0.333 333 333 333 333 314 829 616 256 308 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 308(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
0.333 333 333 333 333 314 829 616 256 308(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: 0.333 333 333 333 333 314 829 616 256 308.

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) 0.333 333 333 333 333 314 829 616 256 308 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 616;
  • 2) 0.666 666 666 666 666 629 659 232 512 616 × 2 = 1 + 0.333 333 333 333 333 259 318 465 025 232;
  • 3) 0.333 333 333 333 333 259 318 465 025 232 × 2 = 0 + 0.666 666 666 666 666 518 636 930 050 464;
  • 4) 0.666 666 666 666 666 518 636 930 050 464 × 2 = 1 + 0.333 333 333 333 333 037 273 860 100 928;
  • 5) 0.333 333 333 333 333 037 273 860 100 928 × 2 = 0 + 0.666 666 666 666 666 074 547 720 201 856;
  • 6) 0.666 666 666 666 666 074 547 720 201 856 × 2 = 1 + 0.333 333 333 333 332 149 095 440 403 712;
  • 7) 0.333 333 333 333 332 149 095 440 403 712 × 2 = 0 + 0.666 666 666 666 664 298 190 880 807 424;
  • 8) 0.666 666 666 666 664 298 190 880 807 424 × 2 = 1 + 0.333 333 333 333 328 596 381 761 614 848;
  • 9) 0.333 333 333 333 328 596 381 761 614 848 × 2 = 0 + 0.666 666 666 666 657 192 763 523 229 696;
  • 10) 0.666 666 666 666 657 192 763 523 229 696 × 2 = 1 + 0.333 333 333 333 314 385 527 046 459 392;
  • 11) 0.333 333 333 333 314 385 527 046 459 392 × 2 = 0 + 0.666 666 666 666 628 771 054 092 918 784;
  • 12) 0.666 666 666 666 628 771 054 092 918 784 × 2 = 1 + 0.333 333 333 333 257 542 108 185 837 568;
  • 13) 0.333 333 333 333 257 542 108 185 837 568 × 2 = 0 + 0.666 666 666 666 515 084 216 371 675 136;
  • 14) 0.666 666 666 666 515 084 216 371 675 136 × 2 = 1 + 0.333 333 333 333 030 168 432 743 350 272;
  • 15) 0.333 333 333 333 030 168 432 743 350 272 × 2 = 0 + 0.666 666 666 666 060 336 865 486 700 544;
  • 16) 0.666 666 666 666 060 336 865 486 700 544 × 2 = 1 + 0.333 333 333 332 120 673 730 973 401 088;
  • 17) 0.333 333 333 332 120 673 730 973 401 088 × 2 = 0 + 0.666 666 666 664 241 347 461 946 802 176;
  • 18) 0.666 666 666 664 241 347 461 946 802 176 × 2 = 1 + 0.333 333 333 328 482 694 923 893 604 352;
  • 19) 0.333 333 333 328 482 694 923 893 604 352 × 2 = 0 + 0.666 666 666 656 965 389 847 787 208 704;
  • 20) 0.666 666 666 656 965 389 847 787 208 704 × 2 = 1 + 0.333 333 333 313 930 779 695 574 417 408;
  • 21) 0.333 333 333 313 930 779 695 574 417 408 × 2 = 0 + 0.666 666 666 627 861 559 391 148 834 816;
  • 22) 0.666 666 666 627 861 559 391 148 834 816 × 2 = 1 + 0.333 333 333 255 723 118 782 297 669 632;
  • 23) 0.333 333 333 255 723 118 782 297 669 632 × 2 = 0 + 0.666 666 666 511 446 237 564 595 339 264;
  • 24) 0.666 666 666 511 446 237 564 595 339 264 × 2 = 1 + 0.333 333 333 022 892 475 129 190 678 528;
  • 25) 0.333 333 333 022 892 475 129 190 678 528 × 2 = 0 + 0.666 666 666 045 784 950 258 381 357 056;
  • 26) 0.666 666 666 045 784 950 258 381 357 056 × 2 = 1 + 0.333 333 332 091 569 900 516 762 714 112;
  • 27) 0.333 333 332 091 569 900 516 762 714 112 × 2 = 0 + 0.666 666 664 183 139 801 033 525 428 224;
  • 28) 0.666 666 664 183 139 801 033 525 428 224 × 2 = 1 + 0.333 333 328 366 279 602 067 050 856 448;
  • 29) 0.333 333 328 366 279 602 067 050 856 448 × 2 = 0 + 0.666 666 656 732 559 204 134 101 712 896;
  • 30) 0.666 666 656 732 559 204 134 101 712 896 × 2 = 1 + 0.333 333 313 465 118 408 268 203 425 792;
  • 31) 0.333 333 313 465 118 408 268 203 425 792 × 2 = 0 + 0.666 666 626 930 236 816 536 406 851 584;
  • 32) 0.666 666 626 930 236 816 536 406 851 584 × 2 = 1 + 0.333 333 253 860 473 633 072 813 703 168;
  • 33) 0.333 333 253 860 473 633 072 813 703 168 × 2 = 0 + 0.666 666 507 720 947 266 145 627 406 336;
  • 34) 0.666 666 507 720 947 266 145 627 406 336 × 2 = 1 + 0.333 333 015 441 894 532 291 254 812 672;
  • 35) 0.333 333 015 441 894 532 291 254 812 672 × 2 = 0 + 0.666 666 030 883 789 064 582 509 625 344;
  • 36) 0.666 666 030 883 789 064 582 509 625 344 × 2 = 1 + 0.333 332 061 767 578 129 165 019 250 688;
  • 37) 0.333 332 061 767 578 129 165 019 250 688 × 2 = 0 + 0.666 664 123 535 156 258 330 038 501 376;
  • 38) 0.666 664 123 535 156 258 330 038 501 376 × 2 = 1 + 0.333 328 247 070 312 516 660 077 002 752;
  • 39) 0.333 328 247 070 312 516 660 077 002 752 × 2 = 0 + 0.666 656 494 140 625 033 320 154 005 504;
  • 40) 0.666 656 494 140 625 033 320 154 005 504 × 2 = 1 + 0.333 312 988 281 250 066 640 308 011 008;
  • 41) 0.333 312 988 281 250 066 640 308 011 008 × 2 = 0 + 0.666 625 976 562 500 133 280 616 022 016;
  • 42) 0.666 625 976 562 500 133 280 616 022 016 × 2 = 1 + 0.333 251 953 125 000 266 561 232 044 032;
  • 43) 0.333 251 953 125 000 266 561 232 044 032 × 2 = 0 + 0.666 503 906 250 000 533 122 464 088 064;
  • 44) 0.666 503 906 250 000 533 122 464 088 064 × 2 = 1 + 0.333 007 812 500 001 066 244 928 176 128;
  • 45) 0.333 007 812 500 001 066 244 928 176 128 × 2 = 0 + 0.666 015 625 000 002 132 489 856 352 256;
  • 46) 0.666 015 625 000 002 132 489 856 352 256 × 2 = 1 + 0.332 031 250 000 004 264 979 712 704 512;
  • 47) 0.332 031 250 000 004 264 979 712 704 512 × 2 = 0 + 0.664 062 500 000 008 529 959 425 409 024;
  • 48) 0.664 062 500 000 008 529 959 425 409 024 × 2 = 1 + 0.328 125 000 000 017 059 918 850 818 048;
  • 49) 0.328 125 000 000 017 059 918 850 818 048 × 2 = 0 + 0.656 250 000 000 034 119 837 701 636 096;
  • 50) 0.656 250 000 000 034 119 837 701 636 096 × 2 = 1 + 0.312 500 000 000 068 239 675 403 272 192;
  • 51) 0.312 500 000 000 068 239 675 403 272 192 × 2 = 0 + 0.625 000 000 000 136 479 350 806 544 384;
  • 52) 0.625 000 000 000 136 479 350 806 544 384 × 2 = 1 + 0.250 000 000 000 272 958 701 613 088 768;
  • 53) 0.250 000 000 000 272 958 701 613 088 768 × 2 = 0 + 0.500 000 000 000 545 917 403 226 177 536;
  • 54) 0.500 000 000 000 545 917 403 226 177 536 × 2 = 1 + 0.000 000 000 001 091 834 806 452 355 072;

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:


0.333 333 333 333 333 314 829 616 256 308(10) =

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

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 308(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:


0.333 333 333 333 333 314 829 616 256 308(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 0.333 333 333 333 333 314 829 616 256 308 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