17.035 499 999 999 998 976 818 701 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 17.035 499 999 999 998 976 818 701(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
17.035 499 999 999 998 976 818 701(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: 0.035 499 999 999 998 976 818 701.

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) 0.035 499 999 999 998 976 818 701 × 2 = 0 + 0.070 999 999 999 997 953 637 402;
  • 2) 0.070 999 999 999 997 953 637 402 × 2 = 0 + 0.141 999 999 999 995 907 274 804;
  • 3) 0.141 999 999 999 995 907 274 804 × 2 = 0 + 0.283 999 999 999 991 814 549 608;
  • 4) 0.283 999 999 999 991 814 549 608 × 2 = 0 + 0.567 999 999 999 983 629 099 216;
  • 5) 0.567 999 999 999 983 629 099 216 × 2 = 1 + 0.135 999 999 999 967 258 198 432;
  • 6) 0.135 999 999 999 967 258 198 432 × 2 = 0 + 0.271 999 999 999 934 516 396 864;
  • 7) 0.271 999 999 999 934 516 396 864 × 2 = 0 + 0.543 999 999 999 869 032 793 728;
  • 8) 0.543 999 999 999 869 032 793 728 × 2 = 1 + 0.087 999 999 999 738 065 587 456;
  • 9) 0.087 999 999 999 738 065 587 456 × 2 = 0 + 0.175 999 999 999 476 131 174 912;
  • 10) 0.175 999 999 999 476 131 174 912 × 2 = 0 + 0.351 999 999 998 952 262 349 824;
  • 11) 0.351 999 999 998 952 262 349 824 × 2 = 0 + 0.703 999 999 997 904 524 699 648;
  • 12) 0.703 999 999 997 904 524 699 648 × 2 = 1 + 0.407 999 999 995 809 049 399 296;
  • 13) 0.407 999 999 995 809 049 399 296 × 2 = 0 + 0.815 999 999 991 618 098 798 592;
  • 14) 0.815 999 999 991 618 098 798 592 × 2 = 1 + 0.631 999 999 983 236 197 597 184;
  • 15) 0.631 999 999 983 236 197 597 184 × 2 = 1 + 0.263 999 999 966 472 395 194 368;
  • 16) 0.263 999 999 966 472 395 194 368 × 2 = 0 + 0.527 999 999 932 944 790 388 736;
  • 17) 0.527 999 999 932 944 790 388 736 × 2 = 1 + 0.055 999 999 865 889 580 777 472;
  • 18) 0.055 999 999 865 889 580 777 472 × 2 = 0 + 0.111 999 999 731 779 161 554 944;
  • 19) 0.111 999 999 731 779 161 554 944 × 2 = 0 + 0.223 999 999 463 558 323 109 888;
  • 20) 0.223 999 999 463 558 323 109 888 × 2 = 0 + 0.447 999 998 927 116 646 219 776;
  • 21) 0.447 999 998 927 116 646 219 776 × 2 = 0 + 0.895 999 997 854 233 292 439 552;
  • 22) 0.895 999 997 854 233 292 439 552 × 2 = 1 + 0.791 999 995 708 466 584 879 104;
  • 23) 0.791 999 995 708 466 584 879 104 × 2 = 1 + 0.583 999 991 416 933 169 758 208;
  • 24) 0.583 999 991 416 933 169 758 208 × 2 = 1 + 0.167 999 982 833 866 339 516 416;
  • 25) 0.167 999 982 833 866 339 516 416 × 2 = 0 + 0.335 999 965 667 732 679 032 832;
  • 26) 0.335 999 965 667 732 679 032 832 × 2 = 0 + 0.671 999 931 335 465 358 065 664;
  • 27) 0.671 999 931 335 465 358 065 664 × 2 = 1 + 0.343 999 862 670 930 716 131 328;
  • 28) 0.343 999 862 670 930 716 131 328 × 2 = 0 + 0.687 999 725 341 861 432 262 656;
  • 29) 0.687 999 725 341 861 432 262 656 × 2 = 1 + 0.375 999 450 683 722 864 525 312;
  • 30) 0.375 999 450 683 722 864 525 312 × 2 = 0 + 0.751 998 901 367 445 729 050 624;
  • 31) 0.751 998 901 367 445 729 050 624 × 2 = 1 + 0.503 997 802 734 891 458 101 248;
  • 32) 0.503 997 802 734 891 458 101 248 × 2 = 1 + 0.007 995 605 469 782 916 202 496;
  • 33) 0.007 995 605 469 782 916 202 496 × 2 = 0 + 0.015 991 210 939 565 832 404 992;
  • 34) 0.015 991 210 939 565 832 404 992 × 2 = 0 + 0.031 982 421 879 131 664 809 984;
  • 35) 0.031 982 421 879 131 664 809 984 × 2 = 0 + 0.063 964 843 758 263 329 619 968;
  • 36) 0.063 964 843 758 263 329 619 968 × 2 = 0 + 0.127 929 687 516 526 659 239 936;
  • 37) 0.127 929 687 516 526 659 239 936 × 2 = 0 + 0.255 859 375 033 053 318 479 872;
  • 38) 0.255 859 375 033 053 318 479 872 × 2 = 0 + 0.511 718 750 066 106 636 959 744;
  • 39) 0.511 718 750 066 106 636 959 744 × 2 = 1 + 0.023 437 500 132 213 273 919 488;
  • 40) 0.023 437 500 132 213 273 919 488 × 2 = 0 + 0.046 875 000 264 426 547 838 976;
  • 41) 0.046 875 000 264 426 547 838 976 × 2 = 0 + 0.093 750 000 528 853 095 677 952;
  • 42) 0.093 750 000 528 853 095 677 952 × 2 = 0 + 0.187 500 001 057 706 191 355 904;
  • 43) 0.187 500 001 057 706 191 355 904 × 2 = 0 + 0.375 000 002 115 412 382 711 808;
  • 44) 0.375 000 002 115 412 382 711 808 × 2 = 0 + 0.750 000 004 230 824 765 423 616;
  • 45) 0.750 000 004 230 824 765 423 616 × 2 = 1 + 0.500 000 008 461 649 530 847 232;
  • 46) 0.500 000 008 461 649 530 847 232 × 2 = 1 + 0.000 000 016 923 299 061 694 464;
  • 47) 0.000 000 016 923 299 061 694 464 × 2 = 0 + 0.000 000 033 846 598 123 388 928;
  • 48) 0.000 000 033 846 598 123 388 928 × 2 = 0 + 0.000 000 067 693 196 246 777 856;
  • 49) 0.000 000 067 693 196 246 777 856 × 2 = 0 + 0.000 000 135 386 392 493 555 712;
  • 50) 0.000 000 135 386 392 493 555 712 × 2 = 0 + 0.000 000 270 772 784 987 111 424;
  • 51) 0.000 000 270 772 784 987 111 424 × 2 = 0 + 0.000 000 541 545 569 974 222 848;
  • 52) 0.000 000 541 545 569 974 222 848 × 2 = 0 + 0.000 001 083 091 139 948 445 696;
  • 53) 0.000 001 083 091 139 948 445 696 × 2 = 0 + 0.000 002 166 182 279 896 891 392;

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:


0.035 499 999 999 998 976 818 701(10) =

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

5. Positive number before normalization:

17.035 499 999 999 998 976 818 701(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:


17.035 499 999 999 998 976 818 701(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 17.035 499 999 999 998 976 818 701 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