0.333 333 333 333 333 314 829 616 256 230 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 230 3(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
0.333 333 333 333 333 314 829 616 256 230 3(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: 0.333 333 333 333 333 314 829 616 256 230 3.

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) 0.333 333 333 333 333 314 829 616 256 230 3 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 460 6;
  • 2) 0.666 666 666 666 666 629 659 232 512 460 6 × 2 = 1 + 0.333 333 333 333 333 259 318 465 024 921 2;
  • 3) 0.333 333 333 333 333 259 318 465 024 921 2 × 2 = 0 + 0.666 666 666 666 666 518 636 930 049 842 4;
  • 4) 0.666 666 666 666 666 518 636 930 049 842 4 × 2 = 1 + 0.333 333 333 333 333 037 273 860 099 684 8;
  • 5) 0.333 333 333 333 333 037 273 860 099 684 8 × 2 = 0 + 0.666 666 666 666 666 074 547 720 199 369 6;
  • 6) 0.666 666 666 666 666 074 547 720 199 369 6 × 2 = 1 + 0.333 333 333 333 332 149 095 440 398 739 2;
  • 7) 0.333 333 333 333 332 149 095 440 398 739 2 × 2 = 0 + 0.666 666 666 666 664 298 190 880 797 478 4;
  • 8) 0.666 666 666 666 664 298 190 880 797 478 4 × 2 = 1 + 0.333 333 333 333 328 596 381 761 594 956 8;
  • 9) 0.333 333 333 333 328 596 381 761 594 956 8 × 2 = 0 + 0.666 666 666 666 657 192 763 523 189 913 6;
  • 10) 0.666 666 666 666 657 192 763 523 189 913 6 × 2 = 1 + 0.333 333 333 333 314 385 527 046 379 827 2;
  • 11) 0.333 333 333 333 314 385 527 046 379 827 2 × 2 = 0 + 0.666 666 666 666 628 771 054 092 759 654 4;
  • 12) 0.666 666 666 666 628 771 054 092 759 654 4 × 2 = 1 + 0.333 333 333 333 257 542 108 185 519 308 8;
  • 13) 0.333 333 333 333 257 542 108 185 519 308 8 × 2 = 0 + 0.666 666 666 666 515 084 216 371 038 617 6;
  • 14) 0.666 666 666 666 515 084 216 371 038 617 6 × 2 = 1 + 0.333 333 333 333 030 168 432 742 077 235 2;
  • 15) 0.333 333 333 333 030 168 432 742 077 235 2 × 2 = 0 + 0.666 666 666 666 060 336 865 484 154 470 4;
  • 16) 0.666 666 666 666 060 336 865 484 154 470 4 × 2 = 1 + 0.333 333 333 332 120 673 730 968 308 940 8;
  • 17) 0.333 333 333 332 120 673 730 968 308 940 8 × 2 = 0 + 0.666 666 666 664 241 347 461 936 617 881 6;
  • 18) 0.666 666 666 664 241 347 461 936 617 881 6 × 2 = 1 + 0.333 333 333 328 482 694 923 873 235 763 2;
  • 19) 0.333 333 333 328 482 694 923 873 235 763 2 × 2 = 0 + 0.666 666 666 656 965 389 847 746 471 526 4;
  • 20) 0.666 666 666 656 965 389 847 746 471 526 4 × 2 = 1 + 0.333 333 333 313 930 779 695 492 943 052 8;
  • 21) 0.333 333 333 313 930 779 695 492 943 052 8 × 2 = 0 + 0.666 666 666 627 861 559 390 985 886 105 6;
  • 22) 0.666 666 666 627 861 559 390 985 886 105 6 × 2 = 1 + 0.333 333 333 255 723 118 781 971 772 211 2;
  • 23) 0.333 333 333 255 723 118 781 971 772 211 2 × 2 = 0 + 0.666 666 666 511 446 237 563 943 544 422 4;
  • 24) 0.666 666 666 511 446 237 563 943 544 422 4 × 2 = 1 + 0.333 333 333 022 892 475 127 887 088 844 8;
  • 25) 0.333 333 333 022 892 475 127 887 088 844 8 × 2 = 0 + 0.666 666 666 045 784 950 255 774 177 689 6;
  • 26) 0.666 666 666 045 784 950 255 774 177 689 6 × 2 = 1 + 0.333 333 332 091 569 900 511 548 355 379 2;
  • 27) 0.333 333 332 091 569 900 511 548 355 379 2 × 2 = 0 + 0.666 666 664 183 139 801 023 096 710 758 4;
  • 28) 0.666 666 664 183 139 801 023 096 710 758 4 × 2 = 1 + 0.333 333 328 366 279 602 046 193 421 516 8;
  • 29) 0.333 333 328 366 279 602 046 193 421 516 8 × 2 = 0 + 0.666 666 656 732 559 204 092 386 843 033 6;
  • 30) 0.666 666 656 732 559 204 092 386 843 033 6 × 2 = 1 + 0.333 333 313 465 118 408 184 773 686 067 2;
  • 31) 0.333 333 313 465 118 408 184 773 686 067 2 × 2 = 0 + 0.666 666 626 930 236 816 369 547 372 134 4;
  • 32) 0.666 666 626 930 236 816 369 547 372 134 4 × 2 = 1 + 0.333 333 253 860 473 632 739 094 744 268 8;
  • 33) 0.333 333 253 860 473 632 739 094 744 268 8 × 2 = 0 + 0.666 666 507 720 947 265 478 189 488 537 6;
  • 34) 0.666 666 507 720 947 265 478 189 488 537 6 × 2 = 1 + 0.333 333 015 441 894 530 956 378 977 075 2;
  • 35) 0.333 333 015 441 894 530 956 378 977 075 2 × 2 = 0 + 0.666 666 030 883 789 061 912 757 954 150 4;
  • 36) 0.666 666 030 883 789 061 912 757 954 150 4 × 2 = 1 + 0.333 332 061 767 578 123 825 515 908 300 8;
  • 37) 0.333 332 061 767 578 123 825 515 908 300 8 × 2 = 0 + 0.666 664 123 535 156 247 651 031 816 601 6;
  • 38) 0.666 664 123 535 156 247 651 031 816 601 6 × 2 = 1 + 0.333 328 247 070 312 495 302 063 633 203 2;
  • 39) 0.333 328 247 070 312 495 302 063 633 203 2 × 2 = 0 + 0.666 656 494 140 624 990 604 127 266 406 4;
  • 40) 0.666 656 494 140 624 990 604 127 266 406 4 × 2 = 1 + 0.333 312 988 281 249 981 208 254 532 812 8;
  • 41) 0.333 312 988 281 249 981 208 254 532 812 8 × 2 = 0 + 0.666 625 976 562 499 962 416 509 065 625 6;
  • 42) 0.666 625 976 562 499 962 416 509 065 625 6 × 2 = 1 + 0.333 251 953 124 999 924 833 018 131 251 2;
  • 43) 0.333 251 953 124 999 924 833 018 131 251 2 × 2 = 0 + 0.666 503 906 249 999 849 666 036 262 502 4;
  • 44) 0.666 503 906 249 999 849 666 036 262 502 4 × 2 = 1 + 0.333 007 812 499 999 699 332 072 525 004 8;
  • 45) 0.333 007 812 499 999 699 332 072 525 004 8 × 2 = 0 + 0.666 015 624 999 999 398 664 145 050 009 6;
  • 46) 0.666 015 624 999 999 398 664 145 050 009 6 × 2 = 1 + 0.332 031 249 999 998 797 328 290 100 019 2;
  • 47) 0.332 031 249 999 998 797 328 290 100 019 2 × 2 = 0 + 0.664 062 499 999 997 594 656 580 200 038 4;
  • 48) 0.664 062 499 999 997 594 656 580 200 038 4 × 2 = 1 + 0.328 124 999 999 995 189 313 160 400 076 8;
  • 49) 0.328 124 999 999 995 189 313 160 400 076 8 × 2 = 0 + 0.656 249 999 999 990 378 626 320 800 153 6;
  • 50) 0.656 249 999 999 990 378 626 320 800 153 6 × 2 = 1 + 0.312 499 999 999 980 757 252 641 600 307 2;
  • 51) 0.312 499 999 999 980 757 252 641 600 307 2 × 2 = 0 + 0.624 999 999 999 961 514 505 283 200 614 4;
  • 52) 0.624 999 999 999 961 514 505 283 200 614 4 × 2 = 1 + 0.249 999 999 999 923 029 010 566 401 228 8;
  • 53) 0.249 999 999 999 923 029 010 566 401 228 8 × 2 = 0 + 0.499 999 999 999 846 058 021 132 802 457 6;
  • 54) 0.499 999 999 999 846 058 021 132 802 457 6 × 2 = 0 + 0.999 999 999 999 692 116 042 265 604 915 2;

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:


0.333 333 333 333 333 314 829 616 256 230 3(10) =

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

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 230 3(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:


0.333 333 333 333 333 314 829 616 256 230 3(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 0.333 333 333 333 333 314 829 616 256 230 3 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