3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3(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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 608 6;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 608 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 217 2;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 217 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 434 4;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 434 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 868 8;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 868 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 737 6;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 737 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 475 2;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 475 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 950 4;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 950 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 900 8;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 900 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 651 801 6;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 651 801 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 303 603 2;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 303 603 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 607 206 4;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 607 206 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 214 412 8;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 214 412 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 428 825 6;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 428 825 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 857 651 2;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 857 651 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 715 302 4;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 715 302 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 430 604 8;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 430 604 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 861 209 6;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 861 209 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 722 419 2;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 722 419 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 651 444 838 4;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 651 444 838 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 889 676 8;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 889 676 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 605 779 353 6;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 605 779 353 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 211 558 707 2;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 211 558 707 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 423 117 414 4;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 423 117 414 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 846 234 828 8;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 846 234 828 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 692 469 657 6;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 692 469 657 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 384 939 315 2;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 384 939 315 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 769 878 630 4;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 769 878 630 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 539 757 260 8;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 539 757 260 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 651 079 514 521 6;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 651 079 514 521 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 159 029 043 2;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 159 029 043 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 604 318 058 086 4;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 604 318 058 086 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 208 636 116 172 8;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 208 636 116 172 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 417 272 232 345 6;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 417 272 232 345 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 834 544 464 691 2;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 834 544 464 691 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 669 088 929 382 4;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 669 088 929 382 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 338 177 858 764 8;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 338 177 858 764 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 676 355 717 529 6;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 676 355 717 529 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 352 711 435 059 2;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 352 711 435 059 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 705 422 870 118 4;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 705 422 870 118 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 410 845 740 236 8;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 410 845 740 236 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 602 821 691 480 473 6;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 602 821 691 480 473 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 205 643 382 960 947 2;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 205 643 382 960 947 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 411 286 765 921 894 4;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 411 286 765 921 894 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 822 573 531 843 788 8;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 822 573 531 843 788 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 645 147 063 687 577 6;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 645 147 063 687 577 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 290 294 127 375 155 2;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 290 294 127 375 155 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 662 580 588 254 750 310 4;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 662 580 588 254 750 310 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 325 161 176 509 500 620 8;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 325 161 176 509 500 620 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 650 322 353 019 001 241 6;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 650 322 353 019 001 241 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 300 644 706 038 002 483 2;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 300 644 706 038 002 483 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 601 289 412 076 004 966 4;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 601 289 412 076 004 966 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 202 578 824 152 009 932 8;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 202 578 824 152 009 932 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 405 157 648 304 019 865 6;

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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 304 3 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