3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8(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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 583 6;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 583 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 167 2;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 167 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 334 4;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 334 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 668 8;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 668 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 337 6;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 337 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 675 2;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 675 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 350 4;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 350 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 322 700 8;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 322 700 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 645 401 6;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 645 401 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 290 803 2;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 290 803 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 581 606 4;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 581 606 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 163 212 8;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 163 212 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 326 425 6;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 326 425 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 652 851 2;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 652 851 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 305 702 4;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 305 702 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 611 404 8;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 611 404 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 222 809 6;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 222 809 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 322 445 619 2;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 322 445 619 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 644 891 238 4;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 644 891 238 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 289 782 476 8;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 289 782 476 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 579 564 953 6;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 579 564 953 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 159 129 907 2;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 159 129 907 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 318 259 814 4;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 318 259 814 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 636 519 628 8;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 636 519 628 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 273 039 257 6;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 273 039 257 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 546 078 515 2;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 546 078 515 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 092 157 030 4;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 092 157 030 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 322 184 314 060 8;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 322 184 314 060 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 644 368 628 121 6;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 644 368 628 121 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 288 737 256 243 2;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 288 737 256 243 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 577 474 512 486 4;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 577 474 512 486 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 154 949 024 972 8;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 154 949 024 972 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 309 898 049 945 6;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 309 898 049 945 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 619 796 099 891 2;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 619 796 099 891 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 239 592 199 782 4;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 239 592 199 782 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 479 184 399 564 8;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 479 184 399 564 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 958 368 799 129 6;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 660 958 368 799 129 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 321 916 737 598 259 2;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 321 916 737 598 259 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 643 833 475 196 518 4;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 643 833 475 196 518 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 287 666 950 393 036 8;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 287 666 950 393 036 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 575 333 900 786 073 6;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 575 333 900 786 073 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 150 667 801 572 147 2;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 150 667 801 572 147 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 301 335 603 144 294 4;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 301 335 603 144 294 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 602 671 206 288 588 8;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 602 671 206 288 588 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 205 342 412 577 177 6;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 205 342 412 577 177 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 410 684 825 154 355 2;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 410 684 825 154 355 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 660 821 369 650 308 710 4;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 660 821 369 650 308 710 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 321 642 739 300 617 420 8;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 321 642 739 300 617 420 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 643 285 478 601 234 841 6;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 643 285 478 601 234 841 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 286 570 957 202 469 683 2;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 286 570 957 202 469 683 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 573 141 914 404 939 366 4;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 573 141 914 404 939 366 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 146 283 828 809 878 732 8;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 146 283 828 809 878 732 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 292 567 657 619 757 465 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 291 8 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