-0.066 298 275 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 298 275 2(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
-0.066 298 275 2(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. Start with the positive version of the number:

|-0.066 298 275 2| = 0.066 298 275 2


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.066 298 275 2.

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) 0.066 298 275 2 × 2 = 0 + 0.132 596 550 4;
  • 2) 0.132 596 550 4 × 2 = 0 + 0.265 193 100 8;
  • 3) 0.265 193 100 8 × 2 = 0 + 0.530 386 201 6;
  • 4) 0.530 386 201 6 × 2 = 1 + 0.060 772 403 2;
  • 5) 0.060 772 403 2 × 2 = 0 + 0.121 544 806 4;
  • 6) 0.121 544 806 4 × 2 = 0 + 0.243 089 612 8;
  • 7) 0.243 089 612 8 × 2 = 0 + 0.486 179 225 6;
  • 8) 0.486 179 225 6 × 2 = 0 + 0.972 358 451 2;
  • 9) 0.972 358 451 2 × 2 = 1 + 0.944 716 902 4;
  • 10) 0.944 716 902 4 × 2 = 1 + 0.889 433 804 8;
  • 11) 0.889 433 804 8 × 2 = 1 + 0.778 867 609 6;
  • 12) 0.778 867 609 6 × 2 = 1 + 0.557 735 219 2;
  • 13) 0.557 735 219 2 × 2 = 1 + 0.115 470 438 4;
  • 14) 0.115 470 438 4 × 2 = 0 + 0.230 940 876 8;
  • 15) 0.230 940 876 8 × 2 = 0 + 0.461 881 753 6;
  • 16) 0.461 881 753 6 × 2 = 0 + 0.923 763 507 2;
  • 17) 0.923 763 507 2 × 2 = 1 + 0.847 527 014 4;
  • 18) 0.847 527 014 4 × 2 = 1 + 0.695 054 028 8;
  • 19) 0.695 054 028 8 × 2 = 1 + 0.390 108 057 6;
  • 20) 0.390 108 057 6 × 2 = 0 + 0.780 216 115 2;
  • 21) 0.780 216 115 2 × 2 = 1 + 0.560 432 230 4;
  • 22) 0.560 432 230 4 × 2 = 1 + 0.120 864 460 8;
  • 23) 0.120 864 460 8 × 2 = 0 + 0.241 728 921 6;
  • 24) 0.241 728 921 6 × 2 = 0 + 0.483 457 843 2;
  • 25) 0.483 457 843 2 × 2 = 0 + 0.966 915 686 4;
  • 26) 0.966 915 686 4 × 2 = 1 + 0.933 831 372 8;
  • 27) 0.933 831 372 8 × 2 = 1 + 0.867 662 745 6;
  • 28) 0.867 662 745 6 × 2 = 1 + 0.735 325 491 2;
  • 29) 0.735 325 491 2 × 2 = 1 + 0.470 650 982 4;
  • 30) 0.470 650 982 4 × 2 = 0 + 0.941 301 964 8;
  • 31) 0.941 301 964 8 × 2 = 1 + 0.882 603 929 6;
  • 32) 0.882 603 929 6 × 2 = 1 + 0.765 207 859 2;
  • 33) 0.765 207 859 2 × 2 = 1 + 0.530 415 718 4;
  • 34) 0.530 415 718 4 × 2 = 1 + 0.060 831 436 8;
  • 35) 0.060 831 436 8 × 2 = 0 + 0.121 662 873 6;
  • 36) 0.121 662 873 6 × 2 = 0 + 0.243 325 747 2;
  • 37) 0.243 325 747 2 × 2 = 0 + 0.486 651 494 4;
  • 38) 0.486 651 494 4 × 2 = 0 + 0.973 302 988 8;
  • 39) 0.973 302 988 8 × 2 = 1 + 0.946 605 977 6;
  • 40) 0.946 605 977 6 × 2 = 1 + 0.893 211 955 2;
  • 41) 0.893 211 955 2 × 2 = 1 + 0.786 423 910 4;
  • 42) 0.786 423 910 4 × 2 = 1 + 0.572 847 820 8;
  • 43) 0.572 847 820 8 × 2 = 1 + 0.145 695 641 6;
  • 44) 0.145 695 641 6 × 2 = 0 + 0.291 391 283 2;
  • 45) 0.291 391 283 2 × 2 = 0 + 0.582 782 566 4;
  • 46) 0.582 782 566 4 × 2 = 1 + 0.165 565 132 8;
  • 47) 0.165 565 132 8 × 2 = 0 + 0.331 130 265 6;
  • 48) 0.331 130 265 6 × 2 = 0 + 0.662 260 531 2;
  • 49) 0.662 260 531 2 × 2 = 1 + 0.324 521 062 4;
  • 50) 0.324 521 062 4 × 2 = 0 + 0.649 042 124 8;
  • 51) 0.649 042 124 8 × 2 = 1 + 0.298 084 249 6;
  • 52) 0.298 084 249 6 × 2 = 0 + 0.596 168 499 2;
  • 53) 0.596 168 499 2 × 2 = 1 + 0.192 336 998 4;
  • 54) 0.192 336 998 4 × 2 = 0 + 0.384 673 996 8;
  • 55) 0.384 673 996 8 × 2 = 0 + 0.769 347 993 6;
  • 56) 0.769 347 993 6 × 2 = 1 + 0.538 695 987 2;

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).


5. 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:


0.066 298 275 2(10) =

0.0001 0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001(2)

6. Positive number before normalization:

0.066 298 275 2(10) =

0.0001 0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001(2)

7. 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:


0.066 298 275 2(10) =

0.0001 0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001(2) =

0.0001 0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001(2) × 20 =

1.0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001(2) × 2-4


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):
1.0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001


9. 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)


10. 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;

11. 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)


12. 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. 0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001 =

0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1011

Mantissa (52 bits) =
0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001


Decimal number -0.066 298 275 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0000 1111 1000 1110 1100 0111 1011 1100 0011 1110 0100 1010 1001

» Convert decimal -0.066 298 266 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