64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.000 000 000 232 830 64 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.000 000 000 232 830 64(10) converted and written in 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: 0.000 000 000 232 830 64.

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) 0.000 000 000 232 830 64 × 2 = 0 + 0.000 000 000 465 661 28;
  • 2) 0.000 000 000 465 661 28 × 2 = 0 + 0.000 000 000 931 322 56;
  • 3) 0.000 000 000 931 322 56 × 2 = 0 + 0.000 000 001 862 645 12;
  • 4) 0.000 000 001 862 645 12 × 2 = 0 + 0.000 000 003 725 290 24;
  • 5) 0.000 000 003 725 290 24 × 2 = 0 + 0.000 000 007 450 580 48;
  • 6) 0.000 000 007 450 580 48 × 2 = 0 + 0.000 000 014 901 160 96;
  • 7) 0.000 000 014 901 160 96 × 2 = 0 + 0.000 000 029 802 321 92;
  • 8) 0.000 000 029 802 321 92 × 2 = 0 + 0.000 000 059 604 643 84;
  • 9) 0.000 000 059 604 643 84 × 2 = 0 + 0.000 000 119 209 287 68;
  • 10) 0.000 000 119 209 287 68 × 2 = 0 + 0.000 000 238 418 575 36;
  • 11) 0.000 000 238 418 575 36 × 2 = 0 + 0.000 000 476 837 150 72;
  • 12) 0.000 000 476 837 150 72 × 2 = 0 + 0.000 000 953 674 301 44;
  • 13) 0.000 000 953 674 301 44 × 2 = 0 + 0.000 001 907 348 602 88;
  • 14) 0.000 001 907 348 602 88 × 2 = 0 + 0.000 003 814 697 205 76;
  • 15) 0.000 003 814 697 205 76 × 2 = 0 + 0.000 007 629 394 411 52;
  • 16) 0.000 007 629 394 411 52 × 2 = 0 + 0.000 015 258 788 823 04;
  • 17) 0.000 015 258 788 823 04 × 2 = 0 + 0.000 030 517 577 646 08;
  • 18) 0.000 030 517 577 646 08 × 2 = 0 + 0.000 061 035 155 292 16;
  • 19) 0.000 061 035 155 292 16 × 2 = 0 + 0.000 122 070 310 584 32;
  • 20) 0.000 122 070 310 584 32 × 2 = 0 + 0.000 244 140 621 168 64;
  • 21) 0.000 244 140 621 168 64 × 2 = 0 + 0.000 488 281 242 337 28;
  • 22) 0.000 488 281 242 337 28 × 2 = 0 + 0.000 976 562 484 674 56;
  • 23) 0.000 976 562 484 674 56 × 2 = 0 + 0.001 953 124 969 349 12;
  • 24) 0.001 953 124 969 349 12 × 2 = 0 + 0.003 906 249 938 698 24;
  • 25) 0.003 906 249 938 698 24 × 2 = 0 + 0.007 812 499 877 396 48;
  • 26) 0.007 812 499 877 396 48 × 2 = 0 + 0.015 624 999 754 792 96;
  • 27) 0.015 624 999 754 792 96 × 2 = 0 + 0.031 249 999 509 585 92;
  • 28) 0.031 249 999 509 585 92 × 2 = 0 + 0.062 499 999 019 171 84;
  • 29) 0.062 499 999 019 171 84 × 2 = 0 + 0.124 999 998 038 343 68;
  • 30) 0.124 999 998 038 343 68 × 2 = 0 + 0.249 999 996 076 687 36;
  • 31) 0.249 999 996 076 687 36 × 2 = 0 + 0.499 999 992 153 374 72;
  • 32) 0.499 999 992 153 374 72 × 2 = 0 + 0.999 999 984 306 749 44;
  • 33) 0.999 999 984 306 749 44 × 2 = 1 + 0.999 999 968 613 498 88;
  • 34) 0.999 999 968 613 498 88 × 2 = 1 + 0.999 999 937 226 997 76;
  • 35) 0.999 999 937 226 997 76 × 2 = 1 + 0.999 999 874 453 995 52;
  • 36) 0.999 999 874 453 995 52 × 2 = 1 + 0.999 999 748 907 991 04;
  • 37) 0.999 999 748 907 991 04 × 2 = 1 + 0.999 999 497 815 982 08;
  • 38) 0.999 999 497 815 982 08 × 2 = 1 + 0.999 998 995 631 964 16;
  • 39) 0.999 998 995 631 964 16 × 2 = 1 + 0.999 997 991 263 928 32;
  • 40) 0.999 997 991 263 928 32 × 2 = 1 + 0.999 995 982 527 856 64;
  • 41) 0.999 995 982 527 856 64 × 2 = 1 + 0.999 991 965 055 713 28;
  • 42) 0.999 991 965 055 713 28 × 2 = 1 + 0.999 983 930 111 426 56;
  • 43) 0.999 983 930 111 426 56 × 2 = 1 + 0.999 967 860 222 853 12;
  • 44) 0.999 967 860 222 853 12 × 2 = 1 + 0.999 935 720 445 706 24;
  • 45) 0.999 935 720 445 706 24 × 2 = 1 + 0.999 871 440 891 412 48;
  • 46) 0.999 871 440 891 412 48 × 2 = 1 + 0.999 742 881 782 824 96;
  • 47) 0.999 742 881 782 824 96 × 2 = 1 + 0.999 485 763 565 649 92;
  • 48) 0.999 485 763 565 649 92 × 2 = 1 + 0.998 971 527 131 299 84;
  • 49) 0.998 971 527 131 299 84 × 2 = 1 + 0.997 943 054 262 599 68;
  • 50) 0.997 943 054 262 599 68 × 2 = 1 + 0.995 886 108 525 199 36;
  • 51) 0.995 886 108 525 199 36 × 2 = 1 + 0.991 772 217 050 398 72;
  • 52) 0.991 772 217 050 398 72 × 2 = 1 + 0.983 544 434 100 797 44;
  • 53) 0.983 544 434 100 797 44 × 2 = 1 + 0.967 088 868 201 594 88;
  • 54) 0.967 088 868 201 594 88 × 2 = 1 + 0.934 177 736 403 189 76;
  • 55) 0.934 177 736 403 189 76 × 2 = 1 + 0.868 355 472 806 379 52;
  • 56) 0.868 355 472 806 379 52 × 2 = 1 + 0.736 710 945 612 759 04;
  • 57) 0.736 710 945 612 759 04 × 2 = 1 + 0.473 421 891 225 518 08;
  • 58) 0.473 421 891 225 518 08 × 2 = 0 + 0.946 843 782 451 036 16;
  • 59) 0.946 843 782 451 036 16 × 2 = 1 + 0.893 687 564 902 072 32;
  • 60) 0.893 687 564 902 072 32 × 2 = 1 + 0.787 375 129 804 144 64;
  • 61) 0.787 375 129 804 144 64 × 2 = 1 + 0.574 750 259 608 289 28;
  • 62) 0.574 750 259 608 289 28 × 2 = 1 + 0.149 500 519 216 578 56;
  • 63) 0.149 500 519 216 578 56 × 2 = 0 + 0.299 001 038 433 157 12;
  • 64) 0.299 001 038 433 157 12 × 2 = 0 + 0.598 002 076 866 314 24;
  • 65) 0.598 002 076 866 314 24 × 2 = 1 + 0.196 004 153 732 628 48;
  • 66) 0.196 004 153 732 628 48 × 2 = 0 + 0.392 008 307 465 256 96;
  • 67) 0.392 008 307 465 256 96 × 2 = 0 + 0.784 016 614 930 513 92;
  • 68) 0.784 016 614 930 513 92 × 2 = 1 + 0.568 033 229 861 027 84;
  • 69) 0.568 033 229 861 027 84 × 2 = 1 + 0.136 066 459 722 055 68;
  • 70) 0.136 066 459 722 055 68 × 2 = 0 + 0.272 132 919 444 111 36;
  • 71) 0.272 132 919 444 111 36 × 2 = 0 + 0.544 265 838 888 222 72;
  • 72) 0.544 265 838 888 222 72 × 2 = 1 + 0.088 531 677 776 445 44;
  • 73) 0.088 531 677 776 445 44 × 2 = 0 + 0.177 063 355 552 890 88;
  • 74) 0.177 063 355 552 890 88 × 2 = 0 + 0.354 126 711 105 781 76;
  • 75) 0.354 126 711 105 781 76 × 2 = 0 + 0.708 253 422 211 563 52;
  • 76) 0.708 253 422 211 563 52 × 2 = 1 + 0.416 506 844 423 127 04;
  • 77) 0.416 506 844 423 127 04 × 2 = 0 + 0.833 013 688 846 254 08;
  • 78) 0.833 013 688 846 254 08 × 2 = 1 + 0.666 027 377 692 508 16;
  • 79) 0.666 027 377 692 508 16 × 2 = 1 + 0.332 054 755 385 016 32;
  • 80) 0.332 054 755 385 016 32 × 2 = 0 + 0.664 109 510 770 032 64;
  • 81) 0.664 109 510 770 032 64 × 2 = 1 + 0.328 219 021 540 065 28;
  • 82) 0.328 219 021 540 065 28 × 2 = 0 + 0.656 438 043 080 130 56;
  • 83) 0.656 438 043 080 130 56 × 2 = 1 + 0.312 876 086 160 261 12;
  • 84) 0.312 876 086 160 261 12 × 2 = 0 + 0.625 752 172 320 522 24;
  • 85) 0.625 752 172 320 522 24 × 2 = 1 + 0.251 504 344 641 044 48;

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...)


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:


0.000 000 000 232 830 64(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1100 1001 1001 0001 0110 1010 1(2)


5. Positive number before normalization:

0.000 000 000 232 830 64(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1100 1001 1001 0001 0110 1010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 33 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 232 830 64(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1100 1001 1001 0001 0110 1010 1(2) =


0.0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1100 1001 1001 0001 0110 1010 1(2) × 20 =


1.1111 1111 1111 1111 1111 1111 0111 1001 0011 0010 0010 1101 0101(2) × 2-33


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): -33


Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 0111 1001 0011 0010 0010 1101 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


-33 + 2(11-1) - 1 =


(-33 + 1 023)(10) =


990(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;
  • 990 ÷ 2 = 495 + 0;
  • 495 ÷ 2 = 247 + 1;
  • 247 ÷ 2 = 123 + 1;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =


990(10) =


011 1101 1110(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. 1111 1111 1111 1111 1111 1111 0111 1001 0011 0010 0010 1101 0101 =


1111 1111 1111 1111 1111 1111 0111 1001 0011 0010 0010 1101 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 1101 1110


Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 0111 1001 0011 0010 0010 1101 0101


The base ten decimal number 0.000 000 000 232 830 64 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1101 1110 - 1111 1111 1111 1111 1111 1111 0111 1001 0011 0010 0010 1101 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