3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 302 7 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 7(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 7(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 7.

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 7 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 605 4;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 605 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 210 8;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 210 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 421 6;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 421 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 843 2;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 843 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 686 4;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 686 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 372 8;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 372 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 745 6;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 745 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 491 2;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 491 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 982 4;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 982 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 964 8;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 964 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 603 929 6;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 603 929 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 207 859 2;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 207 859 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 415 718 4;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 415 718 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 831 436 8;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 831 436 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 662 873 6;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 662 873 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 325 747 2;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 325 747 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 651 494 4;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 651 494 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 302 988 8;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 302 988 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 605 977 6;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 605 977 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 211 955 2;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 301 211 955 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 602 423 910 4;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 602 423 910 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 204 847 820 8;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 204 847 820 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 409 695 641 6;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 409 695 641 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 819 391 283 2;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 819 391 283 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 638 782 566 4;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 638 782 566 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 277 565 132 8;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 277 565 132 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 555 130 265 6;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 555 130 265 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 110 260 531 2;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 325 110 260 531 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 220 521 062 4;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 650 220 521 062 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 300 441 042 124 8;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 300 441 042 124 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 600 882 084 249 6;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 600 882 084 249 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 201 764 168 499 2;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 201 764 168 499 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 403 528 336 998 4;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 403 528 336 998 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 807 056 673 996 8;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 807 056 673 996 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 614 113 347 993 6;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 614 113 347 993 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 228 226 695 987 2;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 228 226 695 987 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 456 453 391 974 4;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 456 453 391 974 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 912 906 783 948 8;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 912 906 783 948 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 649 825 813 567 897 6;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 649 825 813 567 897 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 299 651 627 135 795 2;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 299 651 627 135 795 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 599 303 254 271 590 4;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 599 303 254 271 590 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 198 606 508 543 180 8;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 198 606 508 543 180 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 397 213 017 086 361 6;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 397 213 017 086 361 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 794 426 034 172 723 2;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 794 426 034 172 723 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 588 852 068 345 446 4;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 588 852 068 345 446 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 177 704 136 690 892 8;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 331 177 704 136 690 892 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 662 355 408 273 381 785 6;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 662 355 408 273 381 785 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 324 710 816 546 763 571 2;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 324 710 816 546 763 571 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 649 421 633 093 527 142 4;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 649 421 633 093 527 142 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 298 843 266 187 054 284 8;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 298 843 266 187 054 284 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 597 686 532 374 108 569 6;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 597 686 532 374 108 569 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 195 373 064 748 217 139 2;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 195 373 064 748 217 139 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 390 746 129 496 434 278 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 7(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 7(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 7(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 7 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