20.333 333 333 333 333 333 55 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 20.333 333 333 333 333 333 55(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 55(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 55.

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 55 × 2 = 0 + 0.666 666 666 666 666 667 1;
  • 2) 0.666 666 666 666 666 667 1 × 2 = 1 + 0.333 333 333 333 333 334 2;
  • 3) 0.333 333 333 333 333 334 2 × 2 = 0 + 0.666 666 666 666 666 668 4;
  • 4) 0.666 666 666 666 666 668 4 × 2 = 1 + 0.333 333 333 333 333 336 8;
  • 5) 0.333 333 333 333 333 336 8 × 2 = 0 + 0.666 666 666 666 666 673 6;
  • 6) 0.666 666 666 666 666 673 6 × 2 = 1 + 0.333 333 333 333 333 347 2;
  • 7) 0.333 333 333 333 333 347 2 × 2 = 0 + 0.666 666 666 666 666 694 4;
  • 8) 0.666 666 666 666 666 694 4 × 2 = 1 + 0.333 333 333 333 333 388 8;
  • 9) 0.333 333 333 333 333 388 8 × 2 = 0 + 0.666 666 666 666 666 777 6;
  • 10) 0.666 666 666 666 666 777 6 × 2 = 1 + 0.333 333 333 333 333 555 2;
  • 11) 0.333 333 333 333 333 555 2 × 2 = 0 + 0.666 666 666 666 667 110 4;
  • 12) 0.666 666 666 666 667 110 4 × 2 = 1 + 0.333 333 333 333 334 220 8;
  • 13) 0.333 333 333 333 334 220 8 × 2 = 0 + 0.666 666 666 666 668 441 6;
  • 14) 0.666 666 666 666 668 441 6 × 2 = 1 + 0.333 333 333 333 336 883 2;
  • 15) 0.333 333 333 333 336 883 2 × 2 = 0 + 0.666 666 666 666 673 766 4;
  • 16) 0.666 666 666 666 673 766 4 × 2 = 1 + 0.333 333 333 333 347 532 8;
  • 17) 0.333 333 333 333 347 532 8 × 2 = 0 + 0.666 666 666 666 695 065 6;
  • 18) 0.666 666 666 666 695 065 6 × 2 = 1 + 0.333 333 333 333 390 131 2;
  • 19) 0.333 333 333 333 390 131 2 × 2 = 0 + 0.666 666 666 666 780 262 4;
  • 20) 0.666 666 666 666 780 262 4 × 2 = 1 + 0.333 333 333 333 560 524 8;
  • 21) 0.333 333 333 333 560 524 8 × 2 = 0 + 0.666 666 666 667 121 049 6;
  • 22) 0.666 666 666 667 121 049 6 × 2 = 1 + 0.333 333 333 334 242 099 2;
  • 23) 0.333 333 333 334 242 099 2 × 2 = 0 + 0.666 666 666 668 484 198 4;
  • 24) 0.666 666 666 668 484 198 4 × 2 = 1 + 0.333 333 333 336 968 396 8;
  • 25) 0.333 333 333 336 968 396 8 × 2 = 0 + 0.666 666 666 673 936 793 6;
  • 26) 0.666 666 666 673 936 793 6 × 2 = 1 + 0.333 333 333 347 873 587 2;
  • 27) 0.333 333 333 347 873 587 2 × 2 = 0 + 0.666 666 666 695 747 174 4;
  • 28) 0.666 666 666 695 747 174 4 × 2 = 1 + 0.333 333 333 391 494 348 8;
  • 29) 0.333 333 333 391 494 348 8 × 2 = 0 + 0.666 666 666 782 988 697 6;
  • 30) 0.666 666 666 782 988 697 6 × 2 = 1 + 0.333 333 333 565 977 395 2;
  • 31) 0.333 333 333 565 977 395 2 × 2 = 0 + 0.666 666 667 131 954 790 4;
  • 32) 0.666 666 667 131 954 790 4 × 2 = 1 + 0.333 333 334 263 909 580 8;
  • 33) 0.333 333 334 263 909 580 8 × 2 = 0 + 0.666 666 668 527 819 161 6;
  • 34) 0.666 666 668 527 819 161 6 × 2 = 1 + 0.333 333 337 055 638 323 2;
  • 35) 0.333 333 337 055 638 323 2 × 2 = 0 + 0.666 666 674 111 276 646 4;
  • 36) 0.666 666 674 111 276 646 4 × 2 = 1 + 0.333 333 348 222 553 292 8;
  • 37) 0.333 333 348 222 553 292 8 × 2 = 0 + 0.666 666 696 445 106 585 6;
  • 38) 0.666 666 696 445 106 585 6 × 2 = 1 + 0.333 333 392 890 213 171 2;
  • 39) 0.333 333 392 890 213 171 2 × 2 = 0 + 0.666 666 785 780 426 342 4;
  • 40) 0.666 666 785 780 426 342 4 × 2 = 1 + 0.333 333 571 560 852 684 8;
  • 41) 0.333 333 571 560 852 684 8 × 2 = 0 + 0.666 667 143 121 705 369 6;
  • 42) 0.666 667 143 121 705 369 6 × 2 = 1 + 0.333 334 286 243 410 739 2;
  • 43) 0.333 334 286 243 410 739 2 × 2 = 0 + 0.666 668 572 486 821 478 4;
  • 44) 0.666 668 572 486 821 478 4 × 2 = 1 + 0.333 337 144 973 642 956 8;
  • 45) 0.333 337 144 973 642 956 8 × 2 = 0 + 0.666 674 289 947 285 913 6;
  • 46) 0.666 674 289 947 285 913 6 × 2 = 1 + 0.333 348 579 894 571 827 2;
  • 47) 0.333 348 579 894 571 827 2 × 2 = 0 + 0.666 697 159 789 143 654 4;
  • 48) 0.666 697 159 789 143 654 4 × 2 = 1 + 0.333 394 319 578 287 308 8;
  • 49) 0.333 394 319 578 287 308 8 × 2 = 0 + 0.666 788 639 156 574 617 6;
  • 50) 0.666 788 639 156 574 617 6 × 2 = 1 + 0.333 577 278 313 149 235 2;
  • 51) 0.333 577 278 313 149 235 2 × 2 = 0 + 0.667 154 556 626 298 470 4;
  • 52) 0.667 154 556 626 298 470 4 × 2 = 1 + 0.334 309 113 252 596 940 8;
  • 53) 0.334 309 113 252 596 940 8 × 2 = 0 + 0.668 618 226 505 193 881 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:


0.333 333 333 333 333 333 55(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 55(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 55(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 55 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

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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