0.333 333 333 333 333 314 829 616 256 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 71(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
0.333 333 333 333 333 314 829 616 256 71(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: 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 binary (base 2) the fractional part: 0.333 333 333 333 333 314 829 616 256 71.

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) 0.333 333 333 333 333 314 829 616 256 71 × 2 = 0 + 0.666 666 666 666 666 629 659 232 513 42;
  • 2) 0.666 666 666 666 666 629 659 232 513 42 × 2 = 1 + 0.333 333 333 333 333 259 318 465 026 84;
  • 3) 0.333 333 333 333 333 259 318 465 026 84 × 2 = 0 + 0.666 666 666 666 666 518 636 930 053 68;
  • 4) 0.666 666 666 666 666 518 636 930 053 68 × 2 = 1 + 0.333 333 333 333 333 037 273 860 107 36;
  • 5) 0.333 333 333 333 333 037 273 860 107 36 × 2 = 0 + 0.666 666 666 666 666 074 547 720 214 72;
  • 6) 0.666 666 666 666 666 074 547 720 214 72 × 2 = 1 + 0.333 333 333 333 332 149 095 440 429 44;
  • 7) 0.333 333 333 333 332 149 095 440 429 44 × 2 = 0 + 0.666 666 666 666 664 298 190 880 858 88;
  • 8) 0.666 666 666 666 664 298 190 880 858 88 × 2 = 1 + 0.333 333 333 333 328 596 381 761 717 76;
  • 9) 0.333 333 333 333 328 596 381 761 717 76 × 2 = 0 + 0.666 666 666 666 657 192 763 523 435 52;
  • 10) 0.666 666 666 666 657 192 763 523 435 52 × 2 = 1 + 0.333 333 333 333 314 385 527 046 871 04;
  • 11) 0.333 333 333 333 314 385 527 046 871 04 × 2 = 0 + 0.666 666 666 666 628 771 054 093 742 08;
  • 12) 0.666 666 666 666 628 771 054 093 742 08 × 2 = 1 + 0.333 333 333 333 257 542 108 187 484 16;
  • 13) 0.333 333 333 333 257 542 108 187 484 16 × 2 = 0 + 0.666 666 666 666 515 084 216 374 968 32;
  • 14) 0.666 666 666 666 515 084 216 374 968 32 × 2 = 1 + 0.333 333 333 333 030 168 432 749 936 64;
  • 15) 0.333 333 333 333 030 168 432 749 936 64 × 2 = 0 + 0.666 666 666 666 060 336 865 499 873 28;
  • 16) 0.666 666 666 666 060 336 865 499 873 28 × 2 = 1 + 0.333 333 333 332 120 673 730 999 746 56;
  • 17) 0.333 333 333 332 120 673 730 999 746 56 × 2 = 0 + 0.666 666 666 664 241 347 461 999 493 12;
  • 18) 0.666 666 666 664 241 347 461 999 493 12 × 2 = 1 + 0.333 333 333 328 482 694 923 998 986 24;
  • 19) 0.333 333 333 328 482 694 923 998 986 24 × 2 = 0 + 0.666 666 666 656 965 389 847 997 972 48;
  • 20) 0.666 666 666 656 965 389 847 997 972 48 × 2 = 1 + 0.333 333 333 313 930 779 695 995 944 96;
  • 21) 0.333 333 333 313 930 779 695 995 944 96 × 2 = 0 + 0.666 666 666 627 861 559 391 991 889 92;
  • 22) 0.666 666 666 627 861 559 391 991 889 92 × 2 = 1 + 0.333 333 333 255 723 118 783 983 779 84;
  • 23) 0.333 333 333 255 723 118 783 983 779 84 × 2 = 0 + 0.666 666 666 511 446 237 567 967 559 68;
  • 24) 0.666 666 666 511 446 237 567 967 559 68 × 2 = 1 + 0.333 333 333 022 892 475 135 935 119 36;
  • 25) 0.333 333 333 022 892 475 135 935 119 36 × 2 = 0 + 0.666 666 666 045 784 950 271 870 238 72;
  • 26) 0.666 666 666 045 784 950 271 870 238 72 × 2 = 1 + 0.333 333 332 091 569 900 543 740 477 44;
  • 27) 0.333 333 332 091 569 900 543 740 477 44 × 2 = 0 + 0.666 666 664 183 139 801 087 480 954 88;
  • 28) 0.666 666 664 183 139 801 087 480 954 88 × 2 = 1 + 0.333 333 328 366 279 602 174 961 909 76;
  • 29) 0.333 333 328 366 279 602 174 961 909 76 × 2 = 0 + 0.666 666 656 732 559 204 349 923 819 52;
  • 30) 0.666 666 656 732 559 204 349 923 819 52 × 2 = 1 + 0.333 333 313 465 118 408 699 847 639 04;
  • 31) 0.333 333 313 465 118 408 699 847 639 04 × 2 = 0 + 0.666 666 626 930 236 817 399 695 278 08;
  • 32) 0.666 666 626 930 236 817 399 695 278 08 × 2 = 1 + 0.333 333 253 860 473 634 799 390 556 16;
  • 33) 0.333 333 253 860 473 634 799 390 556 16 × 2 = 0 + 0.666 666 507 720 947 269 598 781 112 32;
  • 34) 0.666 666 507 720 947 269 598 781 112 32 × 2 = 1 + 0.333 333 015 441 894 539 197 562 224 64;
  • 35) 0.333 333 015 441 894 539 197 562 224 64 × 2 = 0 + 0.666 666 030 883 789 078 395 124 449 28;
  • 36) 0.666 666 030 883 789 078 395 124 449 28 × 2 = 1 + 0.333 332 061 767 578 156 790 248 898 56;
  • 37) 0.333 332 061 767 578 156 790 248 898 56 × 2 = 0 + 0.666 664 123 535 156 313 580 497 797 12;
  • 38) 0.666 664 123 535 156 313 580 497 797 12 × 2 = 1 + 0.333 328 247 070 312 627 160 995 594 24;
  • 39) 0.333 328 247 070 312 627 160 995 594 24 × 2 = 0 + 0.666 656 494 140 625 254 321 991 188 48;
  • 40) 0.666 656 494 140 625 254 321 991 188 48 × 2 = 1 + 0.333 312 988 281 250 508 643 982 376 96;
  • 41) 0.333 312 988 281 250 508 643 982 376 96 × 2 = 0 + 0.666 625 976 562 501 017 287 964 753 92;
  • 42) 0.666 625 976 562 501 017 287 964 753 92 × 2 = 1 + 0.333 251 953 125 002 034 575 929 507 84;
  • 43) 0.333 251 953 125 002 034 575 929 507 84 × 2 = 0 + 0.666 503 906 250 004 069 151 859 015 68;
  • 44) 0.666 503 906 250 004 069 151 859 015 68 × 2 = 1 + 0.333 007 812 500 008 138 303 718 031 36;
  • 45) 0.333 007 812 500 008 138 303 718 031 36 × 2 = 0 + 0.666 015 625 000 016 276 607 436 062 72;
  • 46) 0.666 015 625 000 016 276 607 436 062 72 × 2 = 1 + 0.332 031 250 000 032 553 214 872 125 44;
  • 47) 0.332 031 250 000 032 553 214 872 125 44 × 2 = 0 + 0.664 062 500 000 065 106 429 744 250 88;
  • 48) 0.664 062 500 000 065 106 429 744 250 88 × 2 = 1 + 0.328 125 000 000 130 212 859 488 501 76;
  • 49) 0.328 125 000 000 130 212 859 488 501 76 × 2 = 0 + 0.656 250 000 000 260 425 718 977 003 52;
  • 50) 0.656 250 000 000 260 425 718 977 003 52 × 2 = 1 + 0.312 500 000 000 520 851 437 954 007 04;
  • 51) 0.312 500 000 000 520 851 437 954 007 04 × 2 = 0 + 0.625 000 000 001 041 702 875 908 014 08;
  • 52) 0.625 000 000 001 041 702 875 908 014 08 × 2 = 1 + 0.250 000 000 002 083 405 751 816 028 16;
  • 53) 0.250 000 000 002 083 405 751 816 028 16 × 2 = 0 + 0.500 000 000 004 166 811 503 632 056 32;
  • 54) 0.500 000 000 004 166 811 503 632 056 32 × 2 = 1 + 0.000 000 000 008 333 623 007 264 112 64;

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:


0.333 333 333 333 333 314 829 616 256 71(10) =

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

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 71(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:


0.333 333 333 333 333 314 829 616 256 71(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


Decimal number 0.333 333 333 333 333 314 829 616 256 71 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

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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