3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6(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
3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6(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: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 559 2;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 559 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 118 4;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 118 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 236 8;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 236 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 473 6;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 473 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 947 2;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 947 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 894 4;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 894 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 788 8;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 788 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 319 577 6;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 319 577 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 639 155 2;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 639 155 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 278 310 4;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 278 310 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 556 620 8;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 556 620 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 113 241 6;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 113 241 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 226 483 2;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 226 483 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 452 966 4;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 452 966 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 905 932 8;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 905 932 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 811 865 6;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 811 865 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 623 731 2;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 623 731 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 319 247 462 4;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 319 247 462 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 638 494 924 8;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 638 494 924 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 276 989 849 6;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 276 989 849 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 553 979 699 2;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 553 979 699 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 107 959 398 4;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 107 959 398 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 215 918 796 8;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 215 918 796 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 431 837 593 6;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 431 837 593 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 863 675 187 2;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 863 675 187 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 727 350 374 4;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 727 350 374 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 454 700 748 8;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 454 700 748 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 318 909 401 497 6;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 318 909 401 497 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 637 818 802 995 2;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 637 818 802 995 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 275 637 605 990 4;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 275 637 605 990 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 551 275 211 980 8;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 551 275 211 980 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 102 550 423 961 6;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 102 550 423 961 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 205 100 847 923 2;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 205 100 847 923 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 410 201 695 846 4;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 410 201 695 846 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 820 403 391 692 8;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 664 820 403 391 692 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 640 806 783 385 6;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 329 640 806 783 385 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 281 613 566 771 2;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 659 281 613 566 771 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 318 563 227 133 542 4;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 318 563 227 133 542 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 637 126 454 267 084 8;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 637 126 454 267 084 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 274 252 908 534 169 6;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 274 252 908 534 169 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 548 505 817 068 339 2;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 548 505 817 068 339 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 097 011 634 136 678 4;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 097 011 634 136 678 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 194 023 268 273 356 8;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 194 023 268 273 356 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 388 046 536 546 713 6;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 332 388 046 536 546 713 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 664 776 093 073 093 427 2;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 664 776 093 073 093 427 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 329 552 186 146 186 854 4;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 329 552 186 146 186 854 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 659 104 372 292 373 708 8;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 659 104 372 292 373 708 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 318 208 744 584 747 417 6;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 318 208 744 584 747 417 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 636 417 489 169 494 835 2;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 636 417 489 169 494 835 2 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 272 834 978 338 989 670 4;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 272 834 978 338 989 670 4 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 545 669 956 677 979 340 8;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 545 669 956 677 979 340 8 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 091 339 913 355 958 681 6;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 091 339 913 355 958 681 6 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 182 679 826 711 917 363 2;

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:


0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6(10) =

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

5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6(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:


3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6(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 3.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 279 6 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