0.666 666 666 666 673 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.666 666 666 666 673 4(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
0.666 666 666 666 673 4(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: 0.666 666 666 666 673 4.

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) 0.666 666 666 666 673 4 × 2 = 1 + 0.333 333 333 333 346 8;
  • 2) 0.333 333 333 333 346 8 × 2 = 0 + 0.666 666 666 666 693 6;
  • 3) 0.666 666 666 666 693 6 × 2 = 1 + 0.333 333 333 333 387 2;
  • 4) 0.333 333 333 333 387 2 × 2 = 0 + 0.666 666 666 666 774 4;
  • 5) 0.666 666 666 666 774 4 × 2 = 1 + 0.333 333 333 333 548 8;
  • 6) 0.333 333 333 333 548 8 × 2 = 0 + 0.666 666 666 667 097 6;
  • 7) 0.666 666 666 667 097 6 × 2 = 1 + 0.333 333 333 334 195 2;
  • 8) 0.333 333 333 334 195 2 × 2 = 0 + 0.666 666 666 668 390 4;
  • 9) 0.666 666 666 668 390 4 × 2 = 1 + 0.333 333 333 336 780 8;
  • 10) 0.333 333 333 336 780 8 × 2 = 0 + 0.666 666 666 673 561 6;
  • 11) 0.666 666 666 673 561 6 × 2 = 1 + 0.333 333 333 347 123 2;
  • 12) 0.333 333 333 347 123 2 × 2 = 0 + 0.666 666 666 694 246 4;
  • 13) 0.666 666 666 694 246 4 × 2 = 1 + 0.333 333 333 388 492 8;
  • 14) 0.333 333 333 388 492 8 × 2 = 0 + 0.666 666 666 776 985 6;
  • 15) 0.666 666 666 776 985 6 × 2 = 1 + 0.333 333 333 553 971 2;
  • 16) 0.333 333 333 553 971 2 × 2 = 0 + 0.666 666 667 107 942 4;
  • 17) 0.666 666 667 107 942 4 × 2 = 1 + 0.333 333 334 215 884 8;
  • 18) 0.333 333 334 215 884 8 × 2 = 0 + 0.666 666 668 431 769 6;
  • 19) 0.666 666 668 431 769 6 × 2 = 1 + 0.333 333 336 863 539 2;
  • 20) 0.333 333 336 863 539 2 × 2 = 0 + 0.666 666 673 727 078 4;
  • 21) 0.666 666 673 727 078 4 × 2 = 1 + 0.333 333 347 454 156 8;
  • 22) 0.333 333 347 454 156 8 × 2 = 0 + 0.666 666 694 908 313 6;
  • 23) 0.666 666 694 908 313 6 × 2 = 1 + 0.333 333 389 816 627 2;
  • 24) 0.333 333 389 816 627 2 × 2 = 0 + 0.666 666 779 633 254 4;
  • 25) 0.666 666 779 633 254 4 × 2 = 1 + 0.333 333 559 266 508 8;
  • 26) 0.333 333 559 266 508 8 × 2 = 0 + 0.666 667 118 533 017 6;
  • 27) 0.666 667 118 533 017 6 × 2 = 1 + 0.333 334 237 066 035 2;
  • 28) 0.333 334 237 066 035 2 × 2 = 0 + 0.666 668 474 132 070 4;
  • 29) 0.666 668 474 132 070 4 × 2 = 1 + 0.333 336 948 264 140 8;
  • 30) 0.333 336 948 264 140 8 × 2 = 0 + 0.666 673 896 528 281 6;
  • 31) 0.666 673 896 528 281 6 × 2 = 1 + 0.333 347 793 056 563 2;
  • 32) 0.333 347 793 056 563 2 × 2 = 0 + 0.666 695 586 113 126 4;
  • 33) 0.666 695 586 113 126 4 × 2 = 1 + 0.333 391 172 226 252 8;
  • 34) 0.333 391 172 226 252 8 × 2 = 0 + 0.666 782 344 452 505 6;
  • 35) 0.666 782 344 452 505 6 × 2 = 1 + 0.333 564 688 905 011 2;
  • 36) 0.333 564 688 905 011 2 × 2 = 0 + 0.667 129 377 810 022 4;
  • 37) 0.667 129 377 810 022 4 × 2 = 1 + 0.334 258 755 620 044 8;
  • 38) 0.334 258 755 620 044 8 × 2 = 0 + 0.668 517 511 240 089 6;
  • 39) 0.668 517 511 240 089 6 × 2 = 1 + 0.337 035 022 480 179 2;
  • 40) 0.337 035 022 480 179 2 × 2 = 0 + 0.674 070 044 960 358 4;
  • 41) 0.674 070 044 960 358 4 × 2 = 1 + 0.348 140 089 920 716 8;
  • 42) 0.348 140 089 920 716 8 × 2 = 0 + 0.696 280 179 841 433 6;
  • 43) 0.696 280 179 841 433 6 × 2 = 1 + 0.392 560 359 682 867 2;
  • 44) 0.392 560 359 682 867 2 × 2 = 0 + 0.785 120 719 365 734 4;
  • 45) 0.785 120 719 365 734 4 × 2 = 1 + 0.570 241 438 731 468 8;
  • 46) 0.570 241 438 731 468 8 × 2 = 1 + 0.140 482 877 462 937 6;
  • 47) 0.140 482 877 462 937 6 × 2 = 0 + 0.280 965 754 925 875 2;
  • 48) 0.280 965 754 925 875 2 × 2 = 0 + 0.561 931 509 851 750 4;
  • 49) 0.561 931 509 851 750 4 × 2 = 1 + 0.123 863 019 703 500 8;
  • 50) 0.123 863 019 703 500 8 × 2 = 0 + 0.247 726 039 407 001 6;
  • 51) 0.247 726 039 407 001 6 × 2 = 0 + 0.495 452 078 814 003 2;
  • 52) 0.495 452 078 814 003 2 × 2 = 0 + 0.990 904 157 628 006 4;
  • 53) 0.990 904 157 628 006 4 × 2 = 1 + 0.981 808 315 256 012 8;

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:


0.666 666 666 666 673 4(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1100 1000 1(2)

5. Positive number before normalization:

0.666 666 666 666 673 4(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1100 1000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.666 666 666 666 673 4(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1100 1000 1(2) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1100 1000 1(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0001(2) × 2-1


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): -1

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 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) =

1022(10) =

011 1111 1110(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. 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0001 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0001


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 1110

Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0001


Decimal number 0.666 666 666 666 673 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1001 0001

» Convert decimal 0.666 666 666 666 671 5 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