0.333 333 333 333 333 314 829 616 256 247 54 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 247 54(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
0.333 333 333 333 333 314 829 616 256 247 54(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: 0.333 333 333 333 333 314 829 616 256 247 54.

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) 0.333 333 333 333 333 314 829 616 256 247 54 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 495 08;
  • 2) 0.666 666 666 666 666 629 659 232 512 495 08 × 2 = 1 + 0.333 333 333 333 333 259 318 465 024 990 16;
  • 3) 0.333 333 333 333 333 259 318 465 024 990 16 × 2 = 0 + 0.666 666 666 666 666 518 636 930 049 980 32;
  • 4) 0.666 666 666 666 666 518 636 930 049 980 32 × 2 = 1 + 0.333 333 333 333 333 037 273 860 099 960 64;
  • 5) 0.333 333 333 333 333 037 273 860 099 960 64 × 2 = 0 + 0.666 666 666 666 666 074 547 720 199 921 28;
  • 6) 0.666 666 666 666 666 074 547 720 199 921 28 × 2 = 1 + 0.333 333 333 333 332 149 095 440 399 842 56;
  • 7) 0.333 333 333 333 332 149 095 440 399 842 56 × 2 = 0 + 0.666 666 666 666 664 298 190 880 799 685 12;
  • 8) 0.666 666 666 666 664 298 190 880 799 685 12 × 2 = 1 + 0.333 333 333 333 328 596 381 761 599 370 24;
  • 9) 0.333 333 333 333 328 596 381 761 599 370 24 × 2 = 0 + 0.666 666 666 666 657 192 763 523 198 740 48;
  • 10) 0.666 666 666 666 657 192 763 523 198 740 48 × 2 = 1 + 0.333 333 333 333 314 385 527 046 397 480 96;
  • 11) 0.333 333 333 333 314 385 527 046 397 480 96 × 2 = 0 + 0.666 666 666 666 628 771 054 092 794 961 92;
  • 12) 0.666 666 666 666 628 771 054 092 794 961 92 × 2 = 1 + 0.333 333 333 333 257 542 108 185 589 923 84;
  • 13) 0.333 333 333 333 257 542 108 185 589 923 84 × 2 = 0 + 0.666 666 666 666 515 084 216 371 179 847 68;
  • 14) 0.666 666 666 666 515 084 216 371 179 847 68 × 2 = 1 + 0.333 333 333 333 030 168 432 742 359 695 36;
  • 15) 0.333 333 333 333 030 168 432 742 359 695 36 × 2 = 0 + 0.666 666 666 666 060 336 865 484 719 390 72;
  • 16) 0.666 666 666 666 060 336 865 484 719 390 72 × 2 = 1 + 0.333 333 333 332 120 673 730 969 438 781 44;
  • 17) 0.333 333 333 332 120 673 730 969 438 781 44 × 2 = 0 + 0.666 666 666 664 241 347 461 938 877 562 88;
  • 18) 0.666 666 666 664 241 347 461 938 877 562 88 × 2 = 1 + 0.333 333 333 328 482 694 923 877 755 125 76;
  • 19) 0.333 333 333 328 482 694 923 877 755 125 76 × 2 = 0 + 0.666 666 666 656 965 389 847 755 510 251 52;
  • 20) 0.666 666 666 656 965 389 847 755 510 251 52 × 2 = 1 + 0.333 333 333 313 930 779 695 511 020 503 04;
  • 21) 0.333 333 333 313 930 779 695 511 020 503 04 × 2 = 0 + 0.666 666 666 627 861 559 391 022 041 006 08;
  • 22) 0.666 666 666 627 861 559 391 022 041 006 08 × 2 = 1 + 0.333 333 333 255 723 118 782 044 082 012 16;
  • 23) 0.333 333 333 255 723 118 782 044 082 012 16 × 2 = 0 + 0.666 666 666 511 446 237 564 088 164 024 32;
  • 24) 0.666 666 666 511 446 237 564 088 164 024 32 × 2 = 1 + 0.333 333 333 022 892 475 128 176 328 048 64;
  • 25) 0.333 333 333 022 892 475 128 176 328 048 64 × 2 = 0 + 0.666 666 666 045 784 950 256 352 656 097 28;
  • 26) 0.666 666 666 045 784 950 256 352 656 097 28 × 2 = 1 + 0.333 333 332 091 569 900 512 705 312 194 56;
  • 27) 0.333 333 332 091 569 900 512 705 312 194 56 × 2 = 0 + 0.666 666 664 183 139 801 025 410 624 389 12;
  • 28) 0.666 666 664 183 139 801 025 410 624 389 12 × 2 = 1 + 0.333 333 328 366 279 602 050 821 248 778 24;
  • 29) 0.333 333 328 366 279 602 050 821 248 778 24 × 2 = 0 + 0.666 666 656 732 559 204 101 642 497 556 48;
  • 30) 0.666 666 656 732 559 204 101 642 497 556 48 × 2 = 1 + 0.333 333 313 465 118 408 203 284 995 112 96;
  • 31) 0.333 333 313 465 118 408 203 284 995 112 96 × 2 = 0 + 0.666 666 626 930 236 816 406 569 990 225 92;
  • 32) 0.666 666 626 930 236 816 406 569 990 225 92 × 2 = 1 + 0.333 333 253 860 473 632 813 139 980 451 84;
  • 33) 0.333 333 253 860 473 632 813 139 980 451 84 × 2 = 0 + 0.666 666 507 720 947 265 626 279 960 903 68;
  • 34) 0.666 666 507 720 947 265 626 279 960 903 68 × 2 = 1 + 0.333 333 015 441 894 531 252 559 921 807 36;
  • 35) 0.333 333 015 441 894 531 252 559 921 807 36 × 2 = 0 + 0.666 666 030 883 789 062 505 119 843 614 72;
  • 36) 0.666 666 030 883 789 062 505 119 843 614 72 × 2 = 1 + 0.333 332 061 767 578 125 010 239 687 229 44;
  • 37) 0.333 332 061 767 578 125 010 239 687 229 44 × 2 = 0 + 0.666 664 123 535 156 250 020 479 374 458 88;
  • 38) 0.666 664 123 535 156 250 020 479 374 458 88 × 2 = 1 + 0.333 328 247 070 312 500 040 958 748 917 76;
  • 39) 0.333 328 247 070 312 500 040 958 748 917 76 × 2 = 0 + 0.666 656 494 140 625 000 081 917 497 835 52;
  • 40) 0.666 656 494 140 625 000 081 917 497 835 52 × 2 = 1 + 0.333 312 988 281 250 000 163 834 995 671 04;
  • 41) 0.333 312 988 281 250 000 163 834 995 671 04 × 2 = 0 + 0.666 625 976 562 500 000 327 669 991 342 08;
  • 42) 0.666 625 976 562 500 000 327 669 991 342 08 × 2 = 1 + 0.333 251 953 125 000 000 655 339 982 684 16;
  • 43) 0.333 251 953 125 000 000 655 339 982 684 16 × 2 = 0 + 0.666 503 906 250 000 001 310 679 965 368 32;
  • 44) 0.666 503 906 250 000 001 310 679 965 368 32 × 2 = 1 + 0.333 007 812 500 000 002 621 359 930 736 64;
  • 45) 0.333 007 812 500 000 002 621 359 930 736 64 × 2 = 0 + 0.666 015 625 000 000 005 242 719 861 473 28;
  • 46) 0.666 015 625 000 000 005 242 719 861 473 28 × 2 = 1 + 0.332 031 250 000 000 010 485 439 722 946 56;
  • 47) 0.332 031 250 000 000 010 485 439 722 946 56 × 2 = 0 + 0.664 062 500 000 000 020 970 879 445 893 12;
  • 48) 0.664 062 500 000 000 020 970 879 445 893 12 × 2 = 1 + 0.328 125 000 000 000 041 941 758 891 786 24;
  • 49) 0.328 125 000 000 000 041 941 758 891 786 24 × 2 = 0 + 0.656 250 000 000 000 083 883 517 783 572 48;
  • 50) 0.656 250 000 000 000 083 883 517 783 572 48 × 2 = 1 + 0.312 500 000 000 000 167 767 035 567 144 96;
  • 51) 0.312 500 000 000 000 167 767 035 567 144 96 × 2 = 0 + 0.625 000 000 000 000 335 534 071 134 289 92;
  • 52) 0.625 000 000 000 000 335 534 071 134 289 92 × 2 = 1 + 0.250 000 000 000 000 671 068 142 268 579 84;
  • 53) 0.250 000 000 000 000 671 068 142 268 579 84 × 2 = 0 + 0.500 000 000 000 001 342 136 284 537 159 68;
  • 54) 0.500 000 000 000 001 342 136 284 537 159 68 × 2 = 1 + 0.000 000 000 000 002 684 272 569 074 319 36;

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:


0.333 333 333 333 333 314 829 616 256 247 54(10) =

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

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 247 54(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:


0.333 333 333 333 333 314 829 616 256 247 54(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 0.333 333 333 333 333 314 829 616 256 247 54 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