0.520 000 000 002 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.520 000 000 002 3(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
0.520 000 000 002 3(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: 0.520 000 000 002 3.

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) 0.520 000 000 002 3 × 2 = 1 + 0.040 000 000 004 6;
  • 2) 0.040 000 000 004 6 × 2 = 0 + 0.080 000 000 009 2;
  • 3) 0.080 000 000 009 2 × 2 = 0 + 0.160 000 000 018 4;
  • 4) 0.160 000 000 018 4 × 2 = 0 + 0.320 000 000 036 8;
  • 5) 0.320 000 000 036 8 × 2 = 0 + 0.640 000 000 073 6;
  • 6) 0.640 000 000 073 6 × 2 = 1 + 0.280 000 000 147 2;
  • 7) 0.280 000 000 147 2 × 2 = 0 + 0.560 000 000 294 4;
  • 8) 0.560 000 000 294 4 × 2 = 1 + 0.120 000 000 588 8;
  • 9) 0.120 000 000 588 8 × 2 = 0 + 0.240 000 001 177 6;
  • 10) 0.240 000 001 177 6 × 2 = 0 + 0.480 000 002 355 2;
  • 11) 0.480 000 002 355 2 × 2 = 0 + 0.960 000 004 710 4;
  • 12) 0.960 000 004 710 4 × 2 = 1 + 0.920 000 009 420 8;
  • 13) 0.920 000 009 420 8 × 2 = 1 + 0.840 000 018 841 6;
  • 14) 0.840 000 018 841 6 × 2 = 1 + 0.680 000 037 683 2;
  • 15) 0.680 000 037 683 2 × 2 = 1 + 0.360 000 075 366 4;
  • 16) 0.360 000 075 366 4 × 2 = 0 + 0.720 000 150 732 8;
  • 17) 0.720 000 150 732 8 × 2 = 1 + 0.440 000 301 465 6;
  • 18) 0.440 000 301 465 6 × 2 = 0 + 0.880 000 602 931 2;
  • 19) 0.880 000 602 931 2 × 2 = 1 + 0.760 001 205 862 4;
  • 20) 0.760 001 205 862 4 × 2 = 1 + 0.520 002 411 724 8;
  • 21) 0.520 002 411 724 8 × 2 = 1 + 0.040 004 823 449 6;
  • 22) 0.040 004 823 449 6 × 2 = 0 + 0.080 009 646 899 2;
  • 23) 0.080 009 646 899 2 × 2 = 0 + 0.160 019 293 798 4;
  • 24) 0.160 019 293 798 4 × 2 = 0 + 0.320 038 587 596 8;
  • 25) 0.320 038 587 596 8 × 2 = 0 + 0.640 077 175 193 6;
  • 26) 0.640 077 175 193 6 × 2 = 1 + 0.280 154 350 387 2;
  • 27) 0.280 154 350 387 2 × 2 = 0 + 0.560 308 700 774 4;
  • 28) 0.560 308 700 774 4 × 2 = 1 + 0.120 617 401 548 8;
  • 29) 0.120 617 401 548 8 × 2 = 0 + 0.241 234 803 097 6;
  • 30) 0.241 234 803 097 6 × 2 = 0 + 0.482 469 606 195 2;
  • 31) 0.482 469 606 195 2 × 2 = 0 + 0.964 939 212 390 4;
  • 32) 0.964 939 212 390 4 × 2 = 1 + 0.929 878 424 780 8;
  • 33) 0.929 878 424 780 8 × 2 = 1 + 0.859 756 849 561 6;
  • 34) 0.859 756 849 561 6 × 2 = 1 + 0.719 513 699 123 2;
  • 35) 0.719 513 699 123 2 × 2 = 1 + 0.439 027 398 246 4;
  • 36) 0.439 027 398 246 4 × 2 = 0 + 0.878 054 796 492 8;
  • 37) 0.878 054 796 492 8 × 2 = 1 + 0.756 109 592 985 6;
  • 38) 0.756 109 592 985 6 × 2 = 1 + 0.512 219 185 971 2;
  • 39) 0.512 219 185 971 2 × 2 = 1 + 0.024 438 371 942 4;
  • 40) 0.024 438 371 942 4 × 2 = 0 + 0.048 876 743 884 8;
  • 41) 0.048 876 743 884 8 × 2 = 0 + 0.097 753 487 769 6;
  • 42) 0.097 753 487 769 6 × 2 = 0 + 0.195 506 975 539 2;
  • 43) 0.195 506 975 539 2 × 2 = 0 + 0.391 013 951 078 4;
  • 44) 0.391 013 951 078 4 × 2 = 0 + 0.782 027 902 156 8;
  • 45) 0.782 027 902 156 8 × 2 = 1 + 0.564 055 804 313 6;
  • 46) 0.564 055 804 313 6 × 2 = 1 + 0.128 111 608 627 2;
  • 47) 0.128 111 608 627 2 × 2 = 0 + 0.256 223 217 254 4;
  • 48) 0.256 223 217 254 4 × 2 = 0 + 0.512 446 434 508 8;
  • 49) 0.512 446 434 508 8 × 2 = 1 + 0.024 892 869 017 6;
  • 50) 0.024 892 869 017 6 × 2 = 0 + 0.049 785 738 035 2;
  • 51) 0.049 785 738 035 2 × 2 = 0 + 0.099 571 476 070 4;
  • 52) 0.099 571 476 070 4 × 2 = 0 + 0.199 142 952 140 8;
  • 53) 0.199 142 952 140 8 × 2 = 0 + 0.398 285 904 281 6;

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:


0.520 000 000 002 3(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1110 0000 1100 1000 0(2)

5. Positive number before normalization:

0.520 000 000 002 3(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1110 0000 1100 1000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.520 000 000 002 3(10) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1110 0000 1100 1000 0(2) =

0.1000 0101 0001 1110 1011 1000 0101 0001 1110 1110 0000 1100 1000 0(2) × 20 =

1.0000 1010 0011 1101 0111 0000 1010 0011 1101 1100 0001 1001 0000(2) × 2-1


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): -1

Mantissa (not normalized):
1.0000 1010 0011 1101 0111 0000 1010 0011 1101 1100 0001 1001 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 0000 1010 0011 1101 0111 0000 1010 0011 1101 1100 0001 1001 0000 =

0000 1010 0011 1101 0111 0000 1010 0011 1101 1100 0001 1001 0000


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 1111 1110

Mantissa (52 bits) =
0000 1010 0011 1101 0111 0000 1010 0011 1101 1100 0001 1001 0000


Decimal number 0.520 000 000 002 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0000 1010 0011 1101 0111 0000 1010 0011 1101 1100 0001 1001 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