-0.003 226 860 63 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.003 226 860 63(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
-0.003 226 860 63(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:

|-0.003 226 860 63| = 0.003 226 860 63


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: 0.003 226 860 63.

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) 0.003 226 860 63 × 2 = 0 + 0.006 453 721 26;
  • 2) 0.006 453 721 26 × 2 = 0 + 0.012 907 442 52;
  • 3) 0.012 907 442 52 × 2 = 0 + 0.025 814 885 04;
  • 4) 0.025 814 885 04 × 2 = 0 + 0.051 629 770 08;
  • 5) 0.051 629 770 08 × 2 = 0 + 0.103 259 540 16;
  • 6) 0.103 259 540 16 × 2 = 0 + 0.206 519 080 32;
  • 7) 0.206 519 080 32 × 2 = 0 + 0.413 038 160 64;
  • 8) 0.413 038 160 64 × 2 = 0 + 0.826 076 321 28;
  • 9) 0.826 076 321 28 × 2 = 1 + 0.652 152 642 56;
  • 10) 0.652 152 642 56 × 2 = 1 + 0.304 305 285 12;
  • 11) 0.304 305 285 12 × 2 = 0 + 0.608 610 570 24;
  • 12) 0.608 610 570 24 × 2 = 1 + 0.217 221 140 48;
  • 13) 0.217 221 140 48 × 2 = 0 + 0.434 442 280 96;
  • 14) 0.434 442 280 96 × 2 = 0 + 0.868 884 561 92;
  • 15) 0.868 884 561 92 × 2 = 1 + 0.737 769 123 84;
  • 16) 0.737 769 123 84 × 2 = 1 + 0.475 538 247 68;
  • 17) 0.475 538 247 68 × 2 = 0 + 0.951 076 495 36;
  • 18) 0.951 076 495 36 × 2 = 1 + 0.902 152 990 72;
  • 19) 0.902 152 990 72 × 2 = 1 + 0.804 305 981 44;
  • 20) 0.804 305 981 44 × 2 = 1 + 0.608 611 962 88;
  • 21) 0.608 611 962 88 × 2 = 1 + 0.217 223 925 76;
  • 22) 0.217 223 925 76 × 2 = 0 + 0.434 447 851 52;
  • 23) 0.434 447 851 52 × 2 = 0 + 0.868 895 703 04;
  • 24) 0.868 895 703 04 × 2 = 1 + 0.737 791 406 08;
  • 25) 0.737 791 406 08 × 2 = 1 + 0.475 582 812 16;
  • 26) 0.475 582 812 16 × 2 = 0 + 0.951 165 624 32;
  • 27) 0.951 165 624 32 × 2 = 1 + 0.902 331 248 64;
  • 28) 0.902 331 248 64 × 2 = 1 + 0.804 662 497 28;
  • 29) 0.804 662 497 28 × 2 = 1 + 0.609 324 994 56;
  • 30) 0.609 324 994 56 × 2 = 1 + 0.218 649 989 12;
  • 31) 0.218 649 989 12 × 2 = 0 + 0.437 299 978 24;
  • 32) 0.437 299 978 24 × 2 = 0 + 0.874 599 956 48;
  • 33) 0.874 599 956 48 × 2 = 1 + 0.749 199 912 96;
  • 34) 0.749 199 912 96 × 2 = 1 + 0.498 399 825 92;
  • 35) 0.498 399 825 92 × 2 = 0 + 0.996 799 651 84;
  • 36) 0.996 799 651 84 × 2 = 1 + 0.993 599 303 68;
  • 37) 0.993 599 303 68 × 2 = 1 + 0.987 198 607 36;
  • 38) 0.987 198 607 36 × 2 = 1 + 0.974 397 214 72;
  • 39) 0.974 397 214 72 × 2 = 1 + 0.948 794 429 44;
  • 40) 0.948 794 429 44 × 2 = 1 + 0.897 588 858 88;
  • 41) 0.897 588 858 88 × 2 = 1 + 0.795 177 717 76;
  • 42) 0.795 177 717 76 × 2 = 1 + 0.590 355 435 52;
  • 43) 0.590 355 435 52 × 2 = 1 + 0.180 710 871 04;
  • 44) 0.180 710 871 04 × 2 = 0 + 0.361 421 742 08;
  • 45) 0.361 421 742 08 × 2 = 0 + 0.722 843 484 16;
  • 46) 0.722 843 484 16 × 2 = 1 + 0.445 686 968 32;
  • 47) 0.445 686 968 32 × 2 = 0 + 0.891 373 936 64;
  • 48) 0.891 373 936 64 × 2 = 1 + 0.782 747 873 28;
  • 49) 0.782 747 873 28 × 2 = 1 + 0.565 495 746 56;
  • 50) 0.565 495 746 56 × 2 = 1 + 0.130 991 493 12;
  • 51) 0.130 991 493 12 × 2 = 0 + 0.261 982 986 24;
  • 52) 0.261 982 986 24 × 2 = 0 + 0.523 965 972 48;
  • 53) 0.523 965 972 48 × 2 = 1 + 0.047 931 944 96;
  • 54) 0.047 931 944 96 × 2 = 0 + 0.095 863 889 92;
  • 55) 0.095 863 889 92 × 2 = 0 + 0.191 727 779 84;
  • 56) 0.191 727 779 84 × 2 = 0 + 0.383 455 559 68;
  • 57) 0.383 455 559 68 × 2 = 0 + 0.766 911 119 36;
  • 58) 0.766 911 119 36 × 2 = 1 + 0.533 822 238 72;
  • 59) 0.533 822 238 72 × 2 = 1 + 0.067 644 477 44;
  • 60) 0.067 644 477 44 × 2 = 0 + 0.135 288 954 88;
  • 61) 0.135 288 954 88 × 2 = 0 + 0.270 577 909 76;

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:


0.003 226 860 63(10) =

0.0000 0000 1101 0011 0111 1001 1011 1100 1101 1111 1110 0101 1100 1000 0110 0(2)

6. Positive number before normalization:

0.003 226 860 63(10) =

0.0000 0000 1101 0011 0111 1001 1011 1100 1101 1111 1110 0101 1100 1000 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.003 226 860 63(10) =

0.0000 0000 1101 0011 0111 1001 1011 1100 1101 1111 1110 0101 1100 1000 0110 0(2) =

0.0000 0000 1101 0011 0111 1001 1011 1100 1101 1111 1110 0101 1100 1000 0110 0(2) × 20 =

1.1010 0110 1111 0011 0111 1001 1011 1111 1100 1011 1001 0000 1100(2) × 2-9


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): -9

Mantissa (not normalized):
1.1010 0110 1111 0011 0111 1001 1011 1111 1100 1011 1001 0000 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-9 + 2(11-1) - 1 =

(-9 + 1 023)(10) =

1 014(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;
  • 1 014 ÷ 2 = 507 + 0;
  • 507 ÷ 2 = 253 + 1;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1014(10) =

011 1111 0110(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. 1010 0110 1111 0011 0111 1001 1011 1111 1100 1011 1001 0000 1100 =

1010 0110 1111 0011 0111 1001 1011 1111 1100 1011 1001 0000 1100


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 0110

Mantissa (52 bits) =
1010 0110 1111 0011 0111 1001 1011 1111 1100 1011 1001 0000 1100


Decimal number -0.003 226 860 63 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0110 - 1010 0110 1111 0011 0111 1001 1011 1111 1100 1011 1001 0000 1100

» Convert decimal -0.003 226 860 48 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