17.035 499 999 999 998 976 821 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 17.035 499 999 999 998 976 821 57(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
17.035 499 999 999 998 976 821 57(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: 0.035 499 999 999 998 976 821 57.

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) 0.035 499 999 999 998 976 821 57 × 2 = 0 + 0.070 999 999 999 997 953 643 14;
  • 2) 0.070 999 999 999 997 953 643 14 × 2 = 0 + 0.141 999 999 999 995 907 286 28;
  • 3) 0.141 999 999 999 995 907 286 28 × 2 = 0 + 0.283 999 999 999 991 814 572 56;
  • 4) 0.283 999 999 999 991 814 572 56 × 2 = 0 + 0.567 999 999 999 983 629 145 12;
  • 5) 0.567 999 999 999 983 629 145 12 × 2 = 1 + 0.135 999 999 999 967 258 290 24;
  • 6) 0.135 999 999 999 967 258 290 24 × 2 = 0 + 0.271 999 999 999 934 516 580 48;
  • 7) 0.271 999 999 999 934 516 580 48 × 2 = 0 + 0.543 999 999 999 869 033 160 96;
  • 8) 0.543 999 999 999 869 033 160 96 × 2 = 1 + 0.087 999 999 999 738 066 321 92;
  • 9) 0.087 999 999 999 738 066 321 92 × 2 = 0 + 0.175 999 999 999 476 132 643 84;
  • 10) 0.175 999 999 999 476 132 643 84 × 2 = 0 + 0.351 999 999 998 952 265 287 68;
  • 11) 0.351 999 999 998 952 265 287 68 × 2 = 0 + 0.703 999 999 997 904 530 575 36;
  • 12) 0.703 999 999 997 904 530 575 36 × 2 = 1 + 0.407 999 999 995 809 061 150 72;
  • 13) 0.407 999 999 995 809 061 150 72 × 2 = 0 + 0.815 999 999 991 618 122 301 44;
  • 14) 0.815 999 999 991 618 122 301 44 × 2 = 1 + 0.631 999 999 983 236 244 602 88;
  • 15) 0.631 999 999 983 236 244 602 88 × 2 = 1 + 0.263 999 999 966 472 489 205 76;
  • 16) 0.263 999 999 966 472 489 205 76 × 2 = 0 + 0.527 999 999 932 944 978 411 52;
  • 17) 0.527 999 999 932 944 978 411 52 × 2 = 1 + 0.055 999 999 865 889 956 823 04;
  • 18) 0.055 999 999 865 889 956 823 04 × 2 = 0 + 0.111 999 999 731 779 913 646 08;
  • 19) 0.111 999 999 731 779 913 646 08 × 2 = 0 + 0.223 999 999 463 559 827 292 16;
  • 20) 0.223 999 999 463 559 827 292 16 × 2 = 0 + 0.447 999 998 927 119 654 584 32;
  • 21) 0.447 999 998 927 119 654 584 32 × 2 = 0 + 0.895 999 997 854 239 309 168 64;
  • 22) 0.895 999 997 854 239 309 168 64 × 2 = 1 + 0.791 999 995 708 478 618 337 28;
  • 23) 0.791 999 995 708 478 618 337 28 × 2 = 1 + 0.583 999 991 416 957 236 674 56;
  • 24) 0.583 999 991 416 957 236 674 56 × 2 = 1 + 0.167 999 982 833 914 473 349 12;
  • 25) 0.167 999 982 833 914 473 349 12 × 2 = 0 + 0.335 999 965 667 828 946 698 24;
  • 26) 0.335 999 965 667 828 946 698 24 × 2 = 0 + 0.671 999 931 335 657 893 396 48;
  • 27) 0.671 999 931 335 657 893 396 48 × 2 = 1 + 0.343 999 862 671 315 786 792 96;
  • 28) 0.343 999 862 671 315 786 792 96 × 2 = 0 + 0.687 999 725 342 631 573 585 92;
  • 29) 0.687 999 725 342 631 573 585 92 × 2 = 1 + 0.375 999 450 685 263 147 171 84;
  • 30) 0.375 999 450 685 263 147 171 84 × 2 = 0 + 0.751 998 901 370 526 294 343 68;
  • 31) 0.751 998 901 370 526 294 343 68 × 2 = 1 + 0.503 997 802 741 052 588 687 36;
  • 32) 0.503 997 802 741 052 588 687 36 × 2 = 1 + 0.007 995 605 482 105 177 374 72;
  • 33) 0.007 995 605 482 105 177 374 72 × 2 = 0 + 0.015 991 210 964 210 354 749 44;
  • 34) 0.015 991 210 964 210 354 749 44 × 2 = 0 + 0.031 982 421 928 420 709 498 88;
  • 35) 0.031 982 421 928 420 709 498 88 × 2 = 0 + 0.063 964 843 856 841 418 997 76;
  • 36) 0.063 964 843 856 841 418 997 76 × 2 = 0 + 0.127 929 687 713 682 837 995 52;
  • 37) 0.127 929 687 713 682 837 995 52 × 2 = 0 + 0.255 859 375 427 365 675 991 04;
  • 38) 0.255 859 375 427 365 675 991 04 × 2 = 0 + 0.511 718 750 854 731 351 982 08;
  • 39) 0.511 718 750 854 731 351 982 08 × 2 = 1 + 0.023 437 501 709 462 703 964 16;
  • 40) 0.023 437 501 709 462 703 964 16 × 2 = 0 + 0.046 875 003 418 925 407 928 32;
  • 41) 0.046 875 003 418 925 407 928 32 × 2 = 0 + 0.093 750 006 837 850 815 856 64;
  • 42) 0.093 750 006 837 850 815 856 64 × 2 = 0 + 0.187 500 013 675 701 631 713 28;
  • 43) 0.187 500 013 675 701 631 713 28 × 2 = 0 + 0.375 000 027 351 403 263 426 56;
  • 44) 0.375 000 027 351 403 263 426 56 × 2 = 0 + 0.750 000 054 702 806 526 853 12;
  • 45) 0.750 000 054 702 806 526 853 12 × 2 = 1 + 0.500 000 109 405 613 053 706 24;
  • 46) 0.500 000 109 405 613 053 706 24 × 2 = 1 + 0.000 000 218 811 226 107 412 48;
  • 47) 0.000 000 218 811 226 107 412 48 × 2 = 0 + 0.000 000 437 622 452 214 824 96;
  • 48) 0.000 000 437 622 452 214 824 96 × 2 = 0 + 0.000 000 875 244 904 429 649 92;
  • 49) 0.000 000 875 244 904 429 649 92 × 2 = 0 + 0.000 001 750 489 808 859 299 84;
  • 50) 0.000 001 750 489 808 859 299 84 × 2 = 0 + 0.000 003 500 979 617 718 599 68;
  • 51) 0.000 003 500 979 617 718 599 68 × 2 = 0 + 0.000 007 001 959 235 437 199 36;
  • 52) 0.000 007 001 959 235 437 199 36 × 2 = 0 + 0.000 014 003 918 470 874 398 72;
  • 53) 0.000 014 003 918 470 874 398 72 × 2 = 0 + 0.000 028 007 836 941 748 797 44;

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:


0.035 499 999 999 998 976 821 57(10) =

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

5. Positive number before normalization:

17.035 499 999 999 998 976 821 57(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:


17.035 499 999 999 998 976 821 57(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 17.035 499 999 999 998 976 821 57 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

Share

» Convert decimal 17.035 499 999 999 998 976 822 38 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 04, 2026 [April]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: April - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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