3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 572 8;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 572 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 145 6;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 145 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 291 2;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 291 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 582 4;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 582 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 164 8;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 164 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 329 6;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 329 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 659 2;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 659 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 321 318 4;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 321 318 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 642 636 8;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 642 636 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 285 273 6;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 285 273 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 570 547 2;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 570 547 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 141 094 4;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 141 094 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 282 188 8;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 282 188 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 564 377 6;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 564 377 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 128 755 2;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 128 755 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 257 510 4;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 257 510 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 515 020 8;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 515 020 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 321 030 041 6;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 321 030 041 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 642 060 083 2;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 642 060 083 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 284 120 166 4;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 284 120 166 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 568 240 332 8;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 568 240 332 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 136 480 665 6;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 136 480 665 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 272 961 331 2;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 272 961 331 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 545 922 662 4;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 545 922 662 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 091 845 324 8;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 091 845 324 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 183 690 649 6;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 183 690 649 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 367 381 299 2;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 367 381 299 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 320 734 762 598 4;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 320 734 762 598 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 641 469 525 196 8;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 641 469 525 196 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 282 939 050 393 6;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 282 939 050 393 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 565 878 100 787 2;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 565 878 100 787 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 131 756 201 574 4;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 131 756 201 574 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 263 512 403 148 8;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 263 512 403 148 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 527 024 806 297 6;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 527 024 806 297 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 054 049 612 595 2;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 054 049 612 595 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 108 099 225 190 4;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 108 099 225 190 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 216 198 450 380 8;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 216 198 450 380 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 320 432 396 900 761 6;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 320 432 396 900 761 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 640 864 793 801 523 2;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 640 864 793 801 523 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 281 729 587 603 046 4;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 281 729 587 603 046 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 563 459 175 206 092 8;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 563 459 175 206 092 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 126 918 350 412 185 6;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 126 918 350 412 185 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 253 836 700 824 371 2;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 253 836 700 824 371 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 507 673 401 648 742 4;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 507 673 401 648 742 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 015 346 803 297 484 8;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 015 346 803 297 484 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 030 693 606 594 969 6;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 030 693 606 594 969 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 660 061 387 213 189 939 2;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 660 061 387 213 189 939 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 320 122 774 426 379 878 4;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 320 122 774 426 379 878 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 640 245 548 852 759 756 8;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 640 245 548 852 759 756 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 280 491 097 705 519 513 6;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 280 491 097 705 519 513 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 560 982 195 411 039 027 2;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 560 982 195 411 039 027 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 121 964 390 822 078 054 4;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 121 964 390 822 078 054 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 243 928 781 644 156 108 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 4(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 286 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