17.035 499 999 999 998 976 818 771 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 17.035 499 999 999 998 976 818 771(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
17.035 499 999 999 998 976 818 771(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: 0.035 499 999 999 998 976 818 771.

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) 0.035 499 999 999 998 976 818 771 × 2 = 0 + 0.070 999 999 999 997 953 637 542;
  • 2) 0.070 999 999 999 997 953 637 542 × 2 = 0 + 0.141 999 999 999 995 907 275 084;
  • 3) 0.141 999 999 999 995 907 275 084 × 2 = 0 + 0.283 999 999 999 991 814 550 168;
  • 4) 0.283 999 999 999 991 814 550 168 × 2 = 0 + 0.567 999 999 999 983 629 100 336;
  • 5) 0.567 999 999 999 983 629 100 336 × 2 = 1 + 0.135 999 999 999 967 258 200 672;
  • 6) 0.135 999 999 999 967 258 200 672 × 2 = 0 + 0.271 999 999 999 934 516 401 344;
  • 7) 0.271 999 999 999 934 516 401 344 × 2 = 0 + 0.543 999 999 999 869 032 802 688;
  • 8) 0.543 999 999 999 869 032 802 688 × 2 = 1 + 0.087 999 999 999 738 065 605 376;
  • 9) 0.087 999 999 999 738 065 605 376 × 2 = 0 + 0.175 999 999 999 476 131 210 752;
  • 10) 0.175 999 999 999 476 131 210 752 × 2 = 0 + 0.351 999 999 998 952 262 421 504;
  • 11) 0.351 999 999 998 952 262 421 504 × 2 = 0 + 0.703 999 999 997 904 524 843 008;
  • 12) 0.703 999 999 997 904 524 843 008 × 2 = 1 + 0.407 999 999 995 809 049 686 016;
  • 13) 0.407 999 999 995 809 049 686 016 × 2 = 0 + 0.815 999 999 991 618 099 372 032;
  • 14) 0.815 999 999 991 618 099 372 032 × 2 = 1 + 0.631 999 999 983 236 198 744 064;
  • 15) 0.631 999 999 983 236 198 744 064 × 2 = 1 + 0.263 999 999 966 472 397 488 128;
  • 16) 0.263 999 999 966 472 397 488 128 × 2 = 0 + 0.527 999 999 932 944 794 976 256;
  • 17) 0.527 999 999 932 944 794 976 256 × 2 = 1 + 0.055 999 999 865 889 589 952 512;
  • 18) 0.055 999 999 865 889 589 952 512 × 2 = 0 + 0.111 999 999 731 779 179 905 024;
  • 19) 0.111 999 999 731 779 179 905 024 × 2 = 0 + 0.223 999 999 463 558 359 810 048;
  • 20) 0.223 999 999 463 558 359 810 048 × 2 = 0 + 0.447 999 998 927 116 719 620 096;
  • 21) 0.447 999 998 927 116 719 620 096 × 2 = 0 + 0.895 999 997 854 233 439 240 192;
  • 22) 0.895 999 997 854 233 439 240 192 × 2 = 1 + 0.791 999 995 708 466 878 480 384;
  • 23) 0.791 999 995 708 466 878 480 384 × 2 = 1 + 0.583 999 991 416 933 756 960 768;
  • 24) 0.583 999 991 416 933 756 960 768 × 2 = 1 + 0.167 999 982 833 867 513 921 536;
  • 25) 0.167 999 982 833 867 513 921 536 × 2 = 0 + 0.335 999 965 667 735 027 843 072;
  • 26) 0.335 999 965 667 735 027 843 072 × 2 = 0 + 0.671 999 931 335 470 055 686 144;
  • 27) 0.671 999 931 335 470 055 686 144 × 2 = 1 + 0.343 999 862 670 940 111 372 288;
  • 28) 0.343 999 862 670 940 111 372 288 × 2 = 0 + 0.687 999 725 341 880 222 744 576;
  • 29) 0.687 999 725 341 880 222 744 576 × 2 = 1 + 0.375 999 450 683 760 445 489 152;
  • 30) 0.375 999 450 683 760 445 489 152 × 2 = 0 + 0.751 998 901 367 520 890 978 304;
  • 31) 0.751 998 901 367 520 890 978 304 × 2 = 1 + 0.503 997 802 735 041 781 956 608;
  • 32) 0.503 997 802 735 041 781 956 608 × 2 = 1 + 0.007 995 605 470 083 563 913 216;
  • 33) 0.007 995 605 470 083 563 913 216 × 2 = 0 + 0.015 991 210 940 167 127 826 432;
  • 34) 0.015 991 210 940 167 127 826 432 × 2 = 0 + 0.031 982 421 880 334 255 652 864;
  • 35) 0.031 982 421 880 334 255 652 864 × 2 = 0 + 0.063 964 843 760 668 511 305 728;
  • 36) 0.063 964 843 760 668 511 305 728 × 2 = 0 + 0.127 929 687 521 337 022 611 456;
  • 37) 0.127 929 687 521 337 022 611 456 × 2 = 0 + 0.255 859 375 042 674 045 222 912;
  • 38) 0.255 859 375 042 674 045 222 912 × 2 = 0 + 0.511 718 750 085 348 090 445 824;
  • 39) 0.511 718 750 085 348 090 445 824 × 2 = 1 + 0.023 437 500 170 696 180 891 648;
  • 40) 0.023 437 500 170 696 180 891 648 × 2 = 0 + 0.046 875 000 341 392 361 783 296;
  • 41) 0.046 875 000 341 392 361 783 296 × 2 = 0 + 0.093 750 000 682 784 723 566 592;
  • 42) 0.093 750 000 682 784 723 566 592 × 2 = 0 + 0.187 500 001 365 569 447 133 184;
  • 43) 0.187 500 001 365 569 447 133 184 × 2 = 0 + 0.375 000 002 731 138 894 266 368;
  • 44) 0.375 000 002 731 138 894 266 368 × 2 = 0 + 0.750 000 005 462 277 788 532 736;
  • 45) 0.750 000 005 462 277 788 532 736 × 2 = 1 + 0.500 000 010 924 555 577 065 472;
  • 46) 0.500 000 010 924 555 577 065 472 × 2 = 1 + 0.000 000 021 849 111 154 130 944;
  • 47) 0.000 000 021 849 111 154 130 944 × 2 = 0 + 0.000 000 043 698 222 308 261 888;
  • 48) 0.000 000 043 698 222 308 261 888 × 2 = 0 + 0.000 000 087 396 444 616 523 776;
  • 49) 0.000 000 087 396 444 616 523 776 × 2 = 0 + 0.000 000 174 792 889 233 047 552;
  • 50) 0.000 000 174 792 889 233 047 552 × 2 = 0 + 0.000 000 349 585 778 466 095 104;
  • 51) 0.000 000 349 585 778 466 095 104 × 2 = 0 + 0.000 000 699 171 556 932 190 208;
  • 52) 0.000 000 699 171 556 932 190 208 × 2 = 0 + 0.000 001 398 343 113 864 380 416;
  • 53) 0.000 001 398 343 113 864 380 416 × 2 = 0 + 0.000 002 796 686 227 728 760 832;

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:


0.035 499 999 999 998 976 818 771(10) =

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

5. Positive number before normalization:

17.035 499 999 999 998 976 818 771(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:


17.035 499 999 999 998 976 818 771(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 17.035 499 999 999 998 976 818 771 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