1.618 033 990 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.618 033 990 71(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
1.618 033 990 71(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: 0.618 033 990 71.

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) 0.618 033 990 71 × 2 = 1 + 0.236 067 981 42;
  • 2) 0.236 067 981 42 × 2 = 0 + 0.472 135 962 84;
  • 3) 0.472 135 962 84 × 2 = 0 + 0.944 271 925 68;
  • 4) 0.944 271 925 68 × 2 = 1 + 0.888 543 851 36;
  • 5) 0.888 543 851 36 × 2 = 1 + 0.777 087 702 72;
  • 6) 0.777 087 702 72 × 2 = 1 + 0.554 175 405 44;
  • 7) 0.554 175 405 44 × 2 = 1 + 0.108 350 810 88;
  • 8) 0.108 350 810 88 × 2 = 0 + 0.216 701 621 76;
  • 9) 0.216 701 621 76 × 2 = 0 + 0.433 403 243 52;
  • 10) 0.433 403 243 52 × 2 = 0 + 0.866 806 487 04;
  • 11) 0.866 806 487 04 × 2 = 1 + 0.733 612 974 08;
  • 12) 0.733 612 974 08 × 2 = 1 + 0.467 225 948 16;
  • 13) 0.467 225 948 16 × 2 = 0 + 0.934 451 896 32;
  • 14) 0.934 451 896 32 × 2 = 1 + 0.868 903 792 64;
  • 15) 0.868 903 792 64 × 2 = 1 + 0.737 807 585 28;
  • 16) 0.737 807 585 28 × 2 = 1 + 0.475 615 170 56;
  • 17) 0.475 615 170 56 × 2 = 0 + 0.951 230 341 12;
  • 18) 0.951 230 341 12 × 2 = 1 + 0.902 460 682 24;
  • 19) 0.902 460 682 24 × 2 = 1 + 0.804 921 364 48;
  • 20) 0.804 921 364 48 × 2 = 1 + 0.609 842 728 96;
  • 21) 0.609 842 728 96 × 2 = 1 + 0.219 685 457 92;
  • 22) 0.219 685 457 92 × 2 = 0 + 0.439 370 915 84;
  • 23) 0.439 370 915 84 × 2 = 0 + 0.878 741 831 68;
  • 24) 0.878 741 831 68 × 2 = 1 + 0.757 483 663 36;
  • 25) 0.757 483 663 36 × 2 = 1 + 0.514 967 326 72;
  • 26) 0.514 967 326 72 × 2 = 1 + 0.029 934 653 44;
  • 27) 0.029 934 653 44 × 2 = 0 + 0.059 869 306 88;
  • 28) 0.059 869 306 88 × 2 = 0 + 0.119 738 613 76;
  • 29) 0.119 738 613 76 × 2 = 0 + 0.239 477 227 52;
  • 30) 0.239 477 227 52 × 2 = 0 + 0.478 954 455 04;
  • 31) 0.478 954 455 04 × 2 = 0 + 0.957 908 910 08;
  • 32) 0.957 908 910 08 × 2 = 1 + 0.915 817 820 16;
  • 33) 0.915 817 820 16 × 2 = 1 + 0.831 635 640 32;
  • 34) 0.831 635 640 32 × 2 = 1 + 0.663 271 280 64;
  • 35) 0.663 271 280 64 × 2 = 1 + 0.326 542 561 28;
  • 36) 0.326 542 561 28 × 2 = 0 + 0.653 085 122 56;
  • 37) 0.653 085 122 56 × 2 = 1 + 0.306 170 245 12;
  • 38) 0.306 170 245 12 × 2 = 0 + 0.612 340 490 24;
  • 39) 0.612 340 490 24 × 2 = 1 + 0.224 680 980 48;
  • 40) 0.224 680 980 48 × 2 = 0 + 0.449 361 960 96;
  • 41) 0.449 361 960 96 × 2 = 0 + 0.898 723 921 92;
  • 42) 0.898 723 921 92 × 2 = 1 + 0.797 447 843 84;
  • 43) 0.797 447 843 84 × 2 = 1 + 0.594 895 687 68;
  • 44) 0.594 895 687 68 × 2 = 1 + 0.189 791 375 36;
  • 45) 0.189 791 375 36 × 2 = 0 + 0.379 582 750 72;
  • 46) 0.379 582 750 72 × 2 = 0 + 0.759 165 501 44;
  • 47) 0.759 165 501 44 × 2 = 1 + 0.518 331 002 88;
  • 48) 0.518 331 002 88 × 2 = 1 + 0.036 662 005 76;
  • 49) 0.036 662 005 76 × 2 = 0 + 0.073 324 011 52;
  • 50) 0.073 324 011 52 × 2 = 0 + 0.146 648 023 04;
  • 51) 0.146 648 023 04 × 2 = 0 + 0.293 296 046 08;
  • 52) 0.293 296 046 08 × 2 = 0 + 0.586 592 092 16;
  • 53) 0.586 592 092 16 × 2 = 1 + 0.173 184 184 32;

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:


0.618 033 990 71(10) =

0.1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000 1(2)

5. Positive number before normalization:

1.618 033 990 71(10) =

1.1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000 1(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:


1.618 033 990 71(10) =

1.1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000 1(2) =

1.1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000 1(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):
1.1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000 1


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. 1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000 1 =

1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000


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) =
1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000


Decimal number 1.618 033 990 71 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1001 1110 0011 0111 0111 1001 1100 0001 1110 1010 0111 0011 0000

Share

» Convert decimal 1.618 033 990 92 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 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Your Greatest Competitor, Your Most Powerful Ally. Gemini Notebook YouTube Video of 7.54 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