17.035 499 999 999 998 977 01 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 17.035 499 999 999 998 977 01(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
17.035 499 999 999 998 977 01(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: 0.035 499 999 999 998 977 01.

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) 0.035 499 999 999 998 977 01 × 2 = 0 + 0.070 999 999 999 997 954 02;
  • 2) 0.070 999 999 999 997 954 02 × 2 = 0 + 0.141 999 999 999 995 908 04;
  • 3) 0.141 999 999 999 995 908 04 × 2 = 0 + 0.283 999 999 999 991 816 08;
  • 4) 0.283 999 999 999 991 816 08 × 2 = 0 + 0.567 999 999 999 983 632 16;
  • 5) 0.567 999 999 999 983 632 16 × 2 = 1 + 0.135 999 999 999 967 264 32;
  • 6) 0.135 999 999 999 967 264 32 × 2 = 0 + 0.271 999 999 999 934 528 64;
  • 7) 0.271 999 999 999 934 528 64 × 2 = 0 + 0.543 999 999 999 869 057 28;
  • 8) 0.543 999 999 999 869 057 28 × 2 = 1 + 0.087 999 999 999 738 114 56;
  • 9) 0.087 999 999 999 738 114 56 × 2 = 0 + 0.175 999 999 999 476 229 12;
  • 10) 0.175 999 999 999 476 229 12 × 2 = 0 + 0.351 999 999 998 952 458 24;
  • 11) 0.351 999 999 998 952 458 24 × 2 = 0 + 0.703 999 999 997 904 916 48;
  • 12) 0.703 999 999 997 904 916 48 × 2 = 1 + 0.407 999 999 995 809 832 96;
  • 13) 0.407 999 999 995 809 832 96 × 2 = 0 + 0.815 999 999 991 619 665 92;
  • 14) 0.815 999 999 991 619 665 92 × 2 = 1 + 0.631 999 999 983 239 331 84;
  • 15) 0.631 999 999 983 239 331 84 × 2 = 1 + 0.263 999 999 966 478 663 68;
  • 16) 0.263 999 999 966 478 663 68 × 2 = 0 + 0.527 999 999 932 957 327 36;
  • 17) 0.527 999 999 932 957 327 36 × 2 = 1 + 0.055 999 999 865 914 654 72;
  • 18) 0.055 999 999 865 914 654 72 × 2 = 0 + 0.111 999 999 731 829 309 44;
  • 19) 0.111 999 999 731 829 309 44 × 2 = 0 + 0.223 999 999 463 658 618 88;
  • 20) 0.223 999 999 463 658 618 88 × 2 = 0 + 0.447 999 998 927 317 237 76;
  • 21) 0.447 999 998 927 317 237 76 × 2 = 0 + 0.895 999 997 854 634 475 52;
  • 22) 0.895 999 997 854 634 475 52 × 2 = 1 + 0.791 999 995 709 268 951 04;
  • 23) 0.791 999 995 709 268 951 04 × 2 = 1 + 0.583 999 991 418 537 902 08;
  • 24) 0.583 999 991 418 537 902 08 × 2 = 1 + 0.167 999 982 837 075 804 16;
  • 25) 0.167 999 982 837 075 804 16 × 2 = 0 + 0.335 999 965 674 151 608 32;
  • 26) 0.335 999 965 674 151 608 32 × 2 = 0 + 0.671 999 931 348 303 216 64;
  • 27) 0.671 999 931 348 303 216 64 × 2 = 1 + 0.343 999 862 696 606 433 28;
  • 28) 0.343 999 862 696 606 433 28 × 2 = 0 + 0.687 999 725 393 212 866 56;
  • 29) 0.687 999 725 393 212 866 56 × 2 = 1 + 0.375 999 450 786 425 733 12;
  • 30) 0.375 999 450 786 425 733 12 × 2 = 0 + 0.751 998 901 572 851 466 24;
  • 31) 0.751 998 901 572 851 466 24 × 2 = 1 + 0.503 997 803 145 702 932 48;
  • 32) 0.503 997 803 145 702 932 48 × 2 = 1 + 0.007 995 606 291 405 864 96;
  • 33) 0.007 995 606 291 405 864 96 × 2 = 0 + 0.015 991 212 582 811 729 92;
  • 34) 0.015 991 212 582 811 729 92 × 2 = 0 + 0.031 982 425 165 623 459 84;
  • 35) 0.031 982 425 165 623 459 84 × 2 = 0 + 0.063 964 850 331 246 919 68;
  • 36) 0.063 964 850 331 246 919 68 × 2 = 0 + 0.127 929 700 662 493 839 36;
  • 37) 0.127 929 700 662 493 839 36 × 2 = 0 + 0.255 859 401 324 987 678 72;
  • 38) 0.255 859 401 324 987 678 72 × 2 = 0 + 0.511 718 802 649 975 357 44;
  • 39) 0.511 718 802 649 975 357 44 × 2 = 1 + 0.023 437 605 299 950 714 88;
  • 40) 0.023 437 605 299 950 714 88 × 2 = 0 + 0.046 875 210 599 901 429 76;
  • 41) 0.046 875 210 599 901 429 76 × 2 = 0 + 0.093 750 421 199 802 859 52;
  • 42) 0.093 750 421 199 802 859 52 × 2 = 0 + 0.187 500 842 399 605 719 04;
  • 43) 0.187 500 842 399 605 719 04 × 2 = 0 + 0.375 001 684 799 211 438 08;
  • 44) 0.375 001 684 799 211 438 08 × 2 = 0 + 0.750 003 369 598 422 876 16;
  • 45) 0.750 003 369 598 422 876 16 × 2 = 1 + 0.500 006 739 196 845 752 32;
  • 46) 0.500 006 739 196 845 752 32 × 2 = 1 + 0.000 013 478 393 691 504 64;
  • 47) 0.000 013 478 393 691 504 64 × 2 = 0 + 0.000 026 956 787 383 009 28;
  • 48) 0.000 026 956 787 383 009 28 × 2 = 0 + 0.000 053 913 574 766 018 56;
  • 49) 0.000 053 913 574 766 018 56 × 2 = 0 + 0.000 107 827 149 532 037 12;
  • 50) 0.000 107 827 149 532 037 12 × 2 = 0 + 0.000 215 654 299 064 074 24;
  • 51) 0.000 215 654 299 064 074 24 × 2 = 0 + 0.000 431 308 598 128 148 48;
  • 52) 0.000 431 308 598 128 148 48 × 2 = 0 + 0.000 862 617 196 256 296 96;
  • 53) 0.000 862 617 196 256 296 96 × 2 = 0 + 0.001 725 234 392 512 593 92;

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:


0.035 499 999 999 998 977 01(10) =

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

5. Positive number before normalization:

17.035 499 999 999 998 977 01(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:


17.035 499 999 999 998 977 01(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 17.035 499 999 999 998 977 01 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