0.333 333 333 333 333 314 829 616 256 250 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 250 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
0.333 333 333 333 333 314 829 616 256 250 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: 0.333 333 333 333 333 314 829 616 256 250 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) 0.333 333 333 333 333 314 829 616 256 250 3 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 500 6;
  • 2) 0.666 666 666 666 666 629 659 232 512 500 6 × 2 = 1 + 0.333 333 333 333 333 259 318 465 025 001 2;
  • 3) 0.333 333 333 333 333 259 318 465 025 001 2 × 2 = 0 + 0.666 666 666 666 666 518 636 930 050 002 4;
  • 4) 0.666 666 666 666 666 518 636 930 050 002 4 × 2 = 1 + 0.333 333 333 333 333 037 273 860 100 004 8;
  • 5) 0.333 333 333 333 333 037 273 860 100 004 8 × 2 = 0 + 0.666 666 666 666 666 074 547 720 200 009 6;
  • 6) 0.666 666 666 666 666 074 547 720 200 009 6 × 2 = 1 + 0.333 333 333 333 332 149 095 440 400 019 2;
  • 7) 0.333 333 333 333 332 149 095 440 400 019 2 × 2 = 0 + 0.666 666 666 666 664 298 190 880 800 038 4;
  • 8) 0.666 666 666 666 664 298 190 880 800 038 4 × 2 = 1 + 0.333 333 333 333 328 596 381 761 600 076 8;
  • 9) 0.333 333 333 333 328 596 381 761 600 076 8 × 2 = 0 + 0.666 666 666 666 657 192 763 523 200 153 6;
  • 10) 0.666 666 666 666 657 192 763 523 200 153 6 × 2 = 1 + 0.333 333 333 333 314 385 527 046 400 307 2;
  • 11) 0.333 333 333 333 314 385 527 046 400 307 2 × 2 = 0 + 0.666 666 666 666 628 771 054 092 800 614 4;
  • 12) 0.666 666 666 666 628 771 054 092 800 614 4 × 2 = 1 + 0.333 333 333 333 257 542 108 185 601 228 8;
  • 13) 0.333 333 333 333 257 542 108 185 601 228 8 × 2 = 0 + 0.666 666 666 666 515 084 216 371 202 457 6;
  • 14) 0.666 666 666 666 515 084 216 371 202 457 6 × 2 = 1 + 0.333 333 333 333 030 168 432 742 404 915 2;
  • 15) 0.333 333 333 333 030 168 432 742 404 915 2 × 2 = 0 + 0.666 666 666 666 060 336 865 484 809 830 4;
  • 16) 0.666 666 666 666 060 336 865 484 809 830 4 × 2 = 1 + 0.333 333 333 332 120 673 730 969 619 660 8;
  • 17) 0.333 333 333 332 120 673 730 969 619 660 8 × 2 = 0 + 0.666 666 666 664 241 347 461 939 239 321 6;
  • 18) 0.666 666 666 664 241 347 461 939 239 321 6 × 2 = 1 + 0.333 333 333 328 482 694 923 878 478 643 2;
  • 19) 0.333 333 333 328 482 694 923 878 478 643 2 × 2 = 0 + 0.666 666 666 656 965 389 847 756 957 286 4;
  • 20) 0.666 666 666 656 965 389 847 756 957 286 4 × 2 = 1 + 0.333 333 333 313 930 779 695 513 914 572 8;
  • 21) 0.333 333 333 313 930 779 695 513 914 572 8 × 2 = 0 + 0.666 666 666 627 861 559 391 027 829 145 6;
  • 22) 0.666 666 666 627 861 559 391 027 829 145 6 × 2 = 1 + 0.333 333 333 255 723 118 782 055 658 291 2;
  • 23) 0.333 333 333 255 723 118 782 055 658 291 2 × 2 = 0 + 0.666 666 666 511 446 237 564 111 316 582 4;
  • 24) 0.666 666 666 511 446 237 564 111 316 582 4 × 2 = 1 + 0.333 333 333 022 892 475 128 222 633 164 8;
  • 25) 0.333 333 333 022 892 475 128 222 633 164 8 × 2 = 0 + 0.666 666 666 045 784 950 256 445 266 329 6;
  • 26) 0.666 666 666 045 784 950 256 445 266 329 6 × 2 = 1 + 0.333 333 332 091 569 900 512 890 532 659 2;
  • 27) 0.333 333 332 091 569 900 512 890 532 659 2 × 2 = 0 + 0.666 666 664 183 139 801 025 781 065 318 4;
  • 28) 0.666 666 664 183 139 801 025 781 065 318 4 × 2 = 1 + 0.333 333 328 366 279 602 051 562 130 636 8;
  • 29) 0.333 333 328 366 279 602 051 562 130 636 8 × 2 = 0 + 0.666 666 656 732 559 204 103 124 261 273 6;
  • 30) 0.666 666 656 732 559 204 103 124 261 273 6 × 2 = 1 + 0.333 333 313 465 118 408 206 248 522 547 2;
  • 31) 0.333 333 313 465 118 408 206 248 522 547 2 × 2 = 0 + 0.666 666 626 930 236 816 412 497 045 094 4;
  • 32) 0.666 666 626 930 236 816 412 497 045 094 4 × 2 = 1 + 0.333 333 253 860 473 632 824 994 090 188 8;
  • 33) 0.333 333 253 860 473 632 824 994 090 188 8 × 2 = 0 + 0.666 666 507 720 947 265 649 988 180 377 6;
  • 34) 0.666 666 507 720 947 265 649 988 180 377 6 × 2 = 1 + 0.333 333 015 441 894 531 299 976 360 755 2;
  • 35) 0.333 333 015 441 894 531 299 976 360 755 2 × 2 = 0 + 0.666 666 030 883 789 062 599 952 721 510 4;
  • 36) 0.666 666 030 883 789 062 599 952 721 510 4 × 2 = 1 + 0.333 332 061 767 578 125 199 905 443 020 8;
  • 37) 0.333 332 061 767 578 125 199 905 443 020 8 × 2 = 0 + 0.666 664 123 535 156 250 399 810 886 041 6;
  • 38) 0.666 664 123 535 156 250 399 810 886 041 6 × 2 = 1 + 0.333 328 247 070 312 500 799 621 772 083 2;
  • 39) 0.333 328 247 070 312 500 799 621 772 083 2 × 2 = 0 + 0.666 656 494 140 625 001 599 243 544 166 4;
  • 40) 0.666 656 494 140 625 001 599 243 544 166 4 × 2 = 1 + 0.333 312 988 281 250 003 198 487 088 332 8;
  • 41) 0.333 312 988 281 250 003 198 487 088 332 8 × 2 = 0 + 0.666 625 976 562 500 006 396 974 176 665 6;
  • 42) 0.666 625 976 562 500 006 396 974 176 665 6 × 2 = 1 + 0.333 251 953 125 000 012 793 948 353 331 2;
  • 43) 0.333 251 953 125 000 012 793 948 353 331 2 × 2 = 0 + 0.666 503 906 250 000 025 587 896 706 662 4;
  • 44) 0.666 503 906 250 000 025 587 896 706 662 4 × 2 = 1 + 0.333 007 812 500 000 051 175 793 413 324 8;
  • 45) 0.333 007 812 500 000 051 175 793 413 324 8 × 2 = 0 + 0.666 015 625 000 000 102 351 586 826 649 6;
  • 46) 0.666 015 625 000 000 102 351 586 826 649 6 × 2 = 1 + 0.332 031 250 000 000 204 703 173 653 299 2;
  • 47) 0.332 031 250 000 000 204 703 173 653 299 2 × 2 = 0 + 0.664 062 500 000 000 409 406 347 306 598 4;
  • 48) 0.664 062 500 000 000 409 406 347 306 598 4 × 2 = 1 + 0.328 125 000 000 000 818 812 694 613 196 8;
  • 49) 0.328 125 000 000 000 818 812 694 613 196 8 × 2 = 0 + 0.656 250 000 000 001 637 625 389 226 393 6;
  • 50) 0.656 250 000 000 001 637 625 389 226 393 6 × 2 = 1 + 0.312 500 000 000 003 275 250 778 452 787 2;
  • 51) 0.312 500 000 000 003 275 250 778 452 787 2 × 2 = 0 + 0.625 000 000 000 006 550 501 556 905 574 4;
  • 52) 0.625 000 000 000 006 550 501 556 905 574 4 × 2 = 1 + 0.250 000 000 000 013 101 003 113 811 148 8;
  • 53) 0.250 000 000 000 013 101 003 113 811 148 8 × 2 = 0 + 0.500 000 000 000 026 202 006 227 622 297 6;
  • 54) 0.500 000 000 000 026 202 006 227 622 297 6 × 2 = 1 + 0.000 000 000 000 052 404 012 455 244 595 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:


0.333 333 333 333 333 314 829 616 256 250 3(10) =

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

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 250 3(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:


0.333 333 333 333 333 314 829 616 256 250 3(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 0.333 333 333 333 333 314 829 616 256 250 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 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