0.333 333 333 333 333 314 829 616 256 15 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 15(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 15(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 15.

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 15 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 3;
  • 2) 0.666 666 666 666 666 629 659 232 512 3 × 2 = 1 + 0.333 333 333 333 333 259 318 465 024 6;
  • 3) 0.333 333 333 333 333 259 318 465 024 6 × 2 = 0 + 0.666 666 666 666 666 518 636 930 049 2;
  • 4) 0.666 666 666 666 666 518 636 930 049 2 × 2 = 1 + 0.333 333 333 333 333 037 273 860 098 4;
  • 5) 0.333 333 333 333 333 037 273 860 098 4 × 2 = 0 + 0.666 666 666 666 666 074 547 720 196 8;
  • 6) 0.666 666 666 666 666 074 547 720 196 8 × 2 = 1 + 0.333 333 333 333 332 149 095 440 393 6;
  • 7) 0.333 333 333 333 332 149 095 440 393 6 × 2 = 0 + 0.666 666 666 666 664 298 190 880 787 2;
  • 8) 0.666 666 666 666 664 298 190 880 787 2 × 2 = 1 + 0.333 333 333 333 328 596 381 761 574 4;
  • 9) 0.333 333 333 333 328 596 381 761 574 4 × 2 = 0 + 0.666 666 666 666 657 192 763 523 148 8;
  • 10) 0.666 666 666 666 657 192 763 523 148 8 × 2 = 1 + 0.333 333 333 333 314 385 527 046 297 6;
  • 11) 0.333 333 333 333 314 385 527 046 297 6 × 2 = 0 + 0.666 666 666 666 628 771 054 092 595 2;
  • 12) 0.666 666 666 666 628 771 054 092 595 2 × 2 = 1 + 0.333 333 333 333 257 542 108 185 190 4;
  • 13) 0.333 333 333 333 257 542 108 185 190 4 × 2 = 0 + 0.666 666 666 666 515 084 216 370 380 8;
  • 14) 0.666 666 666 666 515 084 216 370 380 8 × 2 = 1 + 0.333 333 333 333 030 168 432 740 761 6;
  • 15) 0.333 333 333 333 030 168 432 740 761 6 × 2 = 0 + 0.666 666 666 666 060 336 865 481 523 2;
  • 16) 0.666 666 666 666 060 336 865 481 523 2 × 2 = 1 + 0.333 333 333 332 120 673 730 963 046 4;
  • 17) 0.333 333 333 332 120 673 730 963 046 4 × 2 = 0 + 0.666 666 666 664 241 347 461 926 092 8;
  • 18) 0.666 666 666 664 241 347 461 926 092 8 × 2 = 1 + 0.333 333 333 328 482 694 923 852 185 6;
  • 19) 0.333 333 333 328 482 694 923 852 185 6 × 2 = 0 + 0.666 666 666 656 965 389 847 704 371 2;
  • 20) 0.666 666 666 656 965 389 847 704 371 2 × 2 = 1 + 0.333 333 333 313 930 779 695 408 742 4;
  • 21) 0.333 333 333 313 930 779 695 408 742 4 × 2 = 0 + 0.666 666 666 627 861 559 390 817 484 8;
  • 22) 0.666 666 666 627 861 559 390 817 484 8 × 2 = 1 + 0.333 333 333 255 723 118 781 634 969 6;
  • 23) 0.333 333 333 255 723 118 781 634 969 6 × 2 = 0 + 0.666 666 666 511 446 237 563 269 939 2;
  • 24) 0.666 666 666 511 446 237 563 269 939 2 × 2 = 1 + 0.333 333 333 022 892 475 126 539 878 4;
  • 25) 0.333 333 333 022 892 475 126 539 878 4 × 2 = 0 + 0.666 666 666 045 784 950 253 079 756 8;
  • 26) 0.666 666 666 045 784 950 253 079 756 8 × 2 = 1 + 0.333 333 332 091 569 900 506 159 513 6;
  • 27) 0.333 333 332 091 569 900 506 159 513 6 × 2 = 0 + 0.666 666 664 183 139 801 012 319 027 2;
  • 28) 0.666 666 664 183 139 801 012 319 027 2 × 2 = 1 + 0.333 333 328 366 279 602 024 638 054 4;
  • 29) 0.333 333 328 366 279 602 024 638 054 4 × 2 = 0 + 0.666 666 656 732 559 204 049 276 108 8;
  • 30) 0.666 666 656 732 559 204 049 276 108 8 × 2 = 1 + 0.333 333 313 465 118 408 098 552 217 6;
  • 31) 0.333 333 313 465 118 408 098 552 217 6 × 2 = 0 + 0.666 666 626 930 236 816 197 104 435 2;
  • 32) 0.666 666 626 930 236 816 197 104 435 2 × 2 = 1 + 0.333 333 253 860 473 632 394 208 870 4;
  • 33) 0.333 333 253 860 473 632 394 208 870 4 × 2 = 0 + 0.666 666 507 720 947 264 788 417 740 8;
  • 34) 0.666 666 507 720 947 264 788 417 740 8 × 2 = 1 + 0.333 333 015 441 894 529 576 835 481 6;
  • 35) 0.333 333 015 441 894 529 576 835 481 6 × 2 = 0 + 0.666 666 030 883 789 059 153 670 963 2;
  • 36) 0.666 666 030 883 789 059 153 670 963 2 × 2 = 1 + 0.333 332 061 767 578 118 307 341 926 4;
  • 37) 0.333 332 061 767 578 118 307 341 926 4 × 2 = 0 + 0.666 664 123 535 156 236 614 683 852 8;
  • 38) 0.666 664 123 535 156 236 614 683 852 8 × 2 = 1 + 0.333 328 247 070 312 473 229 367 705 6;
  • 39) 0.333 328 247 070 312 473 229 367 705 6 × 2 = 0 + 0.666 656 494 140 624 946 458 735 411 2;
  • 40) 0.666 656 494 140 624 946 458 735 411 2 × 2 = 1 + 0.333 312 988 281 249 892 917 470 822 4;
  • 41) 0.333 312 988 281 249 892 917 470 822 4 × 2 = 0 + 0.666 625 976 562 499 785 834 941 644 8;
  • 42) 0.666 625 976 562 499 785 834 941 644 8 × 2 = 1 + 0.333 251 953 124 999 571 669 883 289 6;
  • 43) 0.333 251 953 124 999 571 669 883 289 6 × 2 = 0 + 0.666 503 906 249 999 143 339 766 579 2;
  • 44) 0.666 503 906 249 999 143 339 766 579 2 × 2 = 1 + 0.333 007 812 499 998 286 679 533 158 4;
  • 45) 0.333 007 812 499 998 286 679 533 158 4 × 2 = 0 + 0.666 015 624 999 996 573 359 066 316 8;
  • 46) 0.666 015 624 999 996 573 359 066 316 8 × 2 = 1 + 0.332 031 249 999 993 146 718 132 633 6;
  • 47) 0.332 031 249 999 993 146 718 132 633 6 × 2 = 0 + 0.664 062 499 999 986 293 436 265 267 2;
  • 48) 0.664 062 499 999 986 293 436 265 267 2 × 2 = 1 + 0.328 124 999 999 972 586 872 530 534 4;
  • 49) 0.328 124 999 999 972 586 872 530 534 4 × 2 = 0 + 0.656 249 999 999 945 173 745 061 068 8;
  • 50) 0.656 249 999 999 945 173 745 061 068 8 × 2 = 1 + 0.312 499 999 999 890 347 490 122 137 6;
  • 51) 0.312 499 999 999 890 347 490 122 137 6 × 2 = 0 + 0.624 999 999 999 780 694 980 244 275 2;
  • 52) 0.624 999 999 999 780 694 980 244 275 2 × 2 = 1 + 0.249 999 999 999 561 389 960 488 550 4;
  • 53) 0.249 999 999 999 561 389 960 488 550 4 × 2 = 0 + 0.499 999 999 999 122 779 920 977 100 8;
  • 54) 0.499 999 999 999 122 779 920 977 100 8 × 2 = 0 + 0.999 999 999 998 245 559 841 954 201 6;

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 15(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 15(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 15(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 15 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