-0.066 298 280 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 298 280 8(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
-0.066 298 280 8(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:

|-0.066 298 280 8| = 0.066 298 280 8


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: 0.066 298 280 8.

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) 0.066 298 280 8 × 2 = 0 + 0.132 596 561 6;
  • 2) 0.132 596 561 6 × 2 = 0 + 0.265 193 123 2;
  • 3) 0.265 193 123 2 × 2 = 0 + 0.530 386 246 4;
  • 4) 0.530 386 246 4 × 2 = 1 + 0.060 772 492 8;
  • 5) 0.060 772 492 8 × 2 = 0 + 0.121 544 985 6;
  • 6) 0.121 544 985 6 × 2 = 0 + 0.243 089 971 2;
  • 7) 0.243 089 971 2 × 2 = 0 + 0.486 179 942 4;
  • 8) 0.486 179 942 4 × 2 = 0 + 0.972 359 884 8;
  • 9) 0.972 359 884 8 × 2 = 1 + 0.944 719 769 6;
  • 10) 0.944 719 769 6 × 2 = 1 + 0.889 439 539 2;
  • 11) 0.889 439 539 2 × 2 = 1 + 0.778 879 078 4;
  • 12) 0.778 879 078 4 × 2 = 1 + 0.557 758 156 8;
  • 13) 0.557 758 156 8 × 2 = 1 + 0.115 516 313 6;
  • 14) 0.115 516 313 6 × 2 = 0 + 0.231 032 627 2;
  • 15) 0.231 032 627 2 × 2 = 0 + 0.462 065 254 4;
  • 16) 0.462 065 254 4 × 2 = 0 + 0.924 130 508 8;
  • 17) 0.924 130 508 8 × 2 = 1 + 0.848 261 017 6;
  • 18) 0.848 261 017 6 × 2 = 1 + 0.696 522 035 2;
  • 19) 0.696 522 035 2 × 2 = 1 + 0.393 044 070 4;
  • 20) 0.393 044 070 4 × 2 = 0 + 0.786 088 140 8;
  • 21) 0.786 088 140 8 × 2 = 1 + 0.572 176 281 6;
  • 22) 0.572 176 281 6 × 2 = 1 + 0.144 352 563 2;
  • 23) 0.144 352 563 2 × 2 = 0 + 0.288 705 126 4;
  • 24) 0.288 705 126 4 × 2 = 0 + 0.577 410 252 8;
  • 25) 0.577 410 252 8 × 2 = 1 + 0.154 820 505 6;
  • 26) 0.154 820 505 6 × 2 = 0 + 0.309 641 011 2;
  • 27) 0.309 641 011 2 × 2 = 0 + 0.619 282 022 4;
  • 28) 0.619 282 022 4 × 2 = 1 + 0.238 564 044 8;
  • 29) 0.238 564 044 8 × 2 = 0 + 0.477 128 089 6;
  • 30) 0.477 128 089 6 × 2 = 0 + 0.954 256 179 2;
  • 31) 0.954 256 179 2 × 2 = 1 + 0.908 512 358 4;
  • 32) 0.908 512 358 4 × 2 = 1 + 0.817 024 716 8;
  • 33) 0.817 024 716 8 × 2 = 1 + 0.634 049 433 6;
  • 34) 0.634 049 433 6 × 2 = 1 + 0.268 098 867 2;
  • 35) 0.268 098 867 2 × 2 = 0 + 0.536 197 734 4;
  • 36) 0.536 197 734 4 × 2 = 1 + 0.072 395 468 8;
  • 37) 0.072 395 468 8 × 2 = 0 + 0.144 790 937 6;
  • 38) 0.144 790 937 6 × 2 = 0 + 0.289 581 875 2;
  • 39) 0.289 581 875 2 × 2 = 0 + 0.579 163 750 4;
  • 40) 0.579 163 750 4 × 2 = 1 + 0.158 327 500 8;
  • 41) 0.158 327 500 8 × 2 = 0 + 0.316 655 001 6;
  • 42) 0.316 655 001 6 × 2 = 0 + 0.633 310 003 2;
  • 43) 0.633 310 003 2 × 2 = 1 + 0.266 620 006 4;
  • 44) 0.266 620 006 4 × 2 = 0 + 0.533 240 012 8;
  • 45) 0.533 240 012 8 × 2 = 1 + 0.066 480 025 6;
  • 46) 0.066 480 025 6 × 2 = 0 + 0.132 960 051 2;
  • 47) 0.132 960 051 2 × 2 = 0 + 0.265 920 102 4;
  • 48) 0.265 920 102 4 × 2 = 0 + 0.531 840 204 8;
  • 49) 0.531 840 204 8 × 2 = 1 + 0.063 680 409 6;
  • 50) 0.063 680 409 6 × 2 = 0 + 0.127 360 819 2;
  • 51) 0.127 360 819 2 × 2 = 0 + 0.254 721 638 4;
  • 52) 0.254 721 638 4 × 2 = 0 + 0.509 443 276 8;
  • 53) 0.509 443 276 8 × 2 = 1 + 0.018 886 553 6;
  • 54) 0.018 886 553 6 × 2 = 0 + 0.037 773 107 2;
  • 55) 0.037 773 107 2 × 2 = 0 + 0.075 546 214 4;
  • 56) 0.075 546 214 4 × 2 = 0 + 0.151 092 428 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).


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:


0.066 298 280 8(10) =

0.0001 0000 1111 1000 1110 1100 1001 0011 1101 0001 0010 1000 1000 1000(2)

6. Positive number before normalization:

0.066 298 280 8(10) =

0.0001 0000 1111 1000 1110 1100 1001 0011 1101 0001 0010 1000 1000 1000(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:


0.066 298 280 8(10) =

0.0001 0000 1111 1000 1110 1100 1001 0011 1101 0001 0010 1000 1000 1000(2) =

0.0001 0000 1111 1000 1110 1100 1001 0011 1101 0001 0010 1000 1000 1000(2) × 20 =

1.0000 1111 1000 1110 1100 1001 0011 1101 0001 0010 1000 1000 1000(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 1001 0011 1101 0001 0010 1000 1000 1000


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 1001 0011 1101 0001 0010 1000 1000 1000 =

0000 1111 1000 1110 1100 1001 0011 1101 0001 0010 1000 1000 1000


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 1001 0011 1101 0001 0010 1000 1000 1000


Decimal number -0.066 298 280 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0000 1111 1000 1110 1100 1001 0011 1101 0001 0010 1000 1000 1000

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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