20.333 333 333 333 340 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 20.333 333 333 333 340 15(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
20.333 333 333 333 340 15(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: 0.333 333 333 333 340 15.

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) 0.333 333 333 333 340 15 × 2 = 0 + 0.666 666 666 666 680 3;
  • 2) 0.666 666 666 666 680 3 × 2 = 1 + 0.333 333 333 333 360 6;
  • 3) 0.333 333 333 333 360 6 × 2 = 0 + 0.666 666 666 666 721 2;
  • 4) 0.666 666 666 666 721 2 × 2 = 1 + 0.333 333 333 333 442 4;
  • 5) 0.333 333 333 333 442 4 × 2 = 0 + 0.666 666 666 666 884 8;
  • 6) 0.666 666 666 666 884 8 × 2 = 1 + 0.333 333 333 333 769 6;
  • 7) 0.333 333 333 333 769 6 × 2 = 0 + 0.666 666 666 667 539 2;
  • 8) 0.666 666 666 667 539 2 × 2 = 1 + 0.333 333 333 335 078 4;
  • 9) 0.333 333 333 335 078 4 × 2 = 0 + 0.666 666 666 670 156 8;
  • 10) 0.666 666 666 670 156 8 × 2 = 1 + 0.333 333 333 340 313 6;
  • 11) 0.333 333 333 340 313 6 × 2 = 0 + 0.666 666 666 680 627 2;
  • 12) 0.666 666 666 680 627 2 × 2 = 1 + 0.333 333 333 361 254 4;
  • 13) 0.333 333 333 361 254 4 × 2 = 0 + 0.666 666 666 722 508 8;
  • 14) 0.666 666 666 722 508 8 × 2 = 1 + 0.333 333 333 445 017 6;
  • 15) 0.333 333 333 445 017 6 × 2 = 0 + 0.666 666 666 890 035 2;
  • 16) 0.666 666 666 890 035 2 × 2 = 1 + 0.333 333 333 780 070 4;
  • 17) 0.333 333 333 780 070 4 × 2 = 0 + 0.666 666 667 560 140 8;
  • 18) 0.666 666 667 560 140 8 × 2 = 1 + 0.333 333 335 120 281 6;
  • 19) 0.333 333 335 120 281 6 × 2 = 0 + 0.666 666 670 240 563 2;
  • 20) 0.666 666 670 240 563 2 × 2 = 1 + 0.333 333 340 481 126 4;
  • 21) 0.333 333 340 481 126 4 × 2 = 0 + 0.666 666 680 962 252 8;
  • 22) 0.666 666 680 962 252 8 × 2 = 1 + 0.333 333 361 924 505 6;
  • 23) 0.333 333 361 924 505 6 × 2 = 0 + 0.666 666 723 849 011 2;
  • 24) 0.666 666 723 849 011 2 × 2 = 1 + 0.333 333 447 698 022 4;
  • 25) 0.333 333 447 698 022 4 × 2 = 0 + 0.666 666 895 396 044 8;
  • 26) 0.666 666 895 396 044 8 × 2 = 1 + 0.333 333 790 792 089 6;
  • 27) 0.333 333 790 792 089 6 × 2 = 0 + 0.666 667 581 584 179 2;
  • 28) 0.666 667 581 584 179 2 × 2 = 1 + 0.333 335 163 168 358 4;
  • 29) 0.333 335 163 168 358 4 × 2 = 0 + 0.666 670 326 336 716 8;
  • 30) 0.666 670 326 336 716 8 × 2 = 1 + 0.333 340 652 673 433 6;
  • 31) 0.333 340 652 673 433 6 × 2 = 0 + 0.666 681 305 346 867 2;
  • 32) 0.666 681 305 346 867 2 × 2 = 1 + 0.333 362 610 693 734 4;
  • 33) 0.333 362 610 693 734 4 × 2 = 0 + 0.666 725 221 387 468 8;
  • 34) 0.666 725 221 387 468 8 × 2 = 1 + 0.333 450 442 774 937 6;
  • 35) 0.333 450 442 774 937 6 × 2 = 0 + 0.666 900 885 549 875 2;
  • 36) 0.666 900 885 549 875 2 × 2 = 1 + 0.333 801 771 099 750 4;
  • 37) 0.333 801 771 099 750 4 × 2 = 0 + 0.667 603 542 199 500 8;
  • 38) 0.667 603 542 199 500 8 × 2 = 1 + 0.335 207 084 399 001 6;
  • 39) 0.335 207 084 399 001 6 × 2 = 0 + 0.670 414 168 798 003 2;
  • 40) 0.670 414 168 798 003 2 × 2 = 1 + 0.340 828 337 596 006 4;
  • 41) 0.340 828 337 596 006 4 × 2 = 0 + 0.681 656 675 192 012 8;
  • 42) 0.681 656 675 192 012 8 × 2 = 1 + 0.363 313 350 384 025 6;
  • 43) 0.363 313 350 384 025 6 × 2 = 0 + 0.726 626 700 768 051 2;
  • 44) 0.726 626 700 768 051 2 × 2 = 1 + 0.453 253 401 536 102 4;
  • 45) 0.453 253 401 536 102 4 × 2 = 0 + 0.906 506 803 072 204 8;
  • 46) 0.906 506 803 072 204 8 × 2 = 1 + 0.813 013 606 144 409 6;
  • 47) 0.813 013 606 144 409 6 × 2 = 1 + 0.626 027 212 288 819 2;
  • 48) 0.626 027 212 288 819 2 × 2 = 1 + 0.252 054 424 577 638 4;
  • 49) 0.252 054 424 577 638 4 × 2 = 0 + 0.504 108 849 155 276 8;
  • 50) 0.504 108 849 155 276 8 × 2 = 1 + 0.008 217 698 310 553 6;
  • 51) 0.008 217 698 310 553 6 × 2 = 0 + 0.016 435 396 621 107 2;
  • 52) 0.016 435 396 621 107 2 × 2 = 0 + 0.032 870 793 242 214 4;
  • 53) 0.032 870 793 242 214 4 × 2 = 0 + 0.065 741 586 484 428 8;

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:


0.333 333 333 333 340 15(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0100 0(2)

5. Positive number before normalization:

20.333 333 333 333 340 15(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0100 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:


20.333 333 333 333 340 15(10) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0100 0(2) =

1 0100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0100 0(2) × 20 =

1.0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 0100 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 0111 0100 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 0111 0 1000 =

0100 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111


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 0111


Decimal number 20.333 333 333 333 340 15 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 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