166.666 666 666 667 73 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 166.666 666 666 667 73(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
166.666 666 666 667 73(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: 166.
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;
  • 166 ÷ 2 = 83 + 0;
  • 83 ÷ 2 = 41 + 1;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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.

166(10) =

1010 0110(2)


3. Convert to binary (base 2) the fractional part: 0.666 666 666 667 73.

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) 0.666 666 666 667 73 × 2 = 1 + 0.333 333 333 335 46;
  • 2) 0.333 333 333 335 46 × 2 = 0 + 0.666 666 666 670 92;
  • 3) 0.666 666 666 670 92 × 2 = 1 + 0.333 333 333 341 84;
  • 4) 0.333 333 333 341 84 × 2 = 0 + 0.666 666 666 683 68;
  • 5) 0.666 666 666 683 68 × 2 = 1 + 0.333 333 333 367 36;
  • 6) 0.333 333 333 367 36 × 2 = 0 + 0.666 666 666 734 72;
  • 7) 0.666 666 666 734 72 × 2 = 1 + 0.333 333 333 469 44;
  • 8) 0.333 333 333 469 44 × 2 = 0 + 0.666 666 666 938 88;
  • 9) 0.666 666 666 938 88 × 2 = 1 + 0.333 333 333 877 76;
  • 10) 0.333 333 333 877 76 × 2 = 0 + 0.666 666 667 755 52;
  • 11) 0.666 666 667 755 52 × 2 = 1 + 0.333 333 335 511 04;
  • 12) 0.333 333 335 511 04 × 2 = 0 + 0.666 666 671 022 08;
  • 13) 0.666 666 671 022 08 × 2 = 1 + 0.333 333 342 044 16;
  • 14) 0.333 333 342 044 16 × 2 = 0 + 0.666 666 684 088 32;
  • 15) 0.666 666 684 088 32 × 2 = 1 + 0.333 333 368 176 64;
  • 16) 0.333 333 368 176 64 × 2 = 0 + 0.666 666 736 353 28;
  • 17) 0.666 666 736 353 28 × 2 = 1 + 0.333 333 472 706 56;
  • 18) 0.333 333 472 706 56 × 2 = 0 + 0.666 666 945 413 12;
  • 19) 0.666 666 945 413 12 × 2 = 1 + 0.333 333 890 826 24;
  • 20) 0.333 333 890 826 24 × 2 = 0 + 0.666 667 781 652 48;
  • 21) 0.666 667 781 652 48 × 2 = 1 + 0.333 335 563 304 96;
  • 22) 0.333 335 563 304 96 × 2 = 0 + 0.666 671 126 609 92;
  • 23) 0.666 671 126 609 92 × 2 = 1 + 0.333 342 253 219 84;
  • 24) 0.333 342 253 219 84 × 2 = 0 + 0.666 684 506 439 68;
  • 25) 0.666 684 506 439 68 × 2 = 1 + 0.333 369 012 879 36;
  • 26) 0.333 369 012 879 36 × 2 = 0 + 0.666 738 025 758 72;
  • 27) 0.666 738 025 758 72 × 2 = 1 + 0.333 476 051 517 44;
  • 28) 0.333 476 051 517 44 × 2 = 0 + 0.666 952 103 034 88;
  • 29) 0.666 952 103 034 88 × 2 = 1 + 0.333 904 206 069 76;
  • 30) 0.333 904 206 069 76 × 2 = 0 + 0.667 808 412 139 52;
  • 31) 0.667 808 412 139 52 × 2 = 1 + 0.335 616 824 279 04;
  • 32) 0.335 616 824 279 04 × 2 = 0 + 0.671 233 648 558 08;
  • 33) 0.671 233 648 558 08 × 2 = 1 + 0.342 467 297 116 16;
  • 34) 0.342 467 297 116 16 × 2 = 0 + 0.684 934 594 232 32;
  • 35) 0.684 934 594 232 32 × 2 = 1 + 0.369 869 188 464 64;
  • 36) 0.369 869 188 464 64 × 2 = 0 + 0.739 738 376 929 28;
  • 37) 0.739 738 376 929 28 × 2 = 1 + 0.479 476 753 858 56;
  • 38) 0.479 476 753 858 56 × 2 = 0 + 0.958 953 507 717 12;
  • 39) 0.958 953 507 717 12 × 2 = 1 + 0.917 907 015 434 24;
  • 40) 0.917 907 015 434 24 × 2 = 1 + 0.835 814 030 868 48;
  • 41) 0.835 814 030 868 48 × 2 = 1 + 0.671 628 061 736 96;
  • 42) 0.671 628 061 736 96 × 2 = 1 + 0.343 256 123 473 92;
  • 43) 0.343 256 123 473 92 × 2 = 0 + 0.686 512 246 947 84;
  • 44) 0.686 512 246 947 84 × 2 = 1 + 0.373 024 493 895 68;
  • 45) 0.373 024 493 895 68 × 2 = 0 + 0.746 048 987 791 36;
  • 46) 0.746 048 987 791 36 × 2 = 1 + 0.492 097 975 582 72;
  • 47) 0.492 097 975 582 72 × 2 = 0 + 0.984 195 951 165 44;
  • 48) 0.984 195 951 165 44 × 2 = 1 + 0.968 391 902 330 88;
  • 49) 0.968 391 902 330 88 × 2 = 1 + 0.936 783 804 661 76;
  • 50) 0.936 783 804 661 76 × 2 = 1 + 0.873 567 609 323 52;
  • 51) 0.873 567 609 323 52 × 2 = 1 + 0.747 135 218 647 04;
  • 52) 0.747 135 218 647 04 × 2 = 1 + 0.494 270 437 294 08;
  • 53) 0.494 270 437 294 08 × 2 = 0 + 0.988 540 874 588 16;

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:


0.666 666 666 667 73(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1101 0101 1111 0(2)

5. Positive number before normalization:

166.666 666 666 667 73(10) =

1010 0110.1010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1101 0101 1111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


166.666 666 666 667 73(10) =

1010 0110.1010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1101 0101 1111 0(2) =

1010 0110.1010 1010 1010 1010 1010 1010 1010 1010 1010 1011 1101 0101 1111 0(2) × 20 =

1.0100 1101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 1010 1011 1110(2) × 27


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): 7

Mantissa (not normalized):
1.0100 1101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 1010 1011 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

7 + 2(11-1) - 1 =

(7 + 1 023)(10) =

1 030(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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1030(10) =

100 0000 0110(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. 0100 1101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 1010 1011 1110 =

0100 1101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 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) =
100 0000 0110

Mantissa (52 bits) =
0100 1101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 1010


Decimal number 166.666 666 666 667 73 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0110 - 0100 1101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0111 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