Convert 1.618 033 97 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

1.618 033 97(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.618 033 97.

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 97 × 2 = 1 + 0.236 067 94;
  • 2) 0.236 067 94 × 2 = 0 + 0.472 135 88;
  • 3) 0.472 135 88 × 2 = 0 + 0.944 271 76;
  • 4) 0.944 271 76 × 2 = 1 + 0.888 543 52;
  • 5) 0.888 543 52 × 2 = 1 + 0.777 087 04;
  • 6) 0.777 087 04 × 2 = 1 + 0.554 174 08;
  • 7) 0.554 174 08 × 2 = 1 + 0.108 348 16;
  • 8) 0.108 348 16 × 2 = 0 + 0.216 696 32;
  • 9) 0.216 696 32 × 2 = 0 + 0.433 392 64;
  • 10) 0.433 392 64 × 2 = 0 + 0.866 785 28;
  • 11) 0.866 785 28 × 2 = 1 + 0.733 570 56;
  • 12) 0.733 570 56 × 2 = 1 + 0.467 141 12;
  • 13) 0.467 141 12 × 2 = 0 + 0.934 282 24;
  • 14) 0.934 282 24 × 2 = 1 + 0.868 564 48;
  • 15) 0.868 564 48 × 2 = 1 + 0.737 128 96;
  • 16) 0.737 128 96 × 2 = 1 + 0.474 257 92;
  • 17) 0.474 257 92 × 2 = 0 + 0.948 515 84;
  • 18) 0.948 515 84 × 2 = 1 + 0.897 031 68;
  • 19) 0.897 031 68 × 2 = 1 + 0.794 063 36;
  • 20) 0.794 063 36 × 2 = 1 + 0.588 126 72;
  • 21) 0.588 126 72 × 2 = 1 + 0.176 253 44;
  • 22) 0.176 253 44 × 2 = 0 + 0.352 506 88;
  • 23) 0.352 506 88 × 2 = 0 + 0.705 013 76;
  • 24) 0.705 013 76 × 2 = 1 + 0.410 027 52;
  • 25) 0.410 027 52 × 2 = 0 + 0.820 055 04;
  • 26) 0.820 055 04 × 2 = 1 + 0.640 110 08;
  • 27) 0.640 110 08 × 2 = 1 + 0.280 220 16;
  • 28) 0.280 220 16 × 2 = 0 + 0.560 440 32;
  • 29) 0.560 440 32 × 2 = 1 + 0.120 880 64;
  • 30) 0.120 880 64 × 2 = 0 + 0.241 761 28;
  • 31) 0.241 761 28 × 2 = 0 + 0.483 522 56;
  • 32) 0.483 522 56 × 2 = 0 + 0.967 045 12;
  • 33) 0.967 045 12 × 2 = 1 + 0.934 090 24;
  • 34) 0.934 090 24 × 2 = 1 + 0.868 180 48;
  • 35) 0.868 180 48 × 2 = 1 + 0.736 360 96;
  • 36) 0.736 360 96 × 2 = 1 + 0.472 721 92;
  • 37) 0.472 721 92 × 2 = 0 + 0.945 443 84;
  • 38) 0.945 443 84 × 2 = 1 + 0.890 887 68;
  • 39) 0.890 887 68 × 2 = 1 + 0.781 775 36;
  • 40) 0.781 775 36 × 2 = 1 + 0.563 550 72;
  • 41) 0.563 550 72 × 2 = 1 + 0.127 101 44;
  • 42) 0.127 101 44 × 2 = 0 + 0.254 202 88;
  • 43) 0.254 202 88 × 2 = 0 + 0.508 405 76;
  • 44) 0.508 405 76 × 2 = 1 + 0.016 811 52;
  • 45) 0.016 811 52 × 2 = 0 + 0.033 623 04;
  • 46) 0.033 623 04 × 2 = 0 + 0.067 246 08;
  • 47) 0.067 246 08 × 2 = 0 + 0.134 492 16;
  • 48) 0.134 492 16 × 2 = 0 + 0.268 984 32;
  • 49) 0.268 984 32 × 2 = 0 + 0.537 968 64;
  • 50) 0.537 968 64 × 2 = 1 + 0.075 937 28;
  • 51) 0.075 937 28 × 2 = 0 + 0.151 874 56;
  • 52) 0.151 874 56 × 2 = 0 + 0.303 749 12;
  • 53) 0.303 749 12 × 2 = 0 + 0.607 498 24;

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...)


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 97(10) =


0.1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100 0(2)


5. Positive number before normalization:

1.618 033 97(10) =


1.1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100 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:

1.618 033 97(10) =


1.1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100 0(2) =


1.1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100 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):
1.1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100 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. 1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100 0 =


1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100


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 0110 1000 1111 0111 1001 0000 0100


Number 1.618 033 97 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1111 - 1001 1110 0011 0111 0111 1001 0110 1000 1111 0111 1001 0000 0100

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 1

      47
    • 1

      46
    • 1

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 1

      41
    • 1

      40
    • 0

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 0

      35
    • 1

      34
    • 1

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

1.618 033 96 = ? ... 1.618 033 98 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

1.618 033 97 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:45 UTC (GMT)
1 504 111 578 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:45 UTC (GMT)
-1.062 3 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:45 UTC (GMT)
0.333 333 333 333 333 314 829 616 256 247 390 992 939 472 199 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:44 UTC (GMT)
1.785 714 285 714 285 714 285 714 3 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:44 UTC (GMT)
99 999 999 999 999 988 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:44 UTC (GMT)
3 234 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:43 UTC (GMT)
0.036 862 999 999 999 999 933 830 707 732 340 670 190 751 552 581 787 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:43 UTC (GMT)
-76.14 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:43 UTC (GMT)
-0.000 123 4 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:43 UTC (GMT)
1.797 693 134 9 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:43 UTC (GMT)
0.000 000 476 837 159 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:42 UTC (GMT)
0.722 399 999 to 64 bit double precision IEEE 754 binary floating point = ? Sep 20 02:42 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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