17.035 499 999 999 998 976 818 369 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 17.035 499 999 999 998 976 818 369(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
17.035 499 999 999 998 976 818 369(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: 0.035 499 999 999 998 976 818 369.

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) 0.035 499 999 999 998 976 818 369 × 2 = 0 + 0.070 999 999 999 997 953 636 738;
  • 2) 0.070 999 999 999 997 953 636 738 × 2 = 0 + 0.141 999 999 999 995 907 273 476;
  • 3) 0.141 999 999 999 995 907 273 476 × 2 = 0 + 0.283 999 999 999 991 814 546 952;
  • 4) 0.283 999 999 999 991 814 546 952 × 2 = 0 + 0.567 999 999 999 983 629 093 904;
  • 5) 0.567 999 999 999 983 629 093 904 × 2 = 1 + 0.135 999 999 999 967 258 187 808;
  • 6) 0.135 999 999 999 967 258 187 808 × 2 = 0 + 0.271 999 999 999 934 516 375 616;
  • 7) 0.271 999 999 999 934 516 375 616 × 2 = 0 + 0.543 999 999 999 869 032 751 232;
  • 8) 0.543 999 999 999 869 032 751 232 × 2 = 1 + 0.087 999 999 999 738 065 502 464;
  • 9) 0.087 999 999 999 738 065 502 464 × 2 = 0 + 0.175 999 999 999 476 131 004 928;
  • 10) 0.175 999 999 999 476 131 004 928 × 2 = 0 + 0.351 999 999 998 952 262 009 856;
  • 11) 0.351 999 999 998 952 262 009 856 × 2 = 0 + 0.703 999 999 997 904 524 019 712;
  • 12) 0.703 999 999 997 904 524 019 712 × 2 = 1 + 0.407 999 999 995 809 048 039 424;
  • 13) 0.407 999 999 995 809 048 039 424 × 2 = 0 + 0.815 999 999 991 618 096 078 848;
  • 14) 0.815 999 999 991 618 096 078 848 × 2 = 1 + 0.631 999 999 983 236 192 157 696;
  • 15) 0.631 999 999 983 236 192 157 696 × 2 = 1 + 0.263 999 999 966 472 384 315 392;
  • 16) 0.263 999 999 966 472 384 315 392 × 2 = 0 + 0.527 999 999 932 944 768 630 784;
  • 17) 0.527 999 999 932 944 768 630 784 × 2 = 1 + 0.055 999 999 865 889 537 261 568;
  • 18) 0.055 999 999 865 889 537 261 568 × 2 = 0 + 0.111 999 999 731 779 074 523 136;
  • 19) 0.111 999 999 731 779 074 523 136 × 2 = 0 + 0.223 999 999 463 558 149 046 272;
  • 20) 0.223 999 999 463 558 149 046 272 × 2 = 0 + 0.447 999 998 927 116 298 092 544;
  • 21) 0.447 999 998 927 116 298 092 544 × 2 = 0 + 0.895 999 997 854 232 596 185 088;
  • 22) 0.895 999 997 854 232 596 185 088 × 2 = 1 + 0.791 999 995 708 465 192 370 176;
  • 23) 0.791 999 995 708 465 192 370 176 × 2 = 1 + 0.583 999 991 416 930 384 740 352;
  • 24) 0.583 999 991 416 930 384 740 352 × 2 = 1 + 0.167 999 982 833 860 769 480 704;
  • 25) 0.167 999 982 833 860 769 480 704 × 2 = 0 + 0.335 999 965 667 721 538 961 408;
  • 26) 0.335 999 965 667 721 538 961 408 × 2 = 0 + 0.671 999 931 335 443 077 922 816;
  • 27) 0.671 999 931 335 443 077 922 816 × 2 = 1 + 0.343 999 862 670 886 155 845 632;
  • 28) 0.343 999 862 670 886 155 845 632 × 2 = 0 + 0.687 999 725 341 772 311 691 264;
  • 29) 0.687 999 725 341 772 311 691 264 × 2 = 1 + 0.375 999 450 683 544 623 382 528;
  • 30) 0.375 999 450 683 544 623 382 528 × 2 = 0 + 0.751 998 901 367 089 246 765 056;
  • 31) 0.751 998 901 367 089 246 765 056 × 2 = 1 + 0.503 997 802 734 178 493 530 112;
  • 32) 0.503 997 802 734 178 493 530 112 × 2 = 1 + 0.007 995 605 468 356 987 060 224;
  • 33) 0.007 995 605 468 356 987 060 224 × 2 = 0 + 0.015 991 210 936 713 974 120 448;
  • 34) 0.015 991 210 936 713 974 120 448 × 2 = 0 + 0.031 982 421 873 427 948 240 896;
  • 35) 0.031 982 421 873 427 948 240 896 × 2 = 0 + 0.063 964 843 746 855 896 481 792;
  • 36) 0.063 964 843 746 855 896 481 792 × 2 = 0 + 0.127 929 687 493 711 792 963 584;
  • 37) 0.127 929 687 493 711 792 963 584 × 2 = 0 + 0.255 859 374 987 423 585 927 168;
  • 38) 0.255 859 374 987 423 585 927 168 × 2 = 0 + 0.511 718 749 974 847 171 854 336;
  • 39) 0.511 718 749 974 847 171 854 336 × 2 = 1 + 0.023 437 499 949 694 343 708 672;
  • 40) 0.023 437 499 949 694 343 708 672 × 2 = 0 + 0.046 874 999 899 388 687 417 344;
  • 41) 0.046 874 999 899 388 687 417 344 × 2 = 0 + 0.093 749 999 798 777 374 834 688;
  • 42) 0.093 749 999 798 777 374 834 688 × 2 = 0 + 0.187 499 999 597 554 749 669 376;
  • 43) 0.187 499 999 597 554 749 669 376 × 2 = 0 + 0.374 999 999 195 109 499 338 752;
  • 44) 0.374 999 999 195 109 499 338 752 × 2 = 0 + 0.749 999 998 390 218 998 677 504;
  • 45) 0.749 999 998 390 218 998 677 504 × 2 = 1 + 0.499 999 996 780 437 997 355 008;
  • 46) 0.499 999 996 780 437 997 355 008 × 2 = 0 + 0.999 999 993 560 875 994 710 016;
  • 47) 0.999 999 993 560 875 994 710 016 × 2 = 1 + 0.999 999 987 121 751 989 420 032;
  • 48) 0.999 999 987 121 751 989 420 032 × 2 = 1 + 0.999 999 974 243 503 978 840 064;
  • 49) 0.999 999 974 243 503 978 840 064 × 2 = 1 + 0.999 999 948 487 007 957 680 128;
  • 50) 0.999 999 948 487 007 957 680 128 × 2 = 1 + 0.999 999 896 974 015 915 360 256;
  • 51) 0.999 999 896 974 015 915 360 256 × 2 = 1 + 0.999 999 793 948 031 830 720 512;
  • 52) 0.999 999 793 948 031 830 720 512 × 2 = 1 + 0.999 999 587 896 063 661 441 024;
  • 53) 0.999 999 587 896 063 661 441 024 × 2 = 1 + 0.999 999 175 792 127 322 882 048;

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:


0.035 499 999 999 998 976 818 369(10) =

0.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1011 1111 1(2)

5. Positive number before normalization:

17.035 499 999 999 998 976 818 369(10) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1011 1111 1(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:


17.035 499 999 999 998 976 818 369(10) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1011 1111 1(2) =

1 0001.0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1011 1111 1(2) × 20 =

1.0001 0000 1001 0001 0110 1000 0111 0010 1011 0000 0010 0000 1011 1111 1(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 1011 1111 1


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 1011 1 1111 =

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


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 1011


Decimal number 17.035 499 999 999 998 976 818 369 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 1011

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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