1.732 050 808 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.732 050 808 17(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
1.732 050 808 17(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.732 050 808 17.

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) 0.732 050 808 17 × 2 = 1 + 0.464 101 616 34;
  • 2) 0.464 101 616 34 × 2 = 0 + 0.928 203 232 68;
  • 3) 0.928 203 232 68 × 2 = 1 + 0.856 406 465 36;
  • 4) 0.856 406 465 36 × 2 = 1 + 0.712 812 930 72;
  • 5) 0.712 812 930 72 × 2 = 1 + 0.425 625 861 44;
  • 6) 0.425 625 861 44 × 2 = 0 + 0.851 251 722 88;
  • 7) 0.851 251 722 88 × 2 = 1 + 0.702 503 445 76;
  • 8) 0.702 503 445 76 × 2 = 1 + 0.405 006 891 52;
  • 9) 0.405 006 891 52 × 2 = 0 + 0.810 013 783 04;
  • 10) 0.810 013 783 04 × 2 = 1 + 0.620 027 566 08;
  • 11) 0.620 027 566 08 × 2 = 1 + 0.240 055 132 16;
  • 12) 0.240 055 132 16 × 2 = 0 + 0.480 110 264 32;
  • 13) 0.480 110 264 32 × 2 = 0 + 0.960 220 528 64;
  • 14) 0.960 220 528 64 × 2 = 1 + 0.920 441 057 28;
  • 15) 0.920 441 057 28 × 2 = 1 + 0.840 882 114 56;
  • 16) 0.840 882 114 56 × 2 = 1 + 0.681 764 229 12;
  • 17) 0.681 764 229 12 × 2 = 1 + 0.363 528 458 24;
  • 18) 0.363 528 458 24 × 2 = 0 + 0.727 056 916 48;
  • 19) 0.727 056 916 48 × 2 = 1 + 0.454 113 832 96;
  • 20) 0.454 113 832 96 × 2 = 0 + 0.908 227 665 92;
  • 21) 0.908 227 665 92 × 2 = 1 + 0.816 455 331 84;
  • 22) 0.816 455 331 84 × 2 = 1 + 0.632 910 663 68;
  • 23) 0.632 910 663 68 × 2 = 1 + 0.265 821 327 36;
  • 24) 0.265 821 327 36 × 2 = 0 + 0.531 642 654 72;
  • 25) 0.531 642 654 72 × 2 = 1 + 0.063 285 309 44;
  • 26) 0.063 285 309 44 × 2 = 0 + 0.126 570 618 88;
  • 27) 0.126 570 618 88 × 2 = 0 + 0.253 141 237 76;
  • 28) 0.253 141 237 76 × 2 = 0 + 0.506 282 475 52;
  • 29) 0.506 282 475 52 × 2 = 1 + 0.012 564 951 04;
  • 30) 0.012 564 951 04 × 2 = 0 + 0.025 129 902 08;
  • 31) 0.025 129 902 08 × 2 = 0 + 0.050 259 804 16;
  • 32) 0.050 259 804 16 × 2 = 0 + 0.100 519 608 32;
  • 33) 0.100 519 608 32 × 2 = 0 + 0.201 039 216 64;
  • 34) 0.201 039 216 64 × 2 = 0 + 0.402 078 433 28;
  • 35) 0.402 078 433 28 × 2 = 0 + 0.804 156 866 56;
  • 36) 0.804 156 866 56 × 2 = 1 + 0.608 313 733 12;
  • 37) 0.608 313 733 12 × 2 = 1 + 0.216 627 466 24;
  • 38) 0.216 627 466 24 × 2 = 0 + 0.433 254 932 48;
  • 39) 0.433 254 932 48 × 2 = 0 + 0.866 509 864 96;
  • 40) 0.866 509 864 96 × 2 = 1 + 0.733 019 729 92;
  • 41) 0.733 019 729 92 × 2 = 1 + 0.466 039 459 84;
  • 42) 0.466 039 459 84 × 2 = 0 + 0.932 078 919 68;
  • 43) 0.932 078 919 68 × 2 = 1 + 0.864 157 839 36;
  • 44) 0.864 157 839 36 × 2 = 1 + 0.728 315 678 72;
  • 45) 0.728 315 678 72 × 2 = 1 + 0.456 631 357 44;
  • 46) 0.456 631 357 44 × 2 = 0 + 0.913 262 714 88;
  • 47) 0.913 262 714 88 × 2 = 1 + 0.826 525 429 76;
  • 48) 0.826 525 429 76 × 2 = 1 + 0.653 050 859 52;
  • 49) 0.653 050 859 52 × 2 = 1 + 0.306 101 719 04;
  • 50) 0.306 101 719 04 × 2 = 0 + 0.612 203 438 08;
  • 51) 0.612 203 438 08 × 2 = 1 + 0.224 406 876 16;
  • 52) 0.224 406 876 16 × 2 = 0 + 0.448 813 752 32;
  • 53) 0.448 813 752 32 × 2 = 0 + 0.897 627 504 64;

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:


0.732 050 808 17(10) =

0.1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010 0(2)

5. Positive number before normalization:

1.732 050 808 17(10) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.732 050 808 17(10) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010 0(2) =

1.1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010 0(2) × 20


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): 0

Mantissa (not normalized):
1.1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010 0 =

1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010


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) =
011 1111 1111

Mantissa (52 bits) =
1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010


Decimal number 1.732 050 808 17 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1011 0110 0111 1010 1110 1000 1000 0001 1001 1011 1011 1010

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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