-0.000 282 012 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 012 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
-0.000 282 012 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:

|-0.000 282 012 6| = 0.000 282 012 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: 0.000 282 012 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) 0.000 282 012 6 × 2 = 0 + 0.000 564 025 2;
  • 2) 0.000 564 025 2 × 2 = 0 + 0.001 128 050 4;
  • 3) 0.001 128 050 4 × 2 = 0 + 0.002 256 100 8;
  • 4) 0.002 256 100 8 × 2 = 0 + 0.004 512 201 6;
  • 5) 0.004 512 201 6 × 2 = 0 + 0.009 024 403 2;
  • 6) 0.009 024 403 2 × 2 = 0 + 0.018 048 806 4;
  • 7) 0.018 048 806 4 × 2 = 0 + 0.036 097 612 8;
  • 8) 0.036 097 612 8 × 2 = 0 + 0.072 195 225 6;
  • 9) 0.072 195 225 6 × 2 = 0 + 0.144 390 451 2;
  • 10) 0.144 390 451 2 × 2 = 0 + 0.288 780 902 4;
  • 11) 0.288 780 902 4 × 2 = 0 + 0.577 561 804 8;
  • 12) 0.577 561 804 8 × 2 = 1 + 0.155 123 609 6;
  • 13) 0.155 123 609 6 × 2 = 0 + 0.310 247 219 2;
  • 14) 0.310 247 219 2 × 2 = 0 + 0.620 494 438 4;
  • 15) 0.620 494 438 4 × 2 = 1 + 0.240 988 876 8;
  • 16) 0.240 988 876 8 × 2 = 0 + 0.481 977 753 6;
  • 17) 0.481 977 753 6 × 2 = 0 + 0.963 955 507 2;
  • 18) 0.963 955 507 2 × 2 = 1 + 0.927 911 014 4;
  • 19) 0.927 911 014 4 × 2 = 1 + 0.855 822 028 8;
  • 20) 0.855 822 028 8 × 2 = 1 + 0.711 644 057 6;
  • 21) 0.711 644 057 6 × 2 = 1 + 0.423 288 115 2;
  • 22) 0.423 288 115 2 × 2 = 0 + 0.846 576 230 4;
  • 23) 0.846 576 230 4 × 2 = 1 + 0.693 152 460 8;
  • 24) 0.693 152 460 8 × 2 = 1 + 0.386 304 921 6;
  • 25) 0.386 304 921 6 × 2 = 0 + 0.772 609 843 2;
  • 26) 0.772 609 843 2 × 2 = 1 + 0.545 219 686 4;
  • 27) 0.545 219 686 4 × 2 = 1 + 0.090 439 372 8;
  • 28) 0.090 439 372 8 × 2 = 0 + 0.180 878 745 6;
  • 29) 0.180 878 745 6 × 2 = 0 + 0.361 757 491 2;
  • 30) 0.361 757 491 2 × 2 = 0 + 0.723 514 982 4;
  • 31) 0.723 514 982 4 × 2 = 1 + 0.447 029 964 8;
  • 32) 0.447 029 964 8 × 2 = 0 + 0.894 059 929 6;
  • 33) 0.894 059 929 6 × 2 = 1 + 0.788 119 859 2;
  • 34) 0.788 119 859 2 × 2 = 1 + 0.576 239 718 4;
  • 35) 0.576 239 718 4 × 2 = 1 + 0.152 479 436 8;
  • 36) 0.152 479 436 8 × 2 = 0 + 0.304 958 873 6;
  • 37) 0.304 958 873 6 × 2 = 0 + 0.609 917 747 2;
  • 38) 0.609 917 747 2 × 2 = 1 + 0.219 835 494 4;
  • 39) 0.219 835 494 4 × 2 = 0 + 0.439 670 988 8;
  • 40) 0.439 670 988 8 × 2 = 0 + 0.879 341 977 6;
  • 41) 0.879 341 977 6 × 2 = 1 + 0.758 683 955 2;
  • 42) 0.758 683 955 2 × 2 = 1 + 0.517 367 910 4;
  • 43) 0.517 367 910 4 × 2 = 1 + 0.034 735 820 8;
  • 44) 0.034 735 820 8 × 2 = 0 + 0.069 471 641 6;
  • 45) 0.069 471 641 6 × 2 = 0 + 0.138 943 283 2;
  • 46) 0.138 943 283 2 × 2 = 0 + 0.277 886 566 4;
  • 47) 0.277 886 566 4 × 2 = 0 + 0.555 773 132 8;
  • 48) 0.555 773 132 8 × 2 = 1 + 0.111 546 265 6;
  • 49) 0.111 546 265 6 × 2 = 0 + 0.223 092 531 2;
  • 50) 0.223 092 531 2 × 2 = 0 + 0.446 185 062 4;
  • 51) 0.446 185 062 4 × 2 = 0 + 0.892 370 124 8;
  • 52) 0.892 370 124 8 × 2 = 1 + 0.784 740 249 6;
  • 53) 0.784 740 249 6 × 2 = 1 + 0.569 480 499 2;
  • 54) 0.569 480 499 2 × 2 = 1 + 0.138 960 998 4;
  • 55) 0.138 960 998 4 × 2 = 0 + 0.277 921 996 8;
  • 56) 0.277 921 996 8 × 2 = 0 + 0.555 843 993 6;
  • 57) 0.555 843 993 6 × 2 = 1 + 0.111 687 987 2;
  • 58) 0.111 687 987 2 × 2 = 0 + 0.223 375 974 4;
  • 59) 0.223 375 974 4 × 2 = 0 + 0.446 751 948 8;
  • 60) 0.446 751 948 8 × 2 = 0 + 0.893 503 897 6;
  • 61) 0.893 503 897 6 × 2 = 1 + 0.787 007 795 2;
  • 62) 0.787 007 795 2 × 2 = 1 + 0.574 015 590 4;
  • 63) 0.574 015 590 4 × 2 = 1 + 0.148 031 180 8;
  • 64) 0.148 031 180 8 × 2 = 0 + 0.296 062 361 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:


0.000 282 012 6(10) =

0.0000 0000 0001 0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110(2)

6. Positive number before normalization:

0.000 282 012 6(10) =

0.0000 0000 0001 0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 012 6(10) =

0.0000 0000 0001 0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110(2) =

0.0000 0000 0001 0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110(2) × 20 =

1.0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110 =

0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110


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 0011

Mantissa (52 bits) =
0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110


Decimal number -0.000 282 012 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0110 0010 1110 0100 1110 0001 0001 1100 1000 1110

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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