3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2(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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 604 4;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 604 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 208 8;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 208 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 417 6;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 417 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 835 2;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 835 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 670 4;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 670 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 340 8;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 340 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 681 6;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 681 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 363 2;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 363 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 726 4;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 726 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 452 8;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 452 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 602 905 6;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 602 905 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 205 811 2;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 205 811 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 411 622 4;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 411 622 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 823 244 8;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 823 244 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 646 489 6;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 646 489 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 292 979 2;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 292 979 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 585 958 4;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 585 958 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 171 916 8;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 171 916 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 343 833 6;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 343 833 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 300 687 667 2;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 300 687 667 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 601 375 334 4;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 601 375 334 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 202 750 668 8;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 202 750 668 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 405 501 337 6;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 405 501 337 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 811 002 675 2;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 811 002 675 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 622 005 350 4;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 622 005 350 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 244 010 700 8;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 244 010 700 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 488 021 401 6;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 488 021 401 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 976 042 803 2;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 976 042 803 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 649 952 085 606 4;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 649 952 085 606 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 299 904 171 212 8;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 299 904 171 212 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 599 808 342 425 6;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 599 808 342 425 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 199 616 684 851 2;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 199 616 684 851 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 399 233 369 702 4;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 399 233 369 702 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 798 466 739 404 8;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 798 466 739 404 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 596 933 478 809 6;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 596 933 478 809 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 193 866 957 619 2;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 193 866 957 619 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 387 733 915 238 4;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 387 733 915 238 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 775 467 830 476 8;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 775 467 830 476 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 649 550 935 660 953 6;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 649 550 935 660 953 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 299 101 871 321 907 2;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 299 101 871 321 907 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 598 203 742 643 814 4;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 598 203 742 643 814 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 196 407 485 287 628 8;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 196 407 485 287 628 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 392 814 970 575 257 6;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 392 814 970 575 257 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 785 629 941 150 515 2;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 785 629 941 150 515 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 571 259 882 301 030 4;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 571 259 882 301 030 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 142 519 764 602 060 8;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 142 519 764 602 060 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 662 285 039 529 204 121 6;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 662 285 039 529 204 121 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 324 570 079 058 408 243 2;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 324 570 079 058 408 243 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 649 140 158 116 816 486 4;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 649 140 158 116 816 486 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 298 280 316 233 632 972 8;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 298 280 316 233 632 972 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 596 560 632 467 265 945 6;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 596 560 632 467 265 945 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 193 121 264 934 531 891 2;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 193 121 264 934 531 891 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 386 242 529 869 063 782 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 2 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