20.333 333 333 333 333 333 97 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 20.333 333 333 333 333 333 97(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
20.333 333 333 333 333 333 97(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: 20.
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;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

20(10) =

1 0100(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 333 333 97.

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) 0.333 333 333 333 333 333 97 × 2 = 0 + 0.666 666 666 666 666 667 94;
  • 2) 0.666 666 666 666 666 667 94 × 2 = 1 + 0.333 333 333 333 333 335 88;
  • 3) 0.333 333 333 333 333 335 88 × 2 = 0 + 0.666 666 666 666 666 671 76;
  • 4) 0.666 666 666 666 666 671 76 × 2 = 1 + 0.333 333 333 333 333 343 52;
  • 5) 0.333 333 333 333 333 343 52 × 2 = 0 + 0.666 666 666 666 666 687 04;
  • 6) 0.666 666 666 666 666 687 04 × 2 = 1 + 0.333 333 333 333 333 374 08;
  • 7) 0.333 333 333 333 333 374 08 × 2 = 0 + 0.666 666 666 666 666 748 16;
  • 8) 0.666 666 666 666 666 748 16 × 2 = 1 + 0.333 333 333 333 333 496 32;
  • 9) 0.333 333 333 333 333 496 32 × 2 = 0 + 0.666 666 666 666 666 992 64;
  • 10) 0.666 666 666 666 666 992 64 × 2 = 1 + 0.333 333 333 333 333 985 28;
  • 11) 0.333 333 333 333 333 985 28 × 2 = 0 + 0.666 666 666 666 667 970 56;
  • 12) 0.666 666 666 666 667 970 56 × 2 = 1 + 0.333 333 333 333 335 941 12;
  • 13) 0.333 333 333 333 335 941 12 × 2 = 0 + 0.666 666 666 666 671 882 24;
  • 14) 0.666 666 666 666 671 882 24 × 2 = 1 + 0.333 333 333 333 343 764 48;
  • 15) 0.333 333 333 333 343 764 48 × 2 = 0 + 0.666 666 666 666 687 528 96;
  • 16) 0.666 666 666 666 687 528 96 × 2 = 1 + 0.333 333 333 333 375 057 92;
  • 17) 0.333 333 333 333 375 057 92 × 2 = 0 + 0.666 666 666 666 750 115 84;
  • 18) 0.666 666 666 666 750 115 84 × 2 = 1 + 0.333 333 333 333 500 231 68;
  • 19) 0.333 333 333 333 500 231 68 × 2 = 0 + 0.666 666 666 667 000 463 36;
  • 20) 0.666 666 666 667 000 463 36 × 2 = 1 + 0.333 333 333 334 000 926 72;
  • 21) 0.333 333 333 334 000 926 72 × 2 = 0 + 0.666 666 666 668 001 853 44;
  • 22) 0.666 666 666 668 001 853 44 × 2 = 1 + 0.333 333 333 336 003 706 88;
  • 23) 0.333 333 333 336 003 706 88 × 2 = 0 + 0.666 666 666 672 007 413 76;
  • 24) 0.666 666 666 672 007 413 76 × 2 = 1 + 0.333 333 333 344 014 827 52;
  • 25) 0.333 333 333 344 014 827 52 × 2 = 0 + 0.666 666 666 688 029 655 04;
  • 26) 0.666 666 666 688 029 655 04 × 2 = 1 + 0.333 333 333 376 059 310 08;
  • 27) 0.333 333 333 376 059 310 08 × 2 = 0 + 0.666 666 666 752 118 620 16;
  • 28) 0.666 666 666 752 118 620 16 × 2 = 1 + 0.333 333 333 504 237 240 32;
  • 29) 0.333 333 333 504 237 240 32 × 2 = 0 + 0.666 666 667 008 474 480 64;
  • 30) 0.666 666 667 008 474 480 64 × 2 = 1 + 0.333 333 334 016 948 961 28;
  • 31) 0.333 333 334 016 948 961 28 × 2 = 0 + 0.666 666 668 033 897 922 56;
  • 32) 0.666 666 668 033 897 922 56 × 2 = 1 + 0.333 333 336 067 795 845 12;
  • 33) 0.333 333 336 067 795 845 12 × 2 = 0 + 0.666 666 672 135 591 690 24;
  • 34) 0.666 666 672 135 591 690 24 × 2 = 1 + 0.333 333 344 271 183 380 48;
  • 35) 0.333 333 344 271 183 380 48 × 2 = 0 + 0.666 666 688 542 366 760 96;
  • 36) 0.666 666 688 542 366 760 96 × 2 = 1 + 0.333 333 377 084 733 521 92;
  • 37) 0.333 333 377 084 733 521 92 × 2 = 0 + 0.666 666 754 169 467 043 84;
  • 38) 0.666 666 754 169 467 043 84 × 2 = 1 + 0.333 333 508 338 934 087 68;
  • 39) 0.333 333 508 338 934 087 68 × 2 = 0 + 0.666 667 016 677 868 175 36;
  • 40) 0.666 667 016 677 868 175 36 × 2 = 1 + 0.333 334 033 355 736 350 72;
  • 41) 0.333 334 033 355 736 350 72 × 2 = 0 + 0.666 668 066 711 472 701 44;
  • 42) 0.666 668 066 711 472 701 44 × 2 = 1 + 0.333 336 133 422 945 402 88;
  • 43) 0.333 336 133 422 945 402 88 × 2 = 0 + 0.666 672 266 845 890 805 76;
  • 44) 0.666 672 266 845 890 805 76 × 2 = 1 + 0.333 344 533 691 781 611 52;
  • 45) 0.333 344 533 691 781 611 52 × 2 = 0 + 0.666 689 067 383 563 223 04;
  • 46) 0.666 689 067 383 563 223 04 × 2 = 1 + 0.333 378 134 767 126 446 08;
  • 47) 0.333 378 134 767 126 446 08 × 2 = 0 + 0.666 756 269 534 252 892 16;
  • 48) 0.666 756 269 534 252 892 16 × 2 = 1 + 0.333 512 539 068 505 784 32;
  • 49) 0.333 512 539 068 505 784 32 × 2 = 0 + 0.667 025 078 137 011 568 64;
  • 50) 0.667 025 078 137 011 568 64 × 2 = 1 + 0.334 050 156 274 023 137 28;
  • 51) 0.334 050 156 274 023 137 28 × 2 = 0 + 0.668 100 312 548 046 274 56;
  • 52) 0.668 100 312 548 046 274 56 × 2 = 1 + 0.336 200 625 096 092 549 12;
  • 53) 0.336 200 625 096 092 549 12 × 2 = 0 + 0.672 401 250 192 185 098 24;

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:


0.333 333 333 333 333 333 97(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2)

5. Positive number before normalization:

20.333 333 333 333 333 333 97(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


20.333 333 333 333 333 333 97(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =

1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 24


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): 4

Mantissa (not normalized):
1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1027(10) =

100 0000 0011(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0 1010 =

0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


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) =
100 0000 0011

Mantissa (52 bits) =
0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


Decimal number 20.333 333 333 333 333 333 97 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

Share

» Convert decimal 20.333 333 333 333 333 334 28 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 04, 2026 [April]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: April - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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