3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8(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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 617 6;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 617 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 235 2;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 235 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 470 4;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 470 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 940 8;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 940 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 881 6;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 881 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 763 2;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 763 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 526 4;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 526 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 327 052 8;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 327 052 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 654 105 6;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 654 105 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 211 2;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 211 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 616 422 4;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 616 422 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 232 844 8;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 232 844 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 465 689 6;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 465 689 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 931 379 2;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 931 379 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 862 758 4;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 862 758 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 725 516 8;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 725 516 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 451 033 6;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 451 033 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 902 067 2;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 902 067 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 804 134 4;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 804 134 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 307 608 268 8;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 307 608 268 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 615 216 537 6;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 615 216 537 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 230 433 075 2;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 230 433 075 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 460 866 150 4;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 460 866 150 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 921 732 300 8;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 921 732 300 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 843 464 601 6;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 843 464 601 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 686 929 203 2;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 686 929 203 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 373 858 406 4;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 373 858 406 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 747 716 812 8;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 747 716 812 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 495 433 625 6;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 495 433 625 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 306 990 867 251 2;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 306 990 867 251 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 613 981 734 502 4;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 613 981 734 502 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 227 963 469 004 8;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 227 963 469 004 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 455 926 938 009 6;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 455 926 938 009 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 911 853 876 019 2;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 911 853 876 019 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 823 707 752 038 4;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 823 707 752 038 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 647 415 504 076 8;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 647 415 504 076 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 294 831 008 153 6;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 294 831 008 153 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 589 662 016 307 2;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 589 662 016 307 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 179 324 032 614 4;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 179 324 032 614 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 306 358 648 065 228 8;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 306 358 648 065 228 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 612 717 296 130 457 6;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 612 717 296 130 457 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 225 434 592 260 915 2;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 225 434 592 260 915 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 450 869 184 521 830 4;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 450 869 184 521 830 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 901 738 369 043 660 8;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 901 738 369 043 660 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 803 476 738 087 321 6;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 803 476 738 087 321 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 606 953 476 174 643 2;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 606 953 476 174 643 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 663 213 906 952 349 286 4;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 663 213 906 952 349 286 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 326 427 813 904 698 572 8;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 326 427 813 904 698 572 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 652 855 627 809 397 145 6;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 652 855 627 809 397 145 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 305 711 255 618 794 291 2;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 305 711 255 618 794 291 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 611 422 511 237 588 582 4;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 611 422 511 237 588 582 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 222 845 022 475 177 164 8;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 222 845 022 475 177 164 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 445 690 044 950 354 329 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 8 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