.818 535 122 927 616;
  • 23) 0.818 535 122 927 616 × 2 = 1 + 0.637 070 245 855 232;
  • 24) 0.637 070 245 855 232 × 2 = 1 + 0.274 140 491 710 464;
  • 25) 0.274 140 491 710 464 × 2 = 0 + 0.548 280 983 420 928;
  • 26) 0.548 280 983 420 928 × 2 = 1 + 0.096 561 966 841 856;
  • 27) 0.096 561 966 841 856 × 2 = 0 + 0.193 123 933 683 712;
  • 28) 0.193 123 933 683 712 × 2 = 0 + 0.386 247 867 367 424;
  • 29) 0.386 247 867 367 424 × 2 = 0 + 0.772 495 734 734 848;
  • 30) 0.772 495 734 734 848 × 2 = 1 + 0.544 991 469 469 696;
  • 31) 0.544 991 469 469 696 × 2 = 1 + 0.089 982 938 939 392;
  • 32) 0.089 982 938 939 392 × 2 = 0 + 0.179 965 877 878 784;
  • 33) 0.179 965 877 878 784 × 2 = 0 + 0.359 931 755 757 568;
  • 34) 0.359 931 755 757 568 × 2 = 0 + 0.719 863 511 515 136;
  • 35) 0.719 863 511 515 136 × 2 = 1 + 0.439 727 023 030 272;
  • 36) 0.439 727 023 030 272 × 2 = 0 + 0.879 454 046 060 544;
  • 37) 0.879 454 046 060 544 × 2 = 1 + 0.758 908 092 121 088;
  • 38) 0.758 908 092 121 088 × 2 = 1 + 0.517 816 184 242 176;
  • 39) 0.517 816 184 242 176 × 2 = 1 + 0.035 632 368 484 352;
  • 40) 0.035 632 368 484 352 × 2 = 0 + 0.071 264 736 968 704;
  • 41) 0.071 264 736 968 704 × 2 = 0 + 0.142 529 473 937 408;
  • 42) 0.142 529 473 937 408 × 2 = 0 + 0.285 058 947 874 816;
  • 43) 0.285 058 947 874 816 × 2 = 0 + 0.570 117 895 749 632;
  • 44) 0.570 117 895 749 632 × 2 = 1 + 0.140 235 791 499 264;
  • 45) 0.140 235 791 499 264 × 2 = 0 + 0.280 471 582 998 528;
  • 46) 0.280 471 582 998 528 × 2 = 0 + 0.560 943 165 997 056;
  • 47) 0.560 943 165 997 056 × 2 = 1 + 0.121 886 331 994 112;
  • 48) 0.121 886 331 994 112 × 2 = 0 + 0.243 772 663 988 224;
  • 49) 0.243 772 663 988 224 × 2 = 0 + 0.487 545 327 976 448;
  • 50) 0.487 545 327 976 448 × 2 = 0 + 0.975 090 655 952 896;
  • 51) 0.975 090 655 952 896 × 2 = 1 + 0.950 181 311 905 792;
  • 52) 0.950 181 311 905 792 × 2 = 1 + 0.900 362 623 811 584;
  • 53) 0.900 362 623 811 584 × 2 = 1 + 0.800 725 247 623 168;
  • 54) 0.800 725 247 623 168 × 2 = 1 + 0.601 450 495 246 336;
  • 55) 0.601 450 495 246 336 × 2 = 1 + 0.202 900 990 492 672;
  • 56) 0.202 900 990 492 672 × 2 = 0 + 0.405 801 980 985 344;
  • 57) 0.405 801 980 985 344 × 2 = 0 + 0.811 603 961 970 688;
  • 58) 0.811 603 961 970 688 × 2 = 1 + 0.623 207 923 941 376;
  • 59) 0.623 207 923 941 376 × 2 = 1 + 0.246 415 847 882 752;
  • 60) 0.246 415 847 882 752 × 2 = 0 + 0.492 831 695 765 504;
  • 61) 0.492 831 695 765 504 × 2 = 0 + 0.985 663 391 531 008;
  • 62) 0.985 663 391 531 008 × 2 = 1 + 0.971 326 783 062 016;
  • 63) 0.971 326 783 062 016 × 2 = 1 + 0.942 653 566 124 032;
  • 64) 0.942 653 566 124 032 × 2 = 1 + 0.885 307 132 248 064;

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:


0.000 282 005 914 479(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0010 0011 1110 0110 0111(2)

6. Positive number before normalization:

0.000 282 005 914 479(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0010 0011 1110 0110 0111(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:


0.000 282 005 914 479(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0010 0011 1110 0110 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0010 0011 1110 0110 0111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 0010 0011 1110 0110 0111(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 1110 0001 0010 0011 1110 0110 0111


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 1110 0001 0010 0011 1110 0110 0111 =

0010 0111 1011 0100 0110 0010 1110 0001 0010 0011 1110 0110 0111


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 1110 0001 0010 0011 1110 0110 0111


Decimal number -0.000 282 005 914 479 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 0010 0011 1110 0110 0111

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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