Convert 12.333 333 333 333 333 333 333 333 333 333 333 333 333 337 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

12.333 333 333 333 333 333 333 333 333 333 333 333 333 337(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: 12.
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;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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.

12(10) =


1100(2)


3. Convert to the binary (base 2) the fractional part: 0.333 333 333 333 333 333 333 333 333 333 333 333 333 337.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 337 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 674;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 674 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 348;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 348 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 696;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 696 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 392;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 392 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 666 784;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 666 666 784 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 333 568;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 333 333 568 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 667 136;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 666 667 136 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 334 272;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 333 334 272 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 668 544;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 666 668 544 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 337 088;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 333 337 088 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 674 176;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 666 674 176 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 348 352;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 333 348 352 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 696 704;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 666 666 696 704 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 393 408;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 333 333 393 408 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 786 816;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 666 666 786 816 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 573 632;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 333 333 573 632 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 667 147 264;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 666 667 147 264 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 334 294 528;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 333 334 294 528 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 668 589 056;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 666 668 589 056 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 337 178 112;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 333 337 178 112 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 674 356 224;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 666 674 356 224 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 348 712 448;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 333 348 712 448 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 697 424 896;
  • 24) 0.666 666 666 666 666 666 666 666 666 666 666 697 424 896 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 394 849 792;
  • 25) 0.333 333 333 333 333 333 333 333 333 333 333 394 849 792 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 789 699 584;
  • 26) 0.666 666 666 666 666 666 666 666 666 666 666 789 699 584 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 579 399 168;
  • 27) 0.333 333 333 333 333 333 333 333 333 333 333 579 399 168 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 667 158 798 336;
  • 28) 0.666 666 666 666 666 666 666 666 666 666 667 158 798 336 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 334 317 596 672;
  • 29) 0.333 333 333 333 333 333 333 333 333 333 334 317 596 672 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 668 635 193 344;
  • 30) 0.666 666 666 666 666 666 666 666 666 666 668 635 193 344 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 337 270 386 688;
  • 31) 0.333 333 333 333 333 333 333 333 333 333 337 270 386 688 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 674 540 773 376;
  • 32) 0.666 666 666 666 666 666 666 666 666 666 674 540 773 376 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 349 081 546 752;
  • 33) 0.333 333 333 333 333 333 333 333 333 333 349 081 546 752 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 698 163 093 504;
  • 34) 0.666 666 666 666 666 666 666 666 666 666 698 163 093 504 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 396 326 187 008;
  • 35) 0.333 333 333 333 333 333 333 333 333 333 396 326 187 008 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 792 652 374 016;
  • 36) 0.666 666 666 666 666 666 666 666 666 666 792 652 374 016 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 585 304 748 032;
  • 37) 0.333 333 333 333 333 333 333 333 333 333 585 304 748 032 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 667 170 609 496 064;
  • 38) 0.666 666 666 666 666 666 666 666 666 667 170 609 496 064 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 334 341 218 992 128;
  • 39) 0.333 333 333 333 333 333 333 333 333 334 341 218 992 128 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 668 682 437 984 256;
  • 40) 0.666 666 666 666 666 666 666 666 666 668 682 437 984 256 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 337 364 875 968 512;
  • 41) 0.333 333 333 333 333 333 333 333 333 337 364 875 968 512 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 674 729 751 937 024;
  • 42) 0.666 666 666 666 666 666 666 666 666 674 729 751 937 024 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 349 459 503 874 048;
  • 43) 0.333 333 333 333 333 333 333 333 333 349 459 503 874 048 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 698 919 007 748 096;
  • 44) 0.666 666 666 666 666 666 666 666 666 698 919 007 748 096 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 397 838 015 496 192;
  • 45) 0.333 333 333 333 333 333 333 333 333 397 838 015 496 192 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 795 676 030 992 384;
  • 46) 0.666 666 666 666 666 666 666 666 666 795 676 030 992 384 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 591 352 061 984 768;
  • 47) 0.333 333 333 333 333 333 333 333 333 591 352 061 984 768 × 2 = 0 + 0.666 666 666 666 666 666 666 666 667 182 704 123 969 536;
  • 48) 0.666 666 666 666 666 666 666 666 667 182 704 123 969 536 × 2 = 1 + 0.333 333 333 333 333 333 333 333 334 365 408 247 939 072;
  • 49) 0.333 333 333 333 333 333 333 333 334 365 408 247 939 072 × 2 = 0 + 0.666 666 666 666 666 666 666 666 668 730 816 495 878 144;
  • 50) 0.666 666 666 666 666 666 666 666 668 730 816 495 878 144 × 2 = 1 + 0.333 333 333 333 333 333 333 333 337 461 632 991 756 288;
  • 51) 0.333 333 333 333 333 333 333 333 337 461 632 991 756 288 × 2 = 0 + 0.666 666 666 666 666 666 666 666 674 923 265 983 512 576;
  • 52) 0.666 666 666 666 666 666 666 666 674 923 265 983 512 576 × 2 = 1 + 0.333 333 333 333 333 333 333 333 349 846 531 967 025 152;
  • 53) 0.333 333 333 333 333 333 333 333 349 846 531 967 025 152 × 2 = 0 + 0.666 666 666 666 666 666 666 666 699 693 063 934 050 304;

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:

0.333 333 333 333 333 333 333 333 333 333 333 333 333 337(10) =


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


5. Positive number before normalization:

12.333 333 333 333 333 333 333 333 333 333 333 333 333 337(10) =


1100.0101 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 3 positions to the left so that only one non zero digit remains to the left of it:

12.333 333 333 333 333 333 333 333 333 333 333 333 333 337(10) =


1100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =


1100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =


1.1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010(2) × 23


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): 3


Mantissa (not normalized):
1.1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 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) =


1026(10) =


100 0000 0010(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. 1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 =


1000 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 0010


Mantissa (52 bits) =
1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


Number 12.333 333 333 333 333 333 333 333 333 333 333 333 333 337 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0010 - 1000 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
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      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:

12.333 333 333 333 333 333 333 333 333 333 333 333 333 336 = ? ... 12.333 333 333 333 333 333 333 333 333 333 333 333 333 338 = ?


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

12.333 333 333 333 333 333 333 333 333 333 333 333 333 337 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:12 UTC (GMT)
-1 036.699 999 999 999 818 101 059 645 414 352 416 95 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:12 UTC (GMT)
0.000 000 000 000 755 840 011 110 101 101 010 100 110 000 000 000 001 100 001 010 011 010 011 000 1 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:11 UTC (GMT)
1 619 040 611 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:11 UTC (GMT)
17 231 193 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:11 UTC (GMT)
19.384 765 64 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:10 UTC (GMT)
420.1 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:10 UTC (GMT)
-1.031 24 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:09 UTC (GMT)
371 598 330 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:08 UTC (GMT)
0.343 7 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:07 UTC (GMT)
2.526 32 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:06 UTC (GMT)
170 208 083 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:04 UTC (GMT)
13 827 739 705 887 686 656 to 64 bit double precision IEEE 754 binary floating point = ? Jun 26 20:04 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