0.333 333 333 333 333 314 829 616 256 257 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 257 5(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
0.333 333 333 333 333 314 829 616 256 257 5(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: 0.333 333 333 333 333 314 829 616 256 257 5.

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) 0.333 333 333 333 333 314 829 616 256 257 5 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 515;
  • 2) 0.666 666 666 666 666 629 659 232 512 515 × 2 = 1 + 0.333 333 333 333 333 259 318 465 025 03;
  • 3) 0.333 333 333 333 333 259 318 465 025 03 × 2 = 0 + 0.666 666 666 666 666 518 636 930 050 06;
  • 4) 0.666 666 666 666 666 518 636 930 050 06 × 2 = 1 + 0.333 333 333 333 333 037 273 860 100 12;
  • 5) 0.333 333 333 333 333 037 273 860 100 12 × 2 = 0 + 0.666 666 666 666 666 074 547 720 200 24;
  • 6) 0.666 666 666 666 666 074 547 720 200 24 × 2 = 1 + 0.333 333 333 333 332 149 095 440 400 48;
  • 7) 0.333 333 333 333 332 149 095 440 400 48 × 2 = 0 + 0.666 666 666 666 664 298 190 880 800 96;
  • 8) 0.666 666 666 666 664 298 190 880 800 96 × 2 = 1 + 0.333 333 333 333 328 596 381 761 601 92;
  • 9) 0.333 333 333 333 328 596 381 761 601 92 × 2 = 0 + 0.666 666 666 666 657 192 763 523 203 84;
  • 10) 0.666 666 666 666 657 192 763 523 203 84 × 2 = 1 + 0.333 333 333 333 314 385 527 046 407 68;
  • 11) 0.333 333 333 333 314 385 527 046 407 68 × 2 = 0 + 0.666 666 666 666 628 771 054 092 815 36;
  • 12) 0.666 666 666 666 628 771 054 092 815 36 × 2 = 1 + 0.333 333 333 333 257 542 108 185 630 72;
  • 13) 0.333 333 333 333 257 542 108 185 630 72 × 2 = 0 + 0.666 666 666 666 515 084 216 371 261 44;
  • 14) 0.666 666 666 666 515 084 216 371 261 44 × 2 = 1 + 0.333 333 333 333 030 168 432 742 522 88;
  • 15) 0.333 333 333 333 030 168 432 742 522 88 × 2 = 0 + 0.666 666 666 666 060 336 865 485 045 76;
  • 16) 0.666 666 666 666 060 336 865 485 045 76 × 2 = 1 + 0.333 333 333 332 120 673 730 970 091 52;
  • 17) 0.333 333 333 332 120 673 730 970 091 52 × 2 = 0 + 0.666 666 666 664 241 347 461 940 183 04;
  • 18) 0.666 666 666 664 241 347 461 940 183 04 × 2 = 1 + 0.333 333 333 328 482 694 923 880 366 08;
  • 19) 0.333 333 333 328 482 694 923 880 366 08 × 2 = 0 + 0.666 666 666 656 965 389 847 760 732 16;
  • 20) 0.666 666 666 656 965 389 847 760 732 16 × 2 = 1 + 0.333 333 333 313 930 779 695 521 464 32;
  • 21) 0.333 333 333 313 930 779 695 521 464 32 × 2 = 0 + 0.666 666 666 627 861 559 391 042 928 64;
  • 22) 0.666 666 666 627 861 559 391 042 928 64 × 2 = 1 + 0.333 333 333 255 723 118 782 085 857 28;
  • 23) 0.333 333 333 255 723 118 782 085 857 28 × 2 = 0 + 0.666 666 666 511 446 237 564 171 714 56;
  • 24) 0.666 666 666 511 446 237 564 171 714 56 × 2 = 1 + 0.333 333 333 022 892 475 128 343 429 12;
  • 25) 0.333 333 333 022 892 475 128 343 429 12 × 2 = 0 + 0.666 666 666 045 784 950 256 686 858 24;
  • 26) 0.666 666 666 045 784 950 256 686 858 24 × 2 = 1 + 0.333 333 332 091 569 900 513 373 716 48;
  • 27) 0.333 333 332 091 569 900 513 373 716 48 × 2 = 0 + 0.666 666 664 183 139 801 026 747 432 96;
  • 28) 0.666 666 664 183 139 801 026 747 432 96 × 2 = 1 + 0.333 333 328 366 279 602 053 494 865 92;
  • 29) 0.333 333 328 366 279 602 053 494 865 92 × 2 = 0 + 0.666 666 656 732 559 204 106 989 731 84;
  • 30) 0.666 666 656 732 559 204 106 989 731 84 × 2 = 1 + 0.333 333 313 465 118 408 213 979 463 68;
  • 31) 0.333 333 313 465 118 408 213 979 463 68 × 2 = 0 + 0.666 666 626 930 236 816 427 958 927 36;
  • 32) 0.666 666 626 930 236 816 427 958 927 36 × 2 = 1 + 0.333 333 253 860 473 632 855 917 854 72;
  • 33) 0.333 333 253 860 473 632 855 917 854 72 × 2 = 0 + 0.666 666 507 720 947 265 711 835 709 44;
  • 34) 0.666 666 507 720 947 265 711 835 709 44 × 2 = 1 + 0.333 333 015 441 894 531 423 671 418 88;
  • 35) 0.333 333 015 441 894 531 423 671 418 88 × 2 = 0 + 0.666 666 030 883 789 062 847 342 837 76;
  • 36) 0.666 666 030 883 789 062 847 342 837 76 × 2 = 1 + 0.333 332 061 767 578 125 694 685 675 52;
  • 37) 0.333 332 061 767 578 125 694 685 675 52 × 2 = 0 + 0.666 664 123 535 156 251 389 371 351 04;
  • 38) 0.666 664 123 535 156 251 389 371 351 04 × 2 = 1 + 0.333 328 247 070 312 502 778 742 702 08;
  • 39) 0.333 328 247 070 312 502 778 742 702 08 × 2 = 0 + 0.666 656 494 140 625 005 557 485 404 16;
  • 40) 0.666 656 494 140 625 005 557 485 404 16 × 2 = 1 + 0.333 312 988 281 250 011 114 970 808 32;
  • 41) 0.333 312 988 281 250 011 114 970 808 32 × 2 = 0 + 0.666 625 976 562 500 022 229 941 616 64;
  • 42) 0.666 625 976 562 500 022 229 941 616 64 × 2 = 1 + 0.333 251 953 125 000 044 459 883 233 28;
  • 43) 0.333 251 953 125 000 044 459 883 233 28 × 2 = 0 + 0.666 503 906 250 000 088 919 766 466 56;
  • 44) 0.666 503 906 250 000 088 919 766 466 56 × 2 = 1 + 0.333 007 812 500 000 177 839 532 933 12;
  • 45) 0.333 007 812 500 000 177 839 532 933 12 × 2 = 0 + 0.666 015 625 000 000 355 679 065 866 24;
  • 46) 0.666 015 625 000 000 355 679 065 866 24 × 2 = 1 + 0.332 031 250 000 000 711 358 131 732 48;
  • 47) 0.332 031 250 000 000 711 358 131 732 48 × 2 = 0 + 0.664 062 500 000 001 422 716 263 464 96;
  • 48) 0.664 062 500 000 001 422 716 263 464 96 × 2 = 1 + 0.328 125 000 000 002 845 432 526 929 92;
  • 49) 0.328 125 000 000 002 845 432 526 929 92 × 2 = 0 + 0.656 250 000 000 005 690 865 053 859 84;
  • 50) 0.656 250 000 000 005 690 865 053 859 84 × 2 = 1 + 0.312 500 000 000 011 381 730 107 719 68;
  • 51) 0.312 500 000 000 011 381 730 107 719 68 × 2 = 0 + 0.625 000 000 000 022 763 460 215 439 36;
  • 52) 0.625 000 000 000 022 763 460 215 439 36 × 2 = 1 + 0.250 000 000 000 045 526 920 430 878 72;
  • 53) 0.250 000 000 000 045 526 920 430 878 72 × 2 = 0 + 0.500 000 000 000 091 053 840 861 757 44;
  • 54) 0.500 000 000 000 091 053 840 861 757 44 × 2 = 1 + 0.000 000 000 000 182 107 681 723 514 88;

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:


0.333 333 333 333 333 314 829 616 256 257 5(10) =

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

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 257 5(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:


0.333 333 333 333 333 314 829 616 256 257 5(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 0.333 333 333 333 333 314 829 616 256 257 5 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