1.301 029 995 663 981 359 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.301 029 995 663 981 359(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
1.301 029 995 663 981 359(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: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.301 029 995 663 981 359.

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) 0.301 029 995 663 981 359 × 2 = 0 + 0.602 059 991 327 962 718;
  • 2) 0.602 059 991 327 962 718 × 2 = 1 + 0.204 119 982 655 925 436;
  • 3) 0.204 119 982 655 925 436 × 2 = 0 + 0.408 239 965 311 850 872;
  • 4) 0.408 239 965 311 850 872 × 2 = 0 + 0.816 479 930 623 701 744;
  • 5) 0.816 479 930 623 701 744 × 2 = 1 + 0.632 959 861 247 403 488;
  • 6) 0.632 959 861 247 403 488 × 2 = 1 + 0.265 919 722 494 806 976;
  • 7) 0.265 919 722 494 806 976 × 2 = 0 + 0.531 839 444 989 613 952;
  • 8) 0.531 839 444 989 613 952 × 2 = 1 + 0.063 678 889 979 227 904;
  • 9) 0.063 678 889 979 227 904 × 2 = 0 + 0.127 357 779 958 455 808;
  • 10) 0.127 357 779 958 455 808 × 2 = 0 + 0.254 715 559 916 911 616;
  • 11) 0.254 715 559 916 911 616 × 2 = 0 + 0.509 431 119 833 823 232;
  • 12) 0.509 431 119 833 823 232 × 2 = 1 + 0.018 862 239 667 646 464;
  • 13) 0.018 862 239 667 646 464 × 2 = 0 + 0.037 724 479 335 292 928;
  • 14) 0.037 724 479 335 292 928 × 2 = 0 + 0.075 448 958 670 585 856;
  • 15) 0.075 448 958 670 585 856 × 2 = 0 + 0.150 897 917 341 171 712;
  • 16) 0.150 897 917 341 171 712 × 2 = 0 + 0.301 795 834 682 343 424;
  • 17) 0.301 795 834 682 343 424 × 2 = 0 + 0.603 591 669 364 686 848;
  • 18) 0.603 591 669 364 686 848 × 2 = 1 + 0.207 183 338 729 373 696;
  • 19) 0.207 183 338 729 373 696 × 2 = 0 + 0.414 366 677 458 747 392;
  • 20) 0.414 366 677 458 747 392 × 2 = 0 + 0.828 733 354 917 494 784;
  • 21) 0.828 733 354 917 494 784 × 2 = 1 + 0.657 466 709 834 989 568;
  • 22) 0.657 466 709 834 989 568 × 2 = 1 + 0.314 933 419 669 979 136;
  • 23) 0.314 933 419 669 979 136 × 2 = 0 + 0.629 866 839 339 958 272;
  • 24) 0.629 866 839 339 958 272 × 2 = 1 + 0.259 733 678 679 916 544;
  • 25) 0.259 733 678 679 916 544 × 2 = 0 + 0.519 467 357 359 833 088;
  • 26) 0.519 467 357 359 833 088 × 2 = 1 + 0.038 934 714 719 666 176;
  • 27) 0.038 934 714 719 666 176 × 2 = 0 + 0.077 869 429 439 332 352;
  • 28) 0.077 869 429 439 332 352 × 2 = 0 + 0.155 738 858 878 664 704;
  • 29) 0.155 738 858 878 664 704 × 2 = 0 + 0.311 477 717 757 329 408;
  • 30) 0.311 477 717 757 329 408 × 2 = 0 + 0.622 955 435 514 658 816;
  • 31) 0.622 955 435 514 658 816 × 2 = 1 + 0.245 910 871 029 317 632;
  • 32) 0.245 910 871 029 317 632 × 2 = 0 + 0.491 821 742 058 635 264;
  • 33) 0.491 821 742 058 635 264 × 2 = 0 + 0.983 643 484 117 270 528;
  • 34) 0.983 643 484 117 270 528 × 2 = 1 + 0.967 286 968 234 541 056;
  • 35) 0.967 286 968 234 541 056 × 2 = 1 + 0.934 573 936 469 082 112;
  • 36) 0.934 573 936 469 082 112 × 2 = 1 + 0.869 147 872 938 164 224;
  • 37) 0.869 147 872 938 164 224 × 2 = 1 + 0.738 295 745 876 328 448;
  • 38) 0.738 295 745 876 328 448 × 2 = 1 + 0.476 591 491 752 656 896;
  • 39) 0.476 591 491 752 656 896 × 2 = 0 + 0.953 182 983 505 313 792;
  • 40) 0.953 182 983 505 313 792 × 2 = 1 + 0.906 365 967 010 627 584;
  • 41) 0.906 365 967 010 627 584 × 2 = 1 + 0.812 731 934 021 255 168;
  • 42) 0.812 731 934 021 255 168 × 2 = 1 + 0.625 463 868 042 510 336;
  • 43) 0.625 463 868 042 510 336 × 2 = 1 + 0.250 927 736 085 020 672;
  • 44) 0.250 927 736 085 020 672 × 2 = 0 + 0.501 855 472 170 041 344;
  • 45) 0.501 855 472 170 041 344 × 2 = 1 + 0.003 710 944 340 082 688;
  • 46) 0.003 710 944 340 082 688 × 2 = 0 + 0.007 421 888 680 165 376;
  • 47) 0.007 421 888 680 165 376 × 2 = 0 + 0.014 843 777 360 330 752;
  • 48) 0.014 843 777 360 330 752 × 2 = 0 + 0.029 687 554 720 661 504;
  • 49) 0.029 687 554 720 661 504 × 2 = 0 + 0.059 375 109 441 323 008;
  • 50) 0.059 375 109 441 323 008 × 2 = 0 + 0.118 750 218 882 646 016;
  • 51) 0.118 750 218 882 646 016 × 2 = 0 + 0.237 500 437 765 292 032;
  • 52) 0.237 500 437 765 292 032 × 2 = 0 + 0.475 000 875 530 584 064;
  • 53) 0.475 000 875 530 584 064 × 2 = 0 + 0.950 001 751 061 168 128;

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:


0.301 029 995 663 981 359(10) =

0.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(2)

5. Positive number before normalization:

1.301 029 995 663 981 359(10) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.301 029 995 663 981 359(10) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(2) =

1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0(2) × 20


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): 0

Mantissa (not normalized):
1.0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 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) =

1023(10) =

011 1111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000 0 =

0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000


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 1111

Mantissa (52 bits) =
0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000


Decimal number 1.301 029 995 663 981 359 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0100 1101 0001 0000 0100 1101 0100 0010 0111 1101 1110 1000 0000

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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