0.333 333 333 333 333 314 829 616 256 247 351 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 247 351(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
0.333 333 333 333 333 314 829 616 256 247 351(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: 0.333 333 333 333 333 314 829 616 256 247 351.

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) 0.333 333 333 333 333 314 829 616 256 247 351 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 494 702;
  • 2) 0.666 666 666 666 666 629 659 232 512 494 702 × 2 = 1 + 0.333 333 333 333 333 259 318 465 024 989 404;
  • 3) 0.333 333 333 333 333 259 318 465 024 989 404 × 2 = 0 + 0.666 666 666 666 666 518 636 930 049 978 808;
  • 4) 0.666 666 666 666 666 518 636 930 049 978 808 × 2 = 1 + 0.333 333 333 333 333 037 273 860 099 957 616;
  • 5) 0.333 333 333 333 333 037 273 860 099 957 616 × 2 = 0 + 0.666 666 666 666 666 074 547 720 199 915 232;
  • 6) 0.666 666 666 666 666 074 547 720 199 915 232 × 2 = 1 + 0.333 333 333 333 332 149 095 440 399 830 464;
  • 7) 0.333 333 333 333 332 149 095 440 399 830 464 × 2 = 0 + 0.666 666 666 666 664 298 190 880 799 660 928;
  • 8) 0.666 666 666 666 664 298 190 880 799 660 928 × 2 = 1 + 0.333 333 333 333 328 596 381 761 599 321 856;
  • 9) 0.333 333 333 333 328 596 381 761 599 321 856 × 2 = 0 + 0.666 666 666 666 657 192 763 523 198 643 712;
  • 10) 0.666 666 666 666 657 192 763 523 198 643 712 × 2 = 1 + 0.333 333 333 333 314 385 527 046 397 287 424;
  • 11) 0.333 333 333 333 314 385 527 046 397 287 424 × 2 = 0 + 0.666 666 666 666 628 771 054 092 794 574 848;
  • 12) 0.666 666 666 666 628 771 054 092 794 574 848 × 2 = 1 + 0.333 333 333 333 257 542 108 185 589 149 696;
  • 13) 0.333 333 333 333 257 542 108 185 589 149 696 × 2 = 0 + 0.666 666 666 666 515 084 216 371 178 299 392;
  • 14) 0.666 666 666 666 515 084 216 371 178 299 392 × 2 = 1 + 0.333 333 333 333 030 168 432 742 356 598 784;
  • 15) 0.333 333 333 333 030 168 432 742 356 598 784 × 2 = 0 + 0.666 666 666 666 060 336 865 484 713 197 568;
  • 16) 0.666 666 666 666 060 336 865 484 713 197 568 × 2 = 1 + 0.333 333 333 332 120 673 730 969 426 395 136;
  • 17) 0.333 333 333 332 120 673 730 969 426 395 136 × 2 = 0 + 0.666 666 666 664 241 347 461 938 852 790 272;
  • 18) 0.666 666 666 664 241 347 461 938 852 790 272 × 2 = 1 + 0.333 333 333 328 482 694 923 877 705 580 544;
  • 19) 0.333 333 333 328 482 694 923 877 705 580 544 × 2 = 0 + 0.666 666 666 656 965 389 847 755 411 161 088;
  • 20) 0.666 666 666 656 965 389 847 755 411 161 088 × 2 = 1 + 0.333 333 333 313 930 779 695 510 822 322 176;
  • 21) 0.333 333 333 313 930 779 695 510 822 322 176 × 2 = 0 + 0.666 666 666 627 861 559 391 021 644 644 352;
  • 22) 0.666 666 666 627 861 559 391 021 644 644 352 × 2 = 1 + 0.333 333 333 255 723 118 782 043 289 288 704;
  • 23) 0.333 333 333 255 723 118 782 043 289 288 704 × 2 = 0 + 0.666 666 666 511 446 237 564 086 578 577 408;
  • 24) 0.666 666 666 511 446 237 564 086 578 577 408 × 2 = 1 + 0.333 333 333 022 892 475 128 173 157 154 816;
  • 25) 0.333 333 333 022 892 475 128 173 157 154 816 × 2 = 0 + 0.666 666 666 045 784 950 256 346 314 309 632;
  • 26) 0.666 666 666 045 784 950 256 346 314 309 632 × 2 = 1 + 0.333 333 332 091 569 900 512 692 628 619 264;
  • 27) 0.333 333 332 091 569 900 512 692 628 619 264 × 2 = 0 + 0.666 666 664 183 139 801 025 385 257 238 528;
  • 28) 0.666 666 664 183 139 801 025 385 257 238 528 × 2 = 1 + 0.333 333 328 366 279 602 050 770 514 477 056;
  • 29) 0.333 333 328 366 279 602 050 770 514 477 056 × 2 = 0 + 0.666 666 656 732 559 204 101 541 028 954 112;
  • 30) 0.666 666 656 732 559 204 101 541 028 954 112 × 2 = 1 + 0.333 333 313 465 118 408 203 082 057 908 224;
  • 31) 0.333 333 313 465 118 408 203 082 057 908 224 × 2 = 0 + 0.666 666 626 930 236 816 406 164 115 816 448;
  • 32) 0.666 666 626 930 236 816 406 164 115 816 448 × 2 = 1 + 0.333 333 253 860 473 632 812 328 231 632 896;
  • 33) 0.333 333 253 860 473 632 812 328 231 632 896 × 2 = 0 + 0.666 666 507 720 947 265 624 656 463 265 792;
  • 34) 0.666 666 507 720 947 265 624 656 463 265 792 × 2 = 1 + 0.333 333 015 441 894 531 249 312 926 531 584;
  • 35) 0.333 333 015 441 894 531 249 312 926 531 584 × 2 = 0 + 0.666 666 030 883 789 062 498 625 853 063 168;
  • 36) 0.666 666 030 883 789 062 498 625 853 063 168 × 2 = 1 + 0.333 332 061 767 578 124 997 251 706 126 336;
  • 37) 0.333 332 061 767 578 124 997 251 706 126 336 × 2 = 0 + 0.666 664 123 535 156 249 994 503 412 252 672;
  • 38) 0.666 664 123 535 156 249 994 503 412 252 672 × 2 = 1 + 0.333 328 247 070 312 499 989 006 824 505 344;
  • 39) 0.333 328 247 070 312 499 989 006 824 505 344 × 2 = 0 + 0.666 656 494 140 624 999 978 013 649 010 688;
  • 40) 0.666 656 494 140 624 999 978 013 649 010 688 × 2 = 1 + 0.333 312 988 281 249 999 956 027 298 021 376;
  • 41) 0.333 312 988 281 249 999 956 027 298 021 376 × 2 = 0 + 0.666 625 976 562 499 999 912 054 596 042 752;
  • 42) 0.666 625 976 562 499 999 912 054 596 042 752 × 2 = 1 + 0.333 251 953 124 999 999 824 109 192 085 504;
  • 43) 0.333 251 953 124 999 999 824 109 192 085 504 × 2 = 0 + 0.666 503 906 249 999 999 648 218 384 171 008;
  • 44) 0.666 503 906 249 999 999 648 218 384 171 008 × 2 = 1 + 0.333 007 812 499 999 999 296 436 768 342 016;
  • 45) 0.333 007 812 499 999 999 296 436 768 342 016 × 2 = 0 + 0.666 015 624 999 999 998 592 873 536 684 032;
  • 46) 0.666 015 624 999 999 998 592 873 536 684 032 × 2 = 1 + 0.332 031 249 999 999 997 185 747 073 368 064;
  • 47) 0.332 031 249 999 999 997 185 747 073 368 064 × 2 = 0 + 0.664 062 499 999 999 994 371 494 146 736 128;
  • 48) 0.664 062 499 999 999 994 371 494 146 736 128 × 2 = 1 + 0.328 124 999 999 999 988 742 988 293 472 256;
  • 49) 0.328 124 999 999 999 988 742 988 293 472 256 × 2 = 0 + 0.656 249 999 999 999 977 485 976 586 944 512;
  • 50) 0.656 249 999 999 999 977 485 976 586 944 512 × 2 = 1 + 0.312 499 999 999 999 954 971 953 173 889 024;
  • 51) 0.312 499 999 999 999 954 971 953 173 889 024 × 2 = 0 + 0.624 999 999 999 999 909 943 906 347 778 048;
  • 52) 0.624 999 999 999 999 909 943 906 347 778 048 × 2 = 1 + 0.249 999 999 999 999 819 887 812 695 556 096;
  • 53) 0.249 999 999 999 999 819 887 812 695 556 096 × 2 = 0 + 0.499 999 999 999 999 639 775 625 391 112 192;
  • 54) 0.499 999 999 999 999 639 775 625 391 112 192 × 2 = 0 + 0.999 999 999 999 999 279 551 250 782 224 384;

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:


0.333 333 333 333 333 314 829 616 256 247 351(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(2)

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 247 351(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(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:


0.333 333 333 333 333 314 829 616 256 247 351(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(2) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100(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 0100


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 0100 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100


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 0100


Decimal number 0.333 333 333 333 333 314 829 616 256 247 351 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 0100

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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