-0.000 282 005 914 493 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 493(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
-0.000 282 005 914 493(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:

|-0.000 282 005 914 493| = 0.000 282 005 914 493


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: 0.000 282 005 914 493.

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) 0.000 282 005 914 493 × 2 = 0 + 0.000 564 011 828 986;
  • 2) 0.000 564 011 828 986 × 2 = 0 + 0.001 128 023 657 972;
  • 3) 0.001 128 023 657 972 × 2 = 0 + 0.002 256 047 315 944;
  • 4) 0.002 256 047 315 944 × 2 = 0 + 0.004 512 094 631 888;
  • 5) 0.004 512 094 631 888 × 2 = 0 + 0.009 024 189 263 776;
  • 6) 0.009 024 189 263 776 × 2 = 0 + 0.018 048 378 527 552;
  • 7) 0.018 048 378 527 552 × 2 = 0 + 0.036 096 757 055 104;
  • 8) 0.036 096 757 055 104 × 2 = 0 + 0.072 193 514 110 208;
  • 9) 0.072 193 514 110 208 × 2 = 0 + 0.144 387 028 220 416;
  • 10) 0.144 387 028 220 416 × 2 = 0 + 0.288 774 056 440 832;
  • 11) 0.288 774 056 440 832 × 2 = 0 + 0.577 548 112 881 664;
  • 12) 0.577 548 112 881 664 × 2 = 1 + 0.155 096 225 763 328;
  • 13) 0.155 096 225 763 328 × 2 = 0 + 0.310 192 451 526 656;
  • 14) 0.310 192 451 526 656 × 2 = 0 + 0.620 384 903 053 312;
  • 15) 0.620 384 903 053 312 × 2 = 1 + 0.240 769 806 106 624;
  • 16) 0.240 769 806 106 624 × 2 = 0 + 0.481 539 612 213 248;
  • 17) 0.481 539 612 213 248 × 2 = 0 + 0.963 079 224 426 496;
  • 18) 0.963 079 224 426 496 × 2 = 1 + 0.926 158 448 852 992;
  • 19) 0.926 158 448 852 992 × 2 = 1 + 0.852 316 897 705 984;
  • 20) 0.852 316 897 705 984 × 2 = 1 + 0.704 633 795 411 968;
  • 21) 0.704 633 795 411 968 × 2 = 1 + 0.409 267 590 823 936;
  • 22) 0.409 267 590 823 936 × 2 = 0 + 0.818 535 181 647 872;
  • 23) 0.818 535 181 647 872 × 2 = 1 + 0.637 070 363 295 744;
  • 24) 0.637 070 363 295 744 × 2 = 1 + 0.274 140 726 591 488;
  • 25) 0.274 140 726 591 488 × 2 = 0 + 0.548 281 453 182 976;
  • 26) 0.548 281 453 182 976 × 2 = 1 + 0.096 562 906 365 952;
  • 27) 0.096 562 906 365 952 × 2 = 0 + 0.193 125 812 731 904;
  • 28) 0.193 125 812 731 904 × 2 = 0 + 0.386 251 625 463 808;
  • 29) 0.386 251 625 463 808 × 2 = 0 + 0.772 503 250 927 616;
  • 30) 0.772 503 250 927 616 × 2 = 1 + 0.545 006 501 855 232;
  • 31) 0.545 006 501 855 232 × 2 = 1 + 0.090 013 003 710 464;
  • 32) 0.090 013 003 710 464 × 2 = 0 + 0.180 026 007 420 928;
  • 33) 0.180 026 007 420 928 × 2 = 0 + 0.360 052 014 841 856;
  • 34) 0.360 052 014 841 856 × 2 = 0 + 0.720 104 029 683 712;
  • 35) 0.720 104 029 683 712 × 2 = 1 + 0.440 208 059 367 424;
  • 36) 0.440 208 059 367 424 × 2 = 0 + 0.880 416 118 734 848;
  • 37) 0.880 416 118 734 848 × 2 = 1 + 0.760 832 237 469 696;
  • 38) 0.760 832 237 469 696 × 2 = 1 + 0.521 664 474 939 392;
  • 39) 0.521 664 474 939 392 × 2 = 1 + 0.043 328 949 878 784;
  • 40) 0.043 328 949 878 784 × 2 = 0 + 0.086 657 899 757 568;
  • 41) 0.086 657 899 757 568 × 2 = 0 + 0.173 315 799 515 136;
  • 42) 0.173 315 799 515 136 × 2 = 0 + 0.346 631 599 030 272;
  • 43) 0.346 631 599 030 272 × 2 = 0 + 0.693 263 198 060 544;
  • 44) 0.693 263 198 060 544 × 2 = 1 + 0.386 526 396 121 088;
  • 45) 0.386 526 396 121 088 × 2 = 0 + 0.773 052 792 242 176;
  • 46) 0.773 052 792 242 176 × 2 = 1 + 0.546 105 584 484 352;
  • 47) 0.546 105 584 484 352 × 2 = 1 + 0.092 211 168 968 704;
  • 48) 0.092 211 168 968 704 × 2 = 0 + 0.184 422 337 937 408;
  • 49) 0.184 422 337 937 408 × 2 = 0 + 0.368 844 675 874 816;
  • 50) 0.368 844 675 874 816 × 2 = 0 + 0.737 689 351 749 632;
  • 51) 0.737 689 351 749 632 × 2 = 1 + 0.475 378 703 499 264;
  • 52) 0.475 378 703 499 264 × 2 = 0 + 0.950 757 406 998 528;
  • 53) 0.950 757 406 998 528 × 2 = 1 + 0.901 514 813 997 056;
  • 54) 0.901 514 813 997 056 × 2 = 1 + 0.803 029 627 994 112;
  • 55) 0.803 029 627 994 112 × 2 = 1 + 0.606 059 255 988 224;
  • 56) 0.606 059 255 988 224 × 2 = 1 + 0.212 118 511 976 448;
  • 57) 0.212 118 511 976 448 × 2 = 0 + 0.424 237 023 952 896;
  • 58) 0.424 237 023 952 896 × 2 = 0 + 0.848 474 047 905 792;
  • 59) 0.848 474 047 905 792 × 2 = 1 + 0.696 948 095 811 584;
  • 60) 0.696 948 095 811 584 × 2 = 1 + 0.393 896 191 623 168;
  • 61) 0.393 896 191 623 168 × 2 = 0 + 0.787 792 383 246 336;
  • 62) 0.787 792 383 246 336 × 2 = 1 + 0.575 584 766 492 672;
  • 63) 0.575 584 766 492 672 × 2 = 1 + 0.151 169 532 985 344;
  • 64) 0.151 169 532 985 344 × 2 = 0 + 0.302 339 065 970 688;

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:


0.000 282 005 914 493(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0010 1111 0011 0110(2)

6. Positive number before normalization:

0.000 282 005 914 493(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0010 1111 0011 0110(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:


0.000 282 005 914 493(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0010 1111 0011 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0010 1111 0011 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 0110 0010 1111 0011 0110(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 0110 0010 1111 0011 0110


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 0110 0010 1111 0011 0110 =

0010 0111 1011 0100 0110 0010 1110 0001 0110 0010 1111 0011 0110


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 0110 0010 1111 0011 0110


Decimal number -0.000 282 005 914 493 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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