-0.066 298 264 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 298 264 6(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
-0.066 298 264 6(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:

|-0.066 298 264 6| = 0.066 298 264 6


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: 0.066 298 264 6.

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) 0.066 298 264 6 × 2 = 0 + 0.132 596 529 2;
  • 2) 0.132 596 529 2 × 2 = 0 + 0.265 193 058 4;
  • 3) 0.265 193 058 4 × 2 = 0 + 0.530 386 116 8;
  • 4) 0.530 386 116 8 × 2 = 1 + 0.060 772 233 6;
  • 5) 0.060 772 233 6 × 2 = 0 + 0.121 544 467 2;
  • 6) 0.121 544 467 2 × 2 = 0 + 0.243 088 934 4;
  • 7) 0.243 088 934 4 × 2 = 0 + 0.486 177 868 8;
  • 8) 0.486 177 868 8 × 2 = 0 + 0.972 355 737 6;
  • 9) 0.972 355 737 6 × 2 = 1 + 0.944 711 475 2;
  • 10) 0.944 711 475 2 × 2 = 1 + 0.889 422 950 4;
  • 11) 0.889 422 950 4 × 2 = 1 + 0.778 845 900 8;
  • 12) 0.778 845 900 8 × 2 = 1 + 0.557 691 801 6;
  • 13) 0.557 691 801 6 × 2 = 1 + 0.115 383 603 2;
  • 14) 0.115 383 603 2 × 2 = 0 + 0.230 767 206 4;
  • 15) 0.230 767 206 4 × 2 = 0 + 0.461 534 412 8;
  • 16) 0.461 534 412 8 × 2 = 0 + 0.923 068 825 6;
  • 17) 0.923 068 825 6 × 2 = 1 + 0.846 137 651 2;
  • 18) 0.846 137 651 2 × 2 = 1 + 0.692 275 302 4;
  • 19) 0.692 275 302 4 × 2 = 1 + 0.384 550 604 8;
  • 20) 0.384 550 604 8 × 2 = 0 + 0.769 101 209 6;
  • 21) 0.769 101 209 6 × 2 = 1 + 0.538 202 419 2;
  • 22) 0.538 202 419 2 × 2 = 1 + 0.076 404 838 4;
  • 23) 0.076 404 838 4 × 2 = 0 + 0.152 809 676 8;
  • 24) 0.152 809 676 8 × 2 = 0 + 0.305 619 353 6;
  • 25) 0.305 619 353 6 × 2 = 0 + 0.611 238 707 2;
  • 26) 0.611 238 707 2 × 2 = 1 + 0.222 477 414 4;
  • 27) 0.222 477 414 4 × 2 = 0 + 0.444 954 828 8;
  • 28) 0.444 954 828 8 × 2 = 0 + 0.889 909 657 6;
  • 29) 0.889 909 657 6 × 2 = 1 + 0.779 819 315 2;
  • 30) 0.779 819 315 2 × 2 = 1 + 0.559 638 630 4;
  • 31) 0.559 638 630 4 × 2 = 1 + 0.119 277 260 8;
  • 32) 0.119 277 260 8 × 2 = 0 + 0.238 554 521 6;
  • 33) 0.238 554 521 6 × 2 = 0 + 0.477 109 043 2;
  • 34) 0.477 109 043 2 × 2 = 0 + 0.954 218 086 4;
  • 35) 0.954 218 086 4 × 2 = 1 + 0.908 436 172 8;
  • 36) 0.908 436 172 8 × 2 = 1 + 0.816 872 345 6;
  • 37) 0.816 872 345 6 × 2 = 1 + 0.633 744 691 2;
  • 38) 0.633 744 691 2 × 2 = 1 + 0.267 489 382 4;
  • 39) 0.267 489 382 4 × 2 = 0 + 0.534 978 764 8;
  • 40) 0.534 978 764 8 × 2 = 1 + 0.069 957 529 6;
  • 41) 0.069 957 529 6 × 2 = 0 + 0.139 915 059 2;
  • 42) 0.139 915 059 2 × 2 = 0 + 0.279 830 118 4;
  • 43) 0.279 830 118 4 × 2 = 0 + 0.559 660 236 8;
  • 44) 0.559 660 236 8 × 2 = 1 + 0.119 320 473 6;
  • 45) 0.119 320 473 6 × 2 = 0 + 0.238 640 947 2;
  • 46) 0.238 640 947 2 × 2 = 0 + 0.477 281 894 4;
  • 47) 0.477 281 894 4 × 2 = 0 + 0.954 563 788 8;
  • 48) 0.954 563 788 8 × 2 = 1 + 0.909 127 577 6;
  • 49) 0.909 127 577 6 × 2 = 1 + 0.818 255 155 2;
  • 50) 0.818 255 155 2 × 2 = 1 + 0.636 510 310 4;
  • 51) 0.636 510 310 4 × 2 = 1 + 0.273 020 620 8;
  • 52) 0.273 020 620 8 × 2 = 0 + 0.546 041 241 6;
  • 53) 0.546 041 241 6 × 2 = 1 + 0.092 082 483 2;
  • 54) 0.092 082 483 2 × 2 = 0 + 0.184 164 966 4;
  • 55) 0.184 164 966 4 × 2 = 0 + 0.368 329 932 8;
  • 56) 0.368 329 932 8 × 2 = 0 + 0.736 659 865 6;

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:


0.066 298 264 6(10) =

0.0001 0000 1111 1000 1110 1100 0100 1110 0011 1101 0001 0001 1110 1000(2)

6. Positive number before normalization:

0.066 298 264 6(10) =

0.0001 0000 1111 1000 1110 1100 0100 1110 0011 1101 0001 0001 1110 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:


0.066 298 264 6(10) =

0.0001 0000 1111 1000 1110 1100 0100 1110 0011 1101 0001 0001 1110 1000(2) =

0.0001 0000 1111 1000 1110 1100 0100 1110 0011 1101 0001 0001 1110 1000(2) × 20 =

1.0000 1111 1000 1110 1100 0100 1110 0011 1101 0001 0001 1110 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 0100 1110 0011 1101 0001 0001 1110 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 0100 1110 0011 1101 0001 0001 1110 1000 =

0000 1111 1000 1110 1100 0100 1110 0011 1101 0001 0001 1110 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 0100 1110 0011 1101 0001 0001 1110 1000


Decimal number -0.066 298 264 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0000 1111 1000 1110 1100 0100 1110 0011 1101 0001 0001 1110 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