3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9(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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 595 8;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 595 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 191 6;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 191 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 383 2;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 383 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 766 4;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 766 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 532 8;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 532 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 065 6;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 065 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 131 2;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 131 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 262 4;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 262 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 524 8;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 524 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 049 6;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 049 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 594 099 2;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 594 099 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 188 198 4;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 188 198 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 376 396 8;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 376 396 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 752 793 6;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 752 793 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 505 587 2;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 505 587 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 011 174 4;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 331 011 174 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 022 348 8;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 662 022 348 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 044 697 6;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 324 044 697 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 089 395 2;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 648 089 395 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 296 178 790 4;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 296 178 790 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 592 357 580 8;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 592 357 580 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 184 715 161 6;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 184 715 161 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 369 430 323 2;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 369 430 323 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 738 860 646 4;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 738 860 646 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 477 721 292 8;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 477 721 292 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 955 442 585 6;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 955 442 585 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 910 885 171 2;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 910 885 171 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 821 770 342 4;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 821 770 342 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 643 540 684 8;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 643 540 684 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 295 287 081 369 6;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 295 287 081 369 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 590 574 162 739 2;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 590 574 162 739 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 181 148 325 478 4;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 181 148 325 478 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 362 296 650 956 8;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 362 296 650 956 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 724 593 301 913 6;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 724 593 301 913 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 449 186 603 827 2;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 665 449 186 603 827 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 898 373 207 654 4;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 330 898 373 207 654 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 796 746 415 308 8;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 661 796 746 415 308 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 593 492 830 617 6;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 323 593 492 830 617 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 186 985 661 235 2;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 647 186 985 661 235 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 294 373 971 322 470 4;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 294 373 971 322 470 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 588 747 942 644 940 8;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 588 747 942 644 940 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 177 495 885 289 881 6;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 177 495 885 289 881 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 354 991 770 579 763 2;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 354 991 770 579 763 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 709 983 541 159 526 4;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 709 983 541 159 526 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 419 967 082 319 052 8;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 665 419 967 082 319 052 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 839 934 164 638 105 6;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 330 839 934 164 638 105 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 661 679 868 329 276 211 2;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 661 679 868 329 276 211 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 323 359 736 658 552 422 4;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 323 359 736 658 552 422 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 646 719 473 317 104 844 8;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 646 719 473 317 104 844 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 293 438 946 634 209 689 6;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 293 438 946 634 209 689 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 586 877 893 268 419 379 2;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 586 877 893 268 419 379 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 173 755 786 536 838 758 4;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 173 755 786 536 838 758 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 347 511 573 073 677 516 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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 297 9 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