Convert 0.333 333 333 333 333 333 339 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.333 333 333 333 333 333 339(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =


0(2)


3. Convert to the binary (base 2) the fractional part: 0.333 333 333 333 333 333 339.

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) 0.333 333 333 333 333 333 339 × 2 = 0 + 0.666 666 666 666 666 666 678;
  • 2) 0.666 666 666 666 666 666 678 × 2 = 1 + 0.333 333 333 333 333 333 356;
  • 3) 0.333 333 333 333 333 333 356 × 2 = 0 + 0.666 666 666 666 666 666 712;
  • 4) 0.666 666 666 666 666 666 712 × 2 = 1 + 0.333 333 333 333 333 333 424;
  • 5) 0.333 333 333 333 333 333 424 × 2 = 0 + 0.666 666 666 666 666 666 848;
  • 6) 0.666 666 666 666 666 666 848 × 2 = 1 + 0.333 333 333 333 333 333 696;
  • 7) 0.333 333 333 333 333 333 696 × 2 = 0 + 0.666 666 666 666 666 667 392;
  • 8) 0.666 666 666 666 666 667 392 × 2 = 1 + 0.333 333 333 333 333 334 784;
  • 9) 0.333 333 333 333 333 334 784 × 2 = 0 + 0.666 666 666 666 666 669 568;
  • 10) 0.666 666 666 666 666 669 568 × 2 = 1 + 0.333 333 333 333 333 339 136;
  • 11) 0.333 333 333 333 333 339 136 × 2 = 0 + 0.666 666 666 666 666 678 272;
  • 12) 0.666 666 666 666 666 678 272 × 2 = 1 + 0.333 333 333 333 333 356 544;
  • 13) 0.333 333 333 333 333 356 544 × 2 = 0 + 0.666 666 666 666 666 713 088;
  • 14) 0.666 666 666 666 666 713 088 × 2 = 1 + 0.333 333 333 333 333 426 176;
  • 15) 0.333 333 333 333 333 426 176 × 2 = 0 + 0.666 666 666 666 666 852 352;
  • 16) 0.666 666 666 666 666 852 352 × 2 = 1 + 0.333 333 333 333 333 704 704;
  • 17) 0.333 333 333 333 333 704 704 × 2 = 0 + 0.666 666 666 666 667 409 408;
  • 18) 0.666 666 666 666 667 409 408 × 2 = 1 + 0.333 333 333 333 334 818 816;
  • 19) 0.333 333 333 333 334 818 816 × 2 = 0 + 0.666 666 666 666 669 637 632;
  • 20) 0.666 666 666 666 669 637 632 × 2 = 1 + 0.333 333 333 333 339 275 264;
  • 21) 0.333 333 333 333 339 275 264 × 2 = 0 + 0.666 666 666 666 678 550 528;
  • 22) 0.666 666 666 666 678 550 528 × 2 = 1 + 0.333 333 333 333 357 101 056;
  • 23) 0.333 333 333 333 357 101 056 × 2 = 0 + 0.666 666 666 666 714 202 112;
  • 24) 0.666 666 666 666 714 202 112 × 2 = 1 + 0.333 333 333 333 428 404 224;
  • 25) 0.333 333 333 333 428 404 224 × 2 = 0 + 0.666 666 666 666 856 808 448;
  • 26) 0.666 666 666 666 856 808 448 × 2 = 1 + 0.333 333 333 333 713 616 896;
  • 27) 0.333 333 333 333 713 616 896 × 2 = 0 + 0.666 666 666 667 427 233 792;
  • 28) 0.666 666 666 667 427 233 792 × 2 = 1 + 0.333 333 333 334 854 467 584;
  • 29) 0.333 333 333 334 854 467 584 × 2 = 0 + 0.666 666 666 669 708 935 168;
  • 30) 0.666 666 666 669 708 935 168 × 2 = 1 + 0.333 333 333 339 417 870 336;
  • 31) 0.333 333 333 339 417 870 336 × 2 = 0 + 0.666 666 666 678 835 740 672;
  • 32) 0.666 666 666 678 835 740 672 × 2 = 1 + 0.333 333 333 357 671 481 344;
  • 33) 0.333 333 333 357 671 481 344 × 2 = 0 + 0.666 666 666 715 342 962 688;
  • 34) 0.666 666 666 715 342 962 688 × 2 = 1 + 0.333 333 333 430 685 925 376;
  • 35) 0.333 333 333 430 685 925 376 × 2 = 0 + 0.666 666 666 861 371 850 752;
  • 36) 0.666 666 666 861 371 850 752 × 2 = 1 + 0.333 333 333 722 743 701 504;
  • 37) 0.333 333 333 722 743 701 504 × 2 = 0 + 0.666 666 667 445 487 403 008;
  • 38) 0.666 666 667 445 487 403 008 × 2 = 1 + 0.333 333 334 890 974 806 016;
  • 39) 0.333 333 334 890 974 806 016 × 2 = 0 + 0.666 666 669 781 949 612 032;
  • 40) 0.666 666 669 781 949 612 032 × 2 = 1 + 0.333 333 339 563 899 224 064;
  • 41) 0.333 333 339 563 899 224 064 × 2 = 0 + 0.666 666 679 127 798 448 128;
  • 42) 0.666 666 679 127 798 448 128 × 2 = 1 + 0.333 333 358 255 596 896 256;
  • 43) 0.333 333 358 255 596 896 256 × 2 = 0 + 0.666 666 716 511 193 792 512;
  • 44) 0.666 666 716 511 193 792 512 × 2 = 1 + 0.333 333 433 022 387 585 024;
  • 45) 0.333 333 433 022 387 585 024 × 2 = 0 + 0.666 666 866 044 775 170 048;
  • 46) 0.666 666 866 044 775 170 048 × 2 = 1 + 0.333 333 732 089 550 340 096;
  • 47) 0.333 333 732 089 550 340 096 × 2 = 0 + 0.666 667 464 179 100 680 192;
  • 48) 0.666 667 464 179 100 680 192 × 2 = 1 + 0.333 334 928 358 201 360 384;
  • 49) 0.333 334 928 358 201 360 384 × 2 = 0 + 0.666 669 856 716 402 720 768;
  • 50) 0.666 669 856 716 402 720 768 × 2 = 1 + 0.333 339 713 432 805 441 536;
  • 51) 0.333 339 713 432 805 441 536 × 2 = 0 + 0.666 679 426 865 610 883 072;
  • 52) 0.666 679 426 865 610 883 072 × 2 = 1 + 0.333 358 853 731 221 766 144;
  • 53) 0.333 358 853 731 221 766 144 × 2 = 0 + 0.666 717 707 462 443 532 288;
  • 54) 0.666 717 707 462 443 532 288 × 2 = 1 + 0.333 435 414 924 887 064 576;

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:

0.333 333 333 333 333 333 339(10) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2)


5. Positive number before normalization:

0.333 333 333 333 333 333 339(10) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right so that only one non zero digit remains to the left of it:

0.333 333 333 333 333 333 339(10) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 01(2) × 20 =


1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101(2) × 2-2


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): -2


Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


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


(-2 + 1 023)(10) =


1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =


1021(10) =


011 1111 1101(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, only if necessary (not the case here).

Mantissa (normalized) =


1. 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 =


0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


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) =
011 1111 1101


Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


Number 0.333 333 333 333 333 333 339 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1101 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 1

      42
    • 0

      41
    • 1

      40
    • 0

      39
    • 1

      38
    • 0

      37
    • 1

      36
    • 0

      35
    • 1

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 0

      23
    • 1

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

0.333 333 333 333 333 333 338 = ? ... 0.333 333 333 333 333 333 34 = ?


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

0.333 333 333 333 333 333 339 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:59 UTC (GMT)
6.000 26 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:59 UTC (GMT)
17.16 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:59 UTC (GMT)
1.999 999 940 3 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:59 UTC (GMT)
999 493 928 383 838 999 998 999 989 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:59 UTC (GMT)
85 342 109 779 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:58 UTC (GMT)
0.000 000 178 571 428 571 428 571 428 571 428 58 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:58 UTC (GMT)
0.777 777 777 72 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:58 UTC (GMT)
64.457 565 61 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:58 UTC (GMT)
10 526.762 7 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:58 UTC (GMT)
-0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 199 627 556 688 788 166 568 975 527 661 452 896 255 239 163 182 384 165 639 685 941 175 044 575 939 479 884 726 846 276 314 379 413 890 141 581 497 783 091 301 038 340 662 521 034 907 744 4 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:57 UTC (GMT)
819.9 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:57 UTC (GMT)
860.999 4 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:57 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