1.732 050 808 38 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.732 050 808 38(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 38(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 38.

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 38 × 2 = 1 + 0.464 101 616 76;
  • 2) 0.464 101 616 76 × 2 = 0 + 0.928 203 233 52;
  • 3) 0.928 203 233 52 × 2 = 1 + 0.856 406 467 04;
  • 4) 0.856 406 467 04 × 2 = 1 + 0.712 812 934 08;
  • 5) 0.712 812 934 08 × 2 = 1 + 0.425 625 868 16;
  • 6) 0.425 625 868 16 × 2 = 0 + 0.851 251 736 32;
  • 7) 0.851 251 736 32 × 2 = 1 + 0.702 503 472 64;
  • 8) 0.702 503 472 64 × 2 = 1 + 0.405 006 945 28;
  • 9) 0.405 006 945 28 × 2 = 0 + 0.810 013 890 56;
  • 10) 0.810 013 890 56 × 2 = 1 + 0.620 027 781 12;
  • 11) 0.620 027 781 12 × 2 = 1 + 0.240 055 562 24;
  • 12) 0.240 055 562 24 × 2 = 0 + 0.480 111 124 48;
  • 13) 0.480 111 124 48 × 2 = 0 + 0.960 222 248 96;
  • 14) 0.960 222 248 96 × 2 = 1 + 0.920 444 497 92;
  • 15) 0.920 444 497 92 × 2 = 1 + 0.840 888 995 84;
  • 16) 0.840 888 995 84 × 2 = 1 + 0.681 777 991 68;
  • 17) 0.681 777 991 68 × 2 = 1 + 0.363 555 983 36;
  • 18) 0.363 555 983 36 × 2 = 0 + 0.727 111 966 72;
  • 19) 0.727 111 966 72 × 2 = 1 + 0.454 223 933 44;
  • 20) 0.454 223 933 44 × 2 = 0 + 0.908 447 866 88;
  • 21) 0.908 447 866 88 × 2 = 1 + 0.816 895 733 76;
  • 22) 0.816 895 733 76 × 2 = 1 + 0.633 791 467 52;
  • 23) 0.633 791 467 52 × 2 = 1 + 0.267 582 935 04;
  • 24) 0.267 582 935 04 × 2 = 0 + 0.535 165 870 08;
  • 25) 0.535 165 870 08 × 2 = 1 + 0.070 331 740 16;
  • 26) 0.070 331 740 16 × 2 = 0 + 0.140 663 480 32;
  • 27) 0.140 663 480 32 × 2 = 0 + 0.281 326 960 64;
  • 28) 0.281 326 960 64 × 2 = 0 + 0.562 653 921 28;
  • 29) 0.562 653 921 28 × 2 = 1 + 0.125 307 842 56;
  • 30) 0.125 307 842 56 × 2 = 0 + 0.250 615 685 12;
  • 31) 0.250 615 685 12 × 2 = 0 + 0.501 231 370 24;
  • 32) 0.501 231 370 24 × 2 = 1 + 0.002 462 740 48;
  • 33) 0.002 462 740 48 × 2 = 0 + 0.004 925 480 96;
  • 34) 0.004 925 480 96 × 2 = 0 + 0.009 850 961 92;
  • 35) 0.009 850 961 92 × 2 = 0 + 0.019 701 923 84;
  • 36) 0.019 701 923 84 × 2 = 0 + 0.039 403 847 68;
  • 37) 0.039 403 847 68 × 2 = 0 + 0.078 807 695 36;
  • 38) 0.078 807 695 36 × 2 = 0 + 0.157 615 390 72;
  • 39) 0.157 615 390 72 × 2 = 0 + 0.315 230 781 44;
  • 40) 0.315 230 781 44 × 2 = 0 + 0.630 461 562 88;
  • 41) 0.630 461 562 88 × 2 = 1 + 0.260 923 125 76;
  • 42) 0.260 923 125 76 × 2 = 0 + 0.521 846 251 52;
  • 43) 0.521 846 251 52 × 2 = 1 + 0.043 692 503 04;
  • 44) 0.043 692 503 04 × 2 = 0 + 0.087 385 006 08;
  • 45) 0.087 385 006 08 × 2 = 0 + 0.174 770 012 16;
  • 46) 0.174 770 012 16 × 2 = 0 + 0.349 540 024 32;
  • 47) 0.349 540 024 32 × 2 = 0 + 0.699 080 048 64;
  • 48) 0.699 080 048 64 × 2 = 1 + 0.398 160 097 28;
  • 49) 0.398 160 097 28 × 2 = 0 + 0.796 320 194 56;
  • 50) 0.796 320 194 56 × 2 = 1 + 0.592 640 389 12;
  • 51) 0.592 640 389 12 × 2 = 1 + 0.185 280 778 24;
  • 52) 0.185 280 778 24 × 2 = 0 + 0.370 561 556 48;
  • 53) 0.370 561 556 48 × 2 = 0 + 0.741 123 112 96;

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 38(10) =

0.1011 1011 0110 0111 1010 1110 1000 1001 0000 0000 1010 0001 0110 0(2)

5. Positive number before normalization:

1.732 050 808 38(10) =

1.1011 1011 0110 0111 1010 1110 1000 1001 0000 0000 1010 0001 0110 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 38(10) =

1.1011 1011 0110 0111 1010 1110 1000 1001 0000 0000 1010 0001 0110 0(2) =

1.1011 1011 0110 0111 1010 1110 1000 1001 0000 0000 1010 0001 0110 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 1001 0000 0000 1010 0001 0110 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 1001 0000 0000 1010 0001 0110 0 =

1011 1011 0110 0111 1010 1110 1000 1001 0000 0000 1010 0001 0110


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 1001 0000 0000 1010 0001 0110


Decimal number 1.732 050 808 38 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1011 0110 0111 1010 1110 1000 1001 0000 0000 1010 0001 0110

» Convert decimal 1.732 050 808 11 to 64 bit double precision IEEE 754 binary floating point representation standard

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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