3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 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: 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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 619 2;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 619 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 238 4;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 238 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 476 8;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 476 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 953 6;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 953 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 907 2;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 907 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 814 4;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 814 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 628 8;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 628 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 327 257 6;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 327 257 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 654 515 2;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 654 515 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 030 4;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 030 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 618 060 8;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 618 060 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 236 121 6;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 236 121 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 472 243 2;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 472 243 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 944 486 4;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 944 486 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 888 972 8;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 888 972 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 777 945 6;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 777 945 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 555 891 2;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 555 891 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 327 111 782 4;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 327 111 782 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 654 223 564 8;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 654 223 564 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 447 129 6;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 308 447 129 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 616 894 259 2;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 616 894 259 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 233 788 518 4;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 233 788 518 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 467 577 036 8;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 467 577 036 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 935 154 073 6;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 935 154 073 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 870 308 147 2;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 870 308 147 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 740 616 294 4;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 740 616 294 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 481 232 588 8;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 481 232 588 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 962 465 177 6;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 962 465 177 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 924 930 355 2;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 924 930 355 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 307 849 860 710 4;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 307 849 860 710 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 615 699 721 420 8;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 615 699 721 420 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 231 399 442 841 6;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 231 399 442 841 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 462 798 885 683 2;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 462 798 885 683 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 925 597 771 366 4;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 925 597 771 366 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 851 195 542 732 8;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 851 195 542 732 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 702 391 085 465 6;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 702 391 085 465 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 404 782 170 931 2;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 663 404 782 170 931 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 809 564 341 862 4;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 326 809 564 341 862 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 619 128 683 724 8;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 653 619 128 683 724 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 307 238 257 367 449 6;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 307 238 257 367 449 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 614 476 514 734 899 2;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 614 476 514 734 899 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 228 953 029 469 798 4;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 228 953 029 469 798 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 457 906 058 939 596 8;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 457 906 058 939 596 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 915 812 117 879 193 6;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 915 812 117 879 193 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 831 624 235 758 387 2;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 831 624 235 758 387 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 663 248 471 516 774 4;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 663 248 471 516 774 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 663 326 496 943 033 548 8;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 663 326 496 943 033 548 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 326 652 993 886 067 097 6;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 326 652 993 886 067 097 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 653 305 987 772 134 195 2;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 653 305 987 772 134 195 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 306 611 975 544 268 390 4;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 306 611 975 544 268 390 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 613 223 951 088 536 780 8;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 613 223 951 088 536 780 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 226 447 902 177 073 561 6;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 226 447 902 177 073 561 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 452 895 804 354 147 123 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 6(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 6(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 6(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 309 6 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