0.066 666 666 651 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.066 666 666 651(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
0.066 666 666 651(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.066 666 666 651.

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) 0.066 666 666 651 × 2 = 0 + 0.133 333 333 302;
  • 2) 0.133 333 333 302 × 2 = 0 + 0.266 666 666 604;
  • 3) 0.266 666 666 604 × 2 = 0 + 0.533 333 333 208;
  • 4) 0.533 333 333 208 × 2 = 1 + 0.066 666 666 416;
  • 5) 0.066 666 666 416 × 2 = 0 + 0.133 333 332 832;
  • 6) 0.133 333 332 832 × 2 = 0 + 0.266 666 665 664;
  • 7) 0.266 666 665 664 × 2 = 0 + 0.533 333 331 328;
  • 8) 0.533 333 331 328 × 2 = 1 + 0.066 666 662 656;
  • 9) 0.066 666 662 656 × 2 = 0 + 0.133 333 325 312;
  • 10) 0.133 333 325 312 × 2 = 0 + 0.266 666 650 624;
  • 11) 0.266 666 650 624 × 2 = 0 + 0.533 333 301 248;
  • 12) 0.533 333 301 248 × 2 = 1 + 0.066 666 602 496;
  • 13) 0.066 666 602 496 × 2 = 0 + 0.133 333 204 992;
  • 14) 0.133 333 204 992 × 2 = 0 + 0.266 666 409 984;
  • 15) 0.266 666 409 984 × 2 = 0 + 0.533 332 819 968;
  • 16) 0.533 332 819 968 × 2 = 1 + 0.066 665 639 936;
  • 17) 0.066 665 639 936 × 2 = 0 + 0.133 331 279 872;
  • 18) 0.133 331 279 872 × 2 = 0 + 0.266 662 559 744;
  • 19) 0.266 662 559 744 × 2 = 0 + 0.533 325 119 488;
  • 20) 0.533 325 119 488 × 2 = 1 + 0.066 650 238 976;
  • 21) 0.066 650 238 976 × 2 = 0 + 0.133 300 477 952;
  • 22) 0.133 300 477 952 × 2 = 0 + 0.266 600 955 904;
  • 23) 0.266 600 955 904 × 2 = 0 + 0.533 201 911 808;
  • 24) 0.533 201 911 808 × 2 = 1 + 0.066 403 823 616;
  • 25) 0.066 403 823 616 × 2 = 0 + 0.132 807 647 232;
  • 26) 0.132 807 647 232 × 2 = 0 + 0.265 615 294 464;
  • 27) 0.265 615 294 464 × 2 = 0 + 0.531 230 588 928;
  • 28) 0.531 230 588 928 × 2 = 1 + 0.062 461 177 856;
  • 29) 0.062 461 177 856 × 2 = 0 + 0.124 922 355 712;
  • 30) 0.124 922 355 712 × 2 = 0 + 0.249 844 711 424;
  • 31) 0.249 844 711 424 × 2 = 0 + 0.499 689 422 848;
  • 32) 0.499 689 422 848 × 2 = 0 + 0.999 378 845 696;
  • 33) 0.999 378 845 696 × 2 = 1 + 0.998 757 691 392;
  • 34) 0.998 757 691 392 × 2 = 1 + 0.997 515 382 784;
  • 35) 0.997 515 382 784 × 2 = 1 + 0.995 030 765 568;
  • 36) 0.995 030 765 568 × 2 = 1 + 0.990 061 531 136;
  • 37) 0.990 061 531 136 × 2 = 1 + 0.980 123 062 272;
  • 38) 0.980 123 062 272 × 2 = 1 + 0.960 246 124 544;
  • 39) 0.960 246 124 544 × 2 = 1 + 0.920 492 249 088;
  • 40) 0.920 492 249 088 × 2 = 1 + 0.840 984 498 176;
  • 41) 0.840 984 498 176 × 2 = 1 + 0.681 968 996 352;
  • 42) 0.681 968 996 352 × 2 = 1 + 0.363 937 992 704;
  • 43) 0.363 937 992 704 × 2 = 0 + 0.727 875 985 408;
  • 44) 0.727 875 985 408 × 2 = 1 + 0.455 751 970 816;
  • 45) 0.455 751 970 816 × 2 = 0 + 0.911 503 941 632;
  • 46) 0.911 503 941 632 × 2 = 1 + 0.823 007 883 264;
  • 47) 0.823 007 883 264 × 2 = 1 + 0.646 015 766 528;
  • 48) 0.646 015 766 528 × 2 = 1 + 0.292 031 533 056;
  • 49) 0.292 031 533 056 × 2 = 0 + 0.584 063 066 112;
  • 50) 0.584 063 066 112 × 2 = 1 + 0.168 126 132 224;
  • 51) 0.168 126 132 224 × 2 = 0 + 0.336 252 264 448;
  • 52) 0.336 252 264 448 × 2 = 0 + 0.672 504 528 896;
  • 53) 0.672 504 528 896 × 2 = 1 + 0.345 009 057 792;
  • 54) 0.345 009 057 792 × 2 = 0 + 0.690 018 115 584;
  • 55) 0.690 018 115 584 × 2 = 1 + 0.380 036 231 168;
  • 56) 0.380 036 231 168 × 2 = 0 + 0.760 072 462 336;

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:


0.066 666 666 651(10) =

0.0001 0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010(2)

5. Positive number before normalization:

0.066 666 666 651(10) =

0.0001 0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.066 666 666 651(10) =

0.0001 0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010(2) =

0.0001 0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010(2) × 20 =

1.0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010(2) × 2-4


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 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010


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) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1019(10) =

011 1111 1011(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010 =

0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 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 1011

Mantissa (52 bits) =
0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 1010


Decimal number 0.066 666 666 651 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 0001 0001 0001 0001 0001 0001 0000 1111 1111 1101 0111 0100 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