Convert 2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 8 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 8(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =


10(2)


3. Convert to the binary (base 2) the fractional part: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 6;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 2;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 4;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 340 8;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 340 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 681 6;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 681 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 363 2;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 363 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 726 4;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 726 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 452 8;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 452 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 905 6;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 905 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 811 2;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 811 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 622 4;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 622 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 244 8;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 244 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 489 6;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 489 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 340 979 2;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 340 979 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 681 958 4;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 681 958 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 363 916 8;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 363 916 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 727 833 6;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 727 833 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 455 667 2;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 455 667 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 911 334 4;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 911 334 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 822 668 8;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 822 668 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 645 337 6;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 645 337 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 290 675 2;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 290 675 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 581 350 4;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 581 350 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 341 162 700 8;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 341 162 700 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 682 325 401 6;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 682 325 401 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 364 650 803 2;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 364 650 803 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 729 301 606 4;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 729 301 606 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 458 603 212 8;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 458 603 212 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 917 206 425 6;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 917 206 425 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 834 412 851 2;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 834 412 851 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 668 825 702 4;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 668 825 702 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 337 651 404 8;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 337 651 404 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 675 302 809 6;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 675 302 809 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 341 350 605 619 2;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 341 350 605 619 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 682 701 211 238 4;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 682 701 211 238 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 365 402 422 476 8;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 365 402 422 476 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 730 804 844 953 6;
  • 38) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 730 804 844 953 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 461 609 689 907 2;
  • 39) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 461 609 689 907 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 923 219 379 814 4;
  • 40) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 923 219 379 814 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 846 438 759 628 8;
  • 41) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 846 438 759 628 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 692 877 519 257 6;
  • 42) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 692 877 519 257 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 385 755 038 515 2;
  • 43) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 385 755 038 515 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 771 510 077 030 4;
  • 44) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 771 510 077 030 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 341 543 020 154 060 8;
  • 45) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 341 543 020 154 060 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 683 086 040 308 121 6;
  • 46) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 683 086 040 308 121 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 366 172 080 616 243 2;
  • 47) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 366 172 080 616 243 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 732 344 161 232 486 4;
  • 48) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 732 344 161 232 486 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 464 688 322 464 972 8;
  • 49) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 464 688 322 464 972 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 929 376 644 929 945 6;
  • 50) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 929 376 644 929 945 6 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 858 753 289 859 891 2;
  • 51) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 858 753 289 859 891 2 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 717 506 579 719 782 4;
  • 52) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 667 717 506 579 719 782 4 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 435 013 159 439 564 8;
  • 53) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 335 435 013 159 439 564 8 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 666 666 670 870 026 318 879 129 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...)


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:

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


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


5. Positive number before normalization:

2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 8(10) =


10.0101 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:

2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 8(10) =


10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =


10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =


1.0010 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):
1.0010 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. 0010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 10 =


0010 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) =
0010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


Number 2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 8 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0000 - 0010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 1

      49
    • 0

      48
    • 1

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 0

      40
    • 1

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 0

      32
    • 1

      31
    • 0

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 0

      0

More operations of this kind:

2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 7 = ? ... 2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 9 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 8 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:40 UTC (GMT)
-1 605 282 547 372 794 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:40 UTC (GMT)
0.519 531 25 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:40 UTC (GMT)
0.111 001 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:40 UTC (GMT)
23.893 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:40 UTC (GMT)
0.000 000 000 000 000 000 000 000 009 109 383 5 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:40 UTC (GMT)
1 100 098 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:40 UTC (GMT)
-75.129 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:39 UTC (GMT)
0.049 979 19 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:39 UTC (GMT)
-856.534 9 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:39 UTC (GMT)
73.37 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:39 UTC (GMT)
0.957 603 280 698 573 646 936 305 635 146 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:38 UTC (GMT)
3.14 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 08:38 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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