17.035 499 999 999 998 976 818 747 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 17.035 499 999 999 998 976 818 747(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
17.035 499 999 999 998 976 818 747(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: 17.
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;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

17(10) =

1 0001(2)


3. Convert to binary (base 2) the fractional part: 0.035 499 999 999 998 976 818 747.

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) 0.035 499 999 999 998 976 818 747 × 2 = 0 + 0.070 999 999 999 997 953 637 494;
  • 2) 0.070 999 999 999 997 953 637 494 × 2 = 0 + 0.141 999 999 999 995 907 274 988;
  • 3) 0.141 999 999 999 995 907 274 988 × 2 = 0 + 0.283 999 999 999 991 814 549 976;
  • 4) 0.283 999 999 999 991 814 549 976 × 2 = 0 + 0.567 999 999 999 983 629 099 952;
  • 5) 0.567 999 999 999 983 629 099 952 × 2 = 1 + 0.135 999 999 999 967 258 199 904;
  • 6) 0.135 999 999 999 967 258 199 904 × 2 = 0 + 0.271 999 999 999 934 516 399 808;
  • 7) 0.271 999 999 999 934 516 399 808 × 2 = 0 + 0.543 999 999 999 869 032 799 616;
  • 8) 0.543 999 999 999 869 032 799 616 × 2 = 1 + 0.087 999 999 999 738 065 599 232;
  • 9) 0.087 999 999 999 738 065 599 232 × 2 = 0 + 0.175 999 999 999 476 131 198 464;
  • 10) 0.175 999 999 999 476 131 198 464 × 2 = 0 + 0.351 999 999 998 952 262 396 928;
  • 11) 0.351 999 999 998 952 262 396 928 × 2 = 0 + 0.703 999 999 997 904 524 793 856;
  • 12) 0.703 999 999 997 904 524 793 856 × 2 = 1 + 0.407 999 999 995 809 049 587 712;
  • 13) 0.407 999 999 995 809 049 587 712 × 2 = 0 + 0.815 999 999 991 618 099 175 424;
  • 14) 0.815 999 999 991 618 099 175 424 × 2 = 1 + 0.631 999 999 983 236 198 350 848;
  • 15) 0.631 999 999 983 236 198 350 848 × 2 = 1 + 0.263 999 999 966 472 396 701 696;
  • 16) 0.263 999 999 966 472 396 701 696 × 2 = 0 + 0.527 999 999 932 944 793 403 392;
  • 17) 0.527 999 999 932 944 793 403 392 × 2 = 1 + 0.055 999 999 865 889 586 806 784;
  • 18) 0.055 999 999 865 889 586 806 784 × 2 = 0 + 0.111 999 999 731 779 173 613 568;
  • 19) 0.111 999 999 731 779 173 613 568 × 2 = 0 + 0.223 999 999 463 558 347 227 136;
  • 20) 0.223 999 999 463 558 347 227 136 × 2 = 0 + 0.447 999 998 927 116 694 454 272;
  • 21) 0.447 999 998 927 116 694 454 272 × 2 = 0 + 0.895 999 997 854 233 388 908 544;
  • 22) 0.895 999 997 854 233 388 908 544 × 2 = 1 + 0.791 999 995 708 466 777 817 088;
  • 23) 0.791 999 995 708 466 777 817 088 × 2 = 1 + 0.583 999 991 416 933 555 634 176;
  • 24) 0.583 999 991 416 933 555 634 176 × 2 = 1 + 0.167 999 982 833 867 111 268 352;
  • 25) 0.167 999 982 833 867 111 268 352 × 2 = 0 + 0.335 999 965 667 734 222 536 704;
  • 26) 0.335 999 965 667 734 222 536 704 × 2 = 0 + 0.671 999 931 335 468 445 073 408;
  • 27) 0.671 999 931 335 468 445 073 408 × 2 = 1 + 0.343 999 862 670 936 890 146 816;
  • 28) 0.343 999 862 670 936 890 146 816 × 2 = 0 + 0.687 999 725 341 873 780 293 632;
  • 29) 0.687 999 725 341 873 780 293 632 × 2 = 1 + 0.375 999 450 683 747 560 587 264;
  • 30) 0.375 999 450 683 747 560 587 264 × 2 = 0 + 0.751 998 901 367 495 121 174 528;
  • 31) 0.751 998 901 367 495 121 174 528 × 2 = 1 + 0.503 997 802 734 990 242 349 056;
  • 32) 0.503 997 802 734 990 242 349 056 × 2 = 1 + 0.007 995 605 469 980 484 698 112;
  • 33) 0.007 995 605 469 980 484 698 112 × 2 = 0 + 0.015 991 210 939 960 969 396 224;
  • 34) 0.015 991 210 939 960 969 396 224 × 2 = 0 + 0.031 982 421 879 921 938 792 448;
  • 35) 0.031 982 421 879 921 938 792 448 × 2 = 0 + 0.063 964 843 759 843 877 584 896;
  • 36) 0.063 964 843 759 843 877 584 896 × 2 = 0 + 0.127 929 687 519 687 755 169 792;
  • 37) 0.127 929 687 519 687 755 169 792 × 2 = 0 + 0.255 859 375 039 375 510 339 584;
  • 38) 0.255 859 375 039 375 510 339 584 × 2 = 0 + 0.511 718 750 078 751 020 679 168;
  • 39) 0.511 718 750 078 751 020 679 168 × 2 = 1 + 0.023 437 500 157 502 041 358 336;
  • 40) 0.023 437 500 157 502 041 358 336 × 2 = 0 + 0.046 875 000 315 004 082 716 672;
  • 41) 0.046 875 000 315 004 082 716 672 × 2 = 0 + 0.093 750 000 630 008 165 433 344;
  • 42) 0.093 750 000 630 008 165 433 344 × 2 = 0 + 0.187 500 001 260 016 330 866 688;
  • 43) 0.187 500 001 260 016 330 866 688 × 2 = 0 + 0.375 000 002 520 032 661 733 376;
  • 44) 0.375 000 002 520 032 661 733 376 × 2 = 0 + 0.750 000 005 040 065 323 466 752;
  • 45) 0.750 000 005 040 065 323 466 752 × 2 = 1 + 0.500 000 010 080 130 646 933 504;
  • 46) 0.500 000 010 080 130 646 933 504 × 2 = 1 + 0.000 000 020 160 261 293 867 008;
  • 47) 0.000 000 020 160 261 293 867 008 × 2 = 0 + 0.000 000 040 320 522 587 734 016;
  • 48) 0.000 000 040 320 522 587 734 016 × 2 = 0 + 0.000 000 080 641 045 175 468 032;
  • 49) 0.000 000 080 641 045 175 468 032 × 2 = 0 + 0.000 000 161 282 090 350 936 064;
  • 50) 0.000 000 161 282 090 350 936 064 × 2 = 0 + 0.000 000 322 564 180 701 872 128;
  • 51) 0.000 000 322 564 180 701 872 128 × 2 = 0 + 0.000 000 645 128 361 403 744 256;
  • 52) 0.000 000 645 128 361 403 744 256 × 2 = 0 + 0.000 001 290 256 722 807 488 512;
  • 53) 0.000 001 290 256 722 807 488 512 × 2 = 0 + 0.000 002 580 513 445 614 977 024;

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:


0.035 499 999 999 998 976 818 747(10) =

0.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0000 0(2)

5. Positive number before normalization:

17.035 499 999 999 998 976 818 747(10) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


17.035 499 999 999 998 976 818 747(10) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0000 0(2) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0000 0(2) × 20 =

1.0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0000 0(2) × 24


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): 4

Mantissa (not normalized):
1.0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1027(10) =

100 0000 0011(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. 0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100 0 0000 =

0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100


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) =
100 0000 0011

Mantissa (52 bits) =
0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100


Decimal number 17.035 499 999 999 998 976 818 747 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1100

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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