0.000 020 871 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 871 2(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
0.000 020 871 2(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 020 871 2.

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) 0.000 020 871 2 × 2 = 0 + 0.000 041 742 4;
  • 2) 0.000 041 742 4 × 2 = 0 + 0.000 083 484 8;
  • 3) 0.000 083 484 8 × 2 = 0 + 0.000 166 969 6;
  • 4) 0.000 166 969 6 × 2 = 0 + 0.000 333 939 2;
  • 5) 0.000 333 939 2 × 2 = 0 + 0.000 667 878 4;
  • 6) 0.000 667 878 4 × 2 = 0 + 0.001 335 756 8;
  • 7) 0.001 335 756 8 × 2 = 0 + 0.002 671 513 6;
  • 8) 0.002 671 513 6 × 2 = 0 + 0.005 343 027 2;
  • 9) 0.005 343 027 2 × 2 = 0 + 0.010 686 054 4;
  • 10) 0.010 686 054 4 × 2 = 0 + 0.021 372 108 8;
  • 11) 0.021 372 108 8 × 2 = 0 + 0.042 744 217 6;
  • 12) 0.042 744 217 6 × 2 = 0 + 0.085 488 435 2;
  • 13) 0.085 488 435 2 × 2 = 0 + 0.170 976 870 4;
  • 14) 0.170 976 870 4 × 2 = 0 + 0.341 953 740 8;
  • 15) 0.341 953 740 8 × 2 = 0 + 0.683 907 481 6;
  • 16) 0.683 907 481 6 × 2 = 1 + 0.367 814 963 2;
  • 17) 0.367 814 963 2 × 2 = 0 + 0.735 629 926 4;
  • 18) 0.735 629 926 4 × 2 = 1 + 0.471 259 852 8;
  • 19) 0.471 259 852 8 × 2 = 0 + 0.942 519 705 6;
  • 20) 0.942 519 705 6 × 2 = 1 + 0.885 039 411 2;
  • 21) 0.885 039 411 2 × 2 = 1 + 0.770 078 822 4;
  • 22) 0.770 078 822 4 × 2 = 1 + 0.540 157 644 8;
  • 23) 0.540 157 644 8 × 2 = 1 + 0.080 315 289 6;
  • 24) 0.080 315 289 6 × 2 = 0 + 0.160 630 579 2;
  • 25) 0.160 630 579 2 × 2 = 0 + 0.321 261 158 4;
  • 26) 0.321 261 158 4 × 2 = 0 + 0.642 522 316 8;
  • 27) 0.642 522 316 8 × 2 = 1 + 0.285 044 633 6;
  • 28) 0.285 044 633 6 × 2 = 0 + 0.570 089 267 2;
  • 29) 0.570 089 267 2 × 2 = 1 + 0.140 178 534 4;
  • 30) 0.140 178 534 4 × 2 = 0 + 0.280 357 068 8;
  • 31) 0.280 357 068 8 × 2 = 0 + 0.560 714 137 6;
  • 32) 0.560 714 137 6 × 2 = 1 + 0.121 428 275 2;
  • 33) 0.121 428 275 2 × 2 = 0 + 0.242 856 550 4;
  • 34) 0.242 856 550 4 × 2 = 0 + 0.485 713 100 8;
  • 35) 0.485 713 100 8 × 2 = 0 + 0.971 426 201 6;
  • 36) 0.971 426 201 6 × 2 = 1 + 0.942 852 403 2;
  • 37) 0.942 852 403 2 × 2 = 1 + 0.885 704 806 4;
  • 38) 0.885 704 806 4 × 2 = 1 + 0.771 409 612 8;
  • 39) 0.771 409 612 8 × 2 = 1 + 0.542 819 225 6;
  • 40) 0.542 819 225 6 × 2 = 1 + 0.085 638 451 2;
  • 41) 0.085 638 451 2 × 2 = 0 + 0.171 276 902 4;
  • 42) 0.171 276 902 4 × 2 = 0 + 0.342 553 804 8;
  • 43) 0.342 553 804 8 × 2 = 0 + 0.685 107 609 6;
  • 44) 0.685 107 609 6 × 2 = 1 + 0.370 215 219 2;
  • 45) 0.370 215 219 2 × 2 = 0 + 0.740 430 438 4;
  • 46) 0.740 430 438 4 × 2 = 1 + 0.480 860 876 8;
  • 47) 0.480 860 876 8 × 2 = 0 + 0.961 721 753 6;
  • 48) 0.961 721 753 6 × 2 = 1 + 0.923 443 507 2;
  • 49) 0.923 443 507 2 × 2 = 1 + 0.846 887 014 4;
  • 50) 0.846 887 014 4 × 2 = 1 + 0.693 774 028 8;
  • 51) 0.693 774 028 8 × 2 = 1 + 0.387 548 057 6;
  • 52) 0.387 548 057 6 × 2 = 0 + 0.775 096 115 2;
  • 53) 0.775 096 115 2 × 2 = 1 + 0.550 192 230 4;
  • 54) 0.550 192 230 4 × 2 = 1 + 0.100 384 460 8;
  • 55) 0.100 384 460 8 × 2 = 0 + 0.200 768 921 6;
  • 56) 0.200 768 921 6 × 2 = 0 + 0.401 537 843 2;
  • 57) 0.401 537 843 2 × 2 = 0 + 0.803 075 686 4;
  • 58) 0.803 075 686 4 × 2 = 1 + 0.606 151 372 8;
  • 59) 0.606 151 372 8 × 2 = 1 + 0.212 302 745 6;
  • 60) 0.212 302 745 6 × 2 = 0 + 0.424 605 491 2;
  • 61) 0.424 605 491 2 × 2 = 0 + 0.849 210 982 4;
  • 62) 0.849 210 982 4 × 2 = 1 + 0.698 421 964 8;
  • 63) 0.698 421 964 8 × 2 = 1 + 0.396 843 929 6;
  • 64) 0.396 843 929 6 × 2 = 0 + 0.793 687 859 2;
  • 65) 0.793 687 859 2 × 2 = 1 + 0.587 375 718 4;
  • 66) 0.587 375 718 4 × 2 = 1 + 0.174 751 436 8;
  • 67) 0.174 751 436 8 × 2 = 0 + 0.349 502 873 6;
  • 68) 0.349 502 873 6 × 2 = 0 + 0.699 005 747 2;

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).


4. 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:


0.000 020 871 2(10) =

0.0000 0000 0000 0001 0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100(2)

5. Positive number before normalization:

0.000 020 871 2(10) =

0.0000 0000 0000 0001 0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 020 871 2(10) =

0.0000 0000 0000 0001 0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100(2) =

0.0000 0000 0000 0001 0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100(2) × 20 =

1.0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100(2) × 2-16


7. 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 0 (a positive number)

Exponent (unadjusted): -16

Mantissa (not normalized):
1.0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-16 + 2(11-1) - 1 =

(-16 + 1 023)(10) =

1 007(10)


9. 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;
  • 1 007 ÷ 2 = 503 + 1;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1007(10) =

011 1110 1111(2)


11. 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. 0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100 =

0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 1111

Mantissa (52 bits) =
0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100


Decimal number 0.000 020 871 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1111 - 0101 1110 0010 1001 0001 1111 0001 0101 1110 1100 0110 0110 1100

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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