0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 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
0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 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: 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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 1 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 696 2;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 696 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 392 4;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 392 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 784 8;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 784 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 569 6;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 569 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 139 2;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 139 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 278 4;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 278 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 556 8;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 556 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 113 6;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 113 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 227 2;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 227 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 454 4;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 454 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 696 908 8;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 696 908 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 393 817 6;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 393 817 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 787 635 2;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 787 635 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 575 270 4;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 575 270 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 150 540 8;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 150 540 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 301 081 6;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 301 081 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 602 163 2;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 602 163 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 204 326 4;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 204 326 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 408 652 8;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 408 652 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 817 305 6;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 817 305 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 697 634 611 2;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 697 634 611 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 395 269 222 4;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 395 269 222 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 790 538 444 8;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 790 538 444 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 581 076 889 6;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 581 076 889 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 162 153 779 2;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 162 153 779 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 324 307 558 4;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 324 307 558 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 648 615 116 8;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 648 615 116 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 297 230 233 6;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 297 230 233 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 594 460 467 2;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 594 460 467 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 349 188 920 934 4;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 349 188 920 934 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 698 377 841 868 8;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 698 377 841 868 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 396 755 683 737 6;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 396 755 683 737 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 793 511 367 475 2;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 793 511 367 475 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 587 022 734 950 4;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 587 022 734 950 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 174 045 469 900 8;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 174 045 469 900 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 348 090 939 801 6;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 334 348 090 939 801 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 696 181 879 603 2;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 668 696 181 879 603 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 392 363 759 206 4;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 337 392 363 759 206 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 784 727 518 412 8;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 674 784 727 518 412 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 349 569 455 036 825 6;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 349 569 455 036 825 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 699 138 910 073 651 2;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 699 138 910 073 651 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 398 277 820 147 302 4;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 398 277 820 147 302 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 796 555 640 294 604 8;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 796 555 640 294 604 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 593 111 280 589 209 6;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 593 111 280 589 209 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 667 186 222 561 178 419 2;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 667 186 222 561 178 419 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 334 372 445 122 356 838 4;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 334 372 445 122 356 838 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 668 744 890 244 713 676 8;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 668 744 890 244 713 676 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 337 489 780 489 427 353 6;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 337 489 780 489 427 353 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 674 979 560 978 854 707 2;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 674 979 560 978 854 707 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 349 959 121 957 709 414 4;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 349 959 121 957 709 414 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 699 918 243 915 418 828 8;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 699 918 243 915 418 828 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 399 836 487 830 837 657 6;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 399 836 487 830 837 657 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 799 672 975 661 675 315 2;
  • 54) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 799 672 975 661 675 315 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 599 345 951 323 350 630 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 1(10) =

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

5. Positive number before normalization:

0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 1(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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 1(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 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 348 1 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