0.333 333 333 148 296 162 562 473 909 929 394 721 957 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 148 296 162 562 473 909 929 394 721 957(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
0.333 333 333 148 296 162 562 473 909 929 394 721 957(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: 0.333 333 333 148 296 162 562 473 909 929 394 721 957.

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) 0.333 333 333 148 296 162 562 473 909 929 394 721 957 × 2 = 0 + 0.666 666 666 296 592 325 124 947 819 858 789 443 914;
  • 2) 0.666 666 666 296 592 325 124 947 819 858 789 443 914 × 2 = 1 + 0.333 333 332 593 184 650 249 895 639 717 578 887 828;
  • 3) 0.333 333 332 593 184 650 249 895 639 717 578 887 828 × 2 = 0 + 0.666 666 665 186 369 300 499 791 279 435 157 775 656;
  • 4) 0.666 666 665 186 369 300 499 791 279 435 157 775 656 × 2 = 1 + 0.333 333 330 372 738 600 999 582 558 870 315 551 312;
  • 5) 0.333 333 330 372 738 600 999 582 558 870 315 551 312 × 2 = 0 + 0.666 666 660 745 477 201 999 165 117 740 631 102 624;
  • 6) 0.666 666 660 745 477 201 999 165 117 740 631 102 624 × 2 = 1 + 0.333 333 321 490 954 403 998 330 235 481 262 205 248;
  • 7) 0.333 333 321 490 954 403 998 330 235 481 262 205 248 × 2 = 0 + 0.666 666 642 981 908 807 996 660 470 962 524 410 496;
  • 8) 0.666 666 642 981 908 807 996 660 470 962 524 410 496 × 2 = 1 + 0.333 333 285 963 817 615 993 320 941 925 048 820 992;
  • 9) 0.333 333 285 963 817 615 993 320 941 925 048 820 992 × 2 = 0 + 0.666 666 571 927 635 231 986 641 883 850 097 641 984;
  • 10) 0.666 666 571 927 635 231 986 641 883 850 097 641 984 × 2 = 1 + 0.333 333 143 855 270 463 973 283 767 700 195 283 968;
  • 11) 0.333 333 143 855 270 463 973 283 767 700 195 283 968 × 2 = 0 + 0.666 666 287 710 540 927 946 567 535 400 390 567 936;
  • 12) 0.666 666 287 710 540 927 946 567 535 400 390 567 936 × 2 = 1 + 0.333 332 575 421 081 855 893 135 070 800 781 135 872;
  • 13) 0.333 332 575 421 081 855 893 135 070 800 781 135 872 × 2 = 0 + 0.666 665 150 842 163 711 786 270 141 601 562 271 744;
  • 14) 0.666 665 150 842 163 711 786 270 141 601 562 271 744 × 2 = 1 + 0.333 330 301 684 327 423 572 540 283 203 124 543 488;
  • 15) 0.333 330 301 684 327 423 572 540 283 203 124 543 488 × 2 = 0 + 0.666 660 603 368 654 847 145 080 566 406 249 086 976;
  • 16) 0.666 660 603 368 654 847 145 080 566 406 249 086 976 × 2 = 1 + 0.333 321 206 737 309 694 290 161 132 812 498 173 952;
  • 17) 0.333 321 206 737 309 694 290 161 132 812 498 173 952 × 2 = 0 + 0.666 642 413 474 619 388 580 322 265 624 996 347 904;
  • 18) 0.666 642 413 474 619 388 580 322 265 624 996 347 904 × 2 = 1 + 0.333 284 826 949 238 777 160 644 531 249 992 695 808;
  • 19) 0.333 284 826 949 238 777 160 644 531 249 992 695 808 × 2 = 0 + 0.666 569 653 898 477 554 321 289 062 499 985 391 616;
  • 20) 0.666 569 653 898 477 554 321 289 062 499 985 391 616 × 2 = 1 + 0.333 139 307 796 955 108 642 578 124 999 970 783 232;
  • 21) 0.333 139 307 796 955 108 642 578 124 999 970 783 232 × 2 = 0 + 0.666 278 615 593 910 217 285 156 249 999 941 566 464;
  • 22) 0.666 278 615 593 910 217 285 156 249 999 941 566 464 × 2 = 1 + 0.332 557 231 187 820 434 570 312 499 999 883 132 928;
  • 23) 0.332 557 231 187 820 434 570 312 499 999 883 132 928 × 2 = 0 + 0.665 114 462 375 640 869 140 624 999 999 766 265 856;
  • 24) 0.665 114 462 375 640 869 140 624 999 999 766 265 856 × 2 = 1 + 0.330 228 924 751 281 738 281 249 999 999 532 531 712;
  • 25) 0.330 228 924 751 281 738 281 249 999 999 532 531 712 × 2 = 0 + 0.660 457 849 502 563 476 562 499 999 999 065 063 424;
  • 26) 0.660 457 849 502 563 476 562 499 999 999 065 063 424 × 2 = 1 + 0.320 915 699 005 126 953 124 999 999 998 130 126 848;
  • 27) 0.320 915 699 005 126 953 124 999 999 998 130 126 848 × 2 = 0 + 0.641 831 398 010 253 906 249 999 999 996 260 253 696;
  • 28) 0.641 831 398 010 253 906 249 999 999 996 260 253 696 × 2 = 1 + 0.283 662 796 020 507 812 499 999 999 992 520 507 392;
  • 29) 0.283 662 796 020 507 812 499 999 999 992 520 507 392 × 2 = 0 + 0.567 325 592 041 015 624 999 999 999 985 041 014 784;
  • 30) 0.567 325 592 041 015 624 999 999 999 985 041 014 784 × 2 = 1 + 0.134 651 184 082 031 249 999 999 999 970 082 029 568;
  • 31) 0.134 651 184 082 031 249 999 999 999 970 082 029 568 × 2 = 0 + 0.269 302 368 164 062 499 999 999 999 940 164 059 136;
  • 32) 0.269 302 368 164 062 499 999 999 999 940 164 059 136 × 2 = 0 + 0.538 604 736 328 124 999 999 999 999 880 328 118 272;
  • 33) 0.538 604 736 328 124 999 999 999 999 880 328 118 272 × 2 = 1 + 0.077 209 472 656 249 999 999 999 999 760 656 236 544;
  • 34) 0.077 209 472 656 249 999 999 999 999 760 656 236 544 × 2 = 0 + 0.154 418 945 312 499 999 999 999 999 521 312 473 088;
  • 35) 0.154 418 945 312 499 999 999 999 999 521 312 473 088 × 2 = 0 + 0.308 837 890 624 999 999 999 999 999 042 624 946 176;
  • 36) 0.308 837 890 624 999 999 999 999 999 042 624 946 176 × 2 = 0 + 0.617 675 781 249 999 999 999 999 998 085 249 892 352;
  • 37) 0.617 675 781 249 999 999 999 999 998 085 249 892 352 × 2 = 1 + 0.235 351 562 499 999 999 999 999 996 170 499 784 704;
  • 38) 0.235 351 562 499 999 999 999 999 996 170 499 784 704 × 2 = 0 + 0.470 703 124 999 999 999 999 999 992 340 999 569 408;
  • 39) 0.470 703 124 999 999 999 999 999 992 340 999 569 408 × 2 = 0 + 0.941 406 249 999 999 999 999 999 984 681 999 138 816;
  • 40) 0.941 406 249 999 999 999 999 999 984 681 999 138 816 × 2 = 1 + 0.882 812 499 999 999 999 999 999 969 363 998 277 632;
  • 41) 0.882 812 499 999 999 999 999 999 969 363 998 277 632 × 2 = 1 + 0.765 624 999 999 999 999 999 999 938 727 996 555 264;
  • 42) 0.765 624 999 999 999 999 999 999 938 727 996 555 264 × 2 = 1 + 0.531 249 999 999 999 999 999 999 877 455 993 110 528;
  • 43) 0.531 249 999 999 999 999 999 999 877 455 993 110 528 × 2 = 1 + 0.062 499 999 999 999 999 999 999 754 911 986 221 056;
  • 44) 0.062 499 999 999 999 999 999 999 754 911 986 221 056 × 2 = 0 + 0.124 999 999 999 999 999 999 999 509 823 972 442 112;
  • 45) 0.124 999 999 999 999 999 999 999 509 823 972 442 112 × 2 = 0 + 0.249 999 999 999 999 999 999 999 019 647 944 884 224;
  • 46) 0.249 999 999 999 999 999 999 999 019 647 944 884 224 × 2 = 0 + 0.499 999 999 999 999 999 999 998 039 295 889 768 448;
  • 47) 0.499 999 999 999 999 999 999 998 039 295 889 768 448 × 2 = 0 + 0.999 999 999 999 999 999 999 996 078 591 779 536 896;
  • 48) 0.999 999 999 999 999 999 999 996 078 591 779 536 896 × 2 = 1 + 0.999 999 999 999 999 999 999 992 157 183 559 073 792;
  • 49) 0.999 999 999 999 999 999 999 992 157 183 559 073 792 × 2 = 1 + 0.999 999 999 999 999 999 999 984 314 367 118 147 584;
  • 50) 0.999 999 999 999 999 999 999 984 314 367 118 147 584 × 2 = 1 + 0.999 999 999 999 999 999 999 968 628 734 236 295 168;
  • 51) 0.999 999 999 999 999 999 999 968 628 734 236 295 168 × 2 = 1 + 0.999 999 999 999 999 999 999 937 257 468 472 590 336;
  • 52) 0.999 999 999 999 999 999 999 937 257 468 472 590 336 × 2 = 1 + 0.999 999 999 999 999 999 999 874 514 936 945 180 672;
  • 53) 0.999 999 999 999 999 999 999 874 514 936 945 180 672 × 2 = 1 + 0.999 999 999 999 999 999 999 749 029 873 890 361 344;
  • 54) 0.999 999 999 999 999 999 999 749 029 873 890 361 344 × 2 = 1 + 0.999 999 999 999 999 999 999 498 059 747 780 722 688;

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:


0.333 333 333 148 296 162 562 473 909 929 394 721 957(10) =

0.0101 0101 0101 0101 0101 0101 0101 0100 1000 1001 1110 0001 1111 11(2)

5. Positive number before normalization:

0.333 333 333 148 296 162 562 473 909 929 394 721 957(10) =

0.0101 0101 0101 0101 0101 0101 0101 0100 1000 1001 1110 0001 1111 11(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:


0.333 333 333 148 296 162 562 473 909 929 394 721 957(10) =

0.0101 0101 0101 0101 0101 0101 0101 0100 1000 1001 1110 0001 1111 11(2) =

0.0101 0101 0101 0101 0101 0101 0101 0100 1000 1001 1110 0001 1111 11(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0010 0010 0111 1000 0111 1111(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 0010 0010 0111 1000 0111 1111


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 0010 0010 0111 1000 0111 1111 =

0101 0101 0101 0101 0101 0101 0101 0010 0010 0111 1000 0111 1111


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 0010 0010 0111 1000 0111 1111


Decimal number 0.333 333 333 148 296 162 562 473 909 929 394 721 957 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 0101 0101 0101 0101 0101 0101 0101 0010 0010 0111 1000 0111 1111

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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