3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 2;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 296 4;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 296 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 592 8;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 592 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 185 6;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 185 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 371 2;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 371 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 742 4;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 742 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 484 8;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 484 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 969 6;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 969 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 939 2;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 939 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 878 4;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 878 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 756 8;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 756 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 295 513 6;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 295 513 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 591 027 2;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 591 027 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 182 054 4;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 182 054 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 364 108 8;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 364 108 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 728 217 6;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 728 217 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 456 435 2;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 456 435 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 912 870 4;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 912 870 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 825 740 8;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 825 740 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 651 481 6;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 651 481 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 302 963 2;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 302 963 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 294 605 926 4;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 294 605 926 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 589 211 852 8;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 589 211 852 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 178 423 705 6;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 178 423 705 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 356 847 411 2;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 356 847 411 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 713 694 822 4;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 713 694 822 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 427 389 644 8;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 427 389 644 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 854 779 289 6;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 854 779 289 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 709 558 579 2;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 709 558 579 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 419 117 158 4;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 419 117 158 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 646 838 234 316 8;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 646 838 234 316 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 293 676 468 633 6;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 293 676 468 633 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 587 352 937 267 2;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 587 352 937 267 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 174 705 874 534 4;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 174 705 874 534 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 349 411 749 068 8;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 349 411 749 068 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 698 823 498 137 6;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 698 823 498 137 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 397 646 996 275 2;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 397 646 996 275 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 795 293 992 550 4;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 795 293 992 550 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 590 587 985 100 8;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 590 587 985 100 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 181 175 970 201 6;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 181 175 970 201 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 646 362 351 940 403 2;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 646 362 351 940 403 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 292 724 703 880 806 4;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 292 724 703 880 806 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 585 449 407 761 612 8;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 585 449 407 761 612 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 170 898 815 523 225 6;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 170 898 815 523 225 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 341 797 631 046 451 2;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 341 797 631 046 451 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 683 595 262 092 902 4;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 683 595 262 092 902 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 367 190 524 185 804 8;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 367 190 524 185 804 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 734 381 048 371 609 6;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 734 381 048 371 609 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 661 468 762 096 743 219 2;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 661 468 762 096 743 219 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 322 937 524 193 486 438 4;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 322 937 524 193 486 438 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 645 875 048 386 972 876 8;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 645 875 048 386 972 876 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 291 750 096 773 945 753 6;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 291 750 096 773 945 753 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 583 500 193 547 891 507 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1(10) =

11.0101 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 1 positions to the left, so that only one non zero digit remains to the left of it:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1(10) =

11.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =

11.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =

1.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 10(2) × 21


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):
1.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 10


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 10 =

1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


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 0000

Mantissa (52 bits) =
1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


Decimal number 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010

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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