3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4(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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 596 8;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 596 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 193 6;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 193 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 387 2;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 387 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 774 4;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 774 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 548 8;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 548 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 097 6;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 097 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 195 2;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 195 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 390 4;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 390 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 780 8;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 780 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 561 6;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 561 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 595 123 2;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 595 123 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 190 246 4;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 190 246 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 380 492 8;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 380 492 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 760 985 6;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 760 985 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 521 971 2;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 521 971 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 043 942 4;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 043 942 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 087 884 8;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 087 884 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 175 769 6;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 175 769 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 351 539 2;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 351 539 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 296 703 078 4;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 296 703 078 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 593 406 156 8;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 593 406 156 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 186 812 313 6;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 186 812 313 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 373 624 627 2;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 373 624 627 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 747 249 254 4;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 747 249 254 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 494 498 508 8;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 494 498 508 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 988 997 017 6;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 988 997 017 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 977 994 035 2;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 977 994 035 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 955 988 070 4;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 955 988 070 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 911 976 140 8;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 911 976 140 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 295 823 952 281 6;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 295 823 952 281 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 591 647 904 563 2;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 591 647 904 563 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 183 295 809 126 4;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 183 295 809 126 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 366 591 618 252 8;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 366 591 618 252 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 733 183 236 505 6;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 733 183 236 505 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 466 366 473 011 2;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 466 366 473 011 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 932 732 946 022 4;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 932 732 946 022 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 865 465 892 044 8;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 865 465 892 044 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 730 931 784 089 6;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 730 931 784 089 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 461 863 568 179 2;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 461 863 568 179 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 294 923 727 136 358 4;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 294 923 727 136 358 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 589 847 454 272 716 8;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 589 847 454 272 716 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 179 694 908 545 433 6;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 179 694 908 545 433 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 359 389 817 090 867 2;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 359 389 817 090 867 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 718 779 634 181 734 4;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 718 779 634 181 734 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 437 559 268 363 468 8;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 437 559 268 363 468 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 875 118 536 726 937 6;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 875 118 536 726 937 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 661 750 237 073 453 875 2;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 661 750 237 073 453 875 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 323 500 474 146 907 750 4;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 323 500 474 146 907 750 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 647 000 948 293 815 500 8;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 647 000 948 293 815 500 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 294 001 896 587 631 001 6;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 294 001 896 587 631 001 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 588 003 793 175 262 003 2;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 588 003 793 175 262 003 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 176 007 586 350 524 006 4;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 176 007 586 350 524 006 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 352 015 172 701 048 012 8;

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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 298 4 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