0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6(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
0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 687 2;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 687 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 374 4;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 374 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 748 8;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 748 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 497 6;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 497 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 995 2;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 995 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 990 4;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 990 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 980 8;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 980 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 961 6;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 961 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 671 923 2;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 671 923 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 846 4;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 846 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 687 692 8;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 687 692 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 375 385 6;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 375 385 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 750 771 2;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 750 771 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 501 542 4;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 501 542 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 003 084 8;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 003 084 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 006 169 6;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 006 169 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 012 339 2;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 012 339 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 336 024 678 4;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 336 024 678 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 672 049 356 8;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 672 049 356 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 344 098 713 6;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 344 098 713 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 688 197 427 2;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 688 197 427 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 376 394 854 4;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 376 394 854 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 752 789 708 8;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 752 789 708 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 505 579 417 6;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 505 579 417 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 011 158 835 2;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 011 158 835 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 022 317 670 4;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 022 317 670 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 044 635 340 8;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 044 635 340 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 336 089 270 681 6;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 336 089 270 681 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 672 178 541 363 2;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 672 178 541 363 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 344 357 082 726 4;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 344 357 082 726 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 688 714 165 452 8;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 688 714 165 452 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 377 428 330 905 6;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 377 428 330 905 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 754 856 661 811 2;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 754 856 661 811 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 509 713 323 622 4;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 509 713 323 622 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 019 426 647 244 8;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 019 426 647 244 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 038 853 294 489 6;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 038 853 294 489 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 077 706 588 979 2;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 077 706 588 979 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 336 155 413 177 958 4;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 336 155 413 177 958 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 672 310 826 355 916 8;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 672 310 826 355 916 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 344 621 652 711 833 6;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 344 621 652 711 833 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 689 243 305 423 667 2;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 689 243 305 423 667 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 378 486 610 847 334 4;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 378 486 610 847 334 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 756 973 221 694 668 8;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 756 973 221 694 668 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 513 946 443 389 337 6;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 513 946 443 389 337 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 667 027 892 886 778 675 2;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 667 027 892 886 778 675 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 334 055 785 773 557 350 4;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 334 055 785 773 557 350 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 668 111 571 547 114 700 8;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 668 111 571 547 114 700 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 336 223 143 094 229 401 6;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 336 223 143 094 229 401 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 672 446 286 188 458 803 2;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 672 446 286 188 458 803 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 344 892 572 376 917 606 4;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 344 892 572 376 917 606 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 689 785 144 753 835 212 8;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 689 785 144 753 835 212 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 379 570 289 507 670 425 6;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 379 570 289 507 670 425 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 759 140 579 015 340 851 2;
  • 54) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 759 140 579 015 340 851 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 518 281 158 030 681 702 4;

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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2)

5. Positive number before normalization:

0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101(2) × 2-2


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): -2

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1021(10) =

011 1111 1101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 =

0101 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) =
011 1111 1101

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


Decimal number 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 343 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 0101 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