-0.000 282 009 04 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 04(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
-0.000 282 009 04(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:

|-0.000 282 009 04| = 0.000 282 009 04


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: 0.000 282 009 04.

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) 0.000 282 009 04 × 2 = 0 + 0.000 564 018 08;
  • 2) 0.000 564 018 08 × 2 = 0 + 0.001 128 036 16;
  • 3) 0.001 128 036 16 × 2 = 0 + 0.002 256 072 32;
  • 4) 0.002 256 072 32 × 2 = 0 + 0.004 512 144 64;
  • 5) 0.004 512 144 64 × 2 = 0 + 0.009 024 289 28;
  • 6) 0.009 024 289 28 × 2 = 0 + 0.018 048 578 56;
  • 7) 0.018 048 578 56 × 2 = 0 + 0.036 097 157 12;
  • 8) 0.036 097 157 12 × 2 = 0 + 0.072 194 314 24;
  • 9) 0.072 194 314 24 × 2 = 0 + 0.144 388 628 48;
  • 10) 0.144 388 628 48 × 2 = 0 + 0.288 777 256 96;
  • 11) 0.288 777 256 96 × 2 = 0 + 0.577 554 513 92;
  • 12) 0.577 554 513 92 × 2 = 1 + 0.155 109 027 84;
  • 13) 0.155 109 027 84 × 2 = 0 + 0.310 218 055 68;
  • 14) 0.310 218 055 68 × 2 = 0 + 0.620 436 111 36;
  • 15) 0.620 436 111 36 × 2 = 1 + 0.240 872 222 72;
  • 16) 0.240 872 222 72 × 2 = 0 + 0.481 744 445 44;
  • 17) 0.481 744 445 44 × 2 = 0 + 0.963 488 890 88;
  • 18) 0.963 488 890 88 × 2 = 1 + 0.926 977 781 76;
  • 19) 0.926 977 781 76 × 2 = 1 + 0.853 955 563 52;
  • 20) 0.853 955 563 52 × 2 = 1 + 0.707 911 127 04;
  • 21) 0.707 911 127 04 × 2 = 1 + 0.415 822 254 08;
  • 22) 0.415 822 254 08 × 2 = 0 + 0.831 644 508 16;
  • 23) 0.831 644 508 16 × 2 = 1 + 0.663 289 016 32;
  • 24) 0.663 289 016 32 × 2 = 1 + 0.326 578 032 64;
  • 25) 0.326 578 032 64 × 2 = 0 + 0.653 156 065 28;
  • 26) 0.653 156 065 28 × 2 = 1 + 0.306 312 130 56;
  • 27) 0.306 312 130 56 × 2 = 0 + 0.612 624 261 12;
  • 28) 0.612 624 261 12 × 2 = 1 + 0.225 248 522 24;
  • 29) 0.225 248 522 24 × 2 = 0 + 0.450 497 044 48;
  • 30) 0.450 497 044 48 × 2 = 0 + 0.900 994 088 96;
  • 31) 0.900 994 088 96 × 2 = 1 + 0.801 988 177 92;
  • 32) 0.801 988 177 92 × 2 = 1 + 0.603 976 355 84;
  • 33) 0.603 976 355 84 × 2 = 1 + 0.207 952 711 68;
  • 34) 0.207 952 711 68 × 2 = 0 + 0.415 905 423 36;
  • 35) 0.415 905 423 36 × 2 = 0 + 0.831 810 846 72;
  • 36) 0.831 810 846 72 × 2 = 1 + 0.663 621 693 44;
  • 37) 0.663 621 693 44 × 2 = 1 + 0.327 243 386 88;
  • 38) 0.327 243 386 88 × 2 = 0 + 0.654 486 773 76;
  • 39) 0.654 486 773 76 × 2 = 1 + 0.308 973 547 52;
  • 40) 0.308 973 547 52 × 2 = 0 + 0.617 947 095 04;
  • 41) 0.617 947 095 04 × 2 = 1 + 0.235 894 190 08;
  • 42) 0.235 894 190 08 × 2 = 0 + 0.471 788 380 16;
  • 43) 0.471 788 380 16 × 2 = 0 + 0.943 576 760 32;
  • 44) 0.943 576 760 32 × 2 = 1 + 0.887 153 520 64;
  • 45) 0.887 153 520 64 × 2 = 1 + 0.774 307 041 28;
  • 46) 0.774 307 041 28 × 2 = 1 + 0.548 614 082 56;
  • 47) 0.548 614 082 56 × 2 = 1 + 0.097 228 165 12;
  • 48) 0.097 228 165 12 × 2 = 0 + 0.194 456 330 24;
  • 49) 0.194 456 330 24 × 2 = 0 + 0.388 912 660 48;
  • 50) 0.388 912 660 48 × 2 = 0 + 0.777 825 320 96;
  • 51) 0.777 825 320 96 × 2 = 1 + 0.555 650 641 92;
  • 52) 0.555 650 641 92 × 2 = 1 + 0.111 301 283 84;
  • 53) 0.111 301 283 84 × 2 = 0 + 0.222 602 567 68;
  • 54) 0.222 602 567 68 × 2 = 0 + 0.445 205 135 36;
  • 55) 0.445 205 135 36 × 2 = 0 + 0.890 410 270 72;
  • 56) 0.890 410 270 72 × 2 = 1 + 0.780 820 541 44;
  • 57) 0.780 820 541 44 × 2 = 1 + 0.561 641 082 88;
  • 58) 0.561 641 082 88 × 2 = 1 + 0.123 282 165 76;
  • 59) 0.123 282 165 76 × 2 = 0 + 0.246 564 331 52;
  • 60) 0.246 564 331 52 × 2 = 0 + 0.493 128 663 04;
  • 61) 0.493 128 663 04 × 2 = 0 + 0.986 257 326 08;
  • 62) 0.986 257 326 08 × 2 = 1 + 0.972 514 652 16;
  • 63) 0.972 514 652 16 × 2 = 1 + 0.945 029 304 32;
  • 64) 0.945 029 304 32 × 2 = 1 + 0.890 058 608 64;

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:


0.000 282 009 04(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111(2)

6. Positive number before normalization:

0.000 282 009 04(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111(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:


0.000 282 009 04(10) =

0.0000 0000 0001 0010 0111 1011 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111(2) =

0.0000 0000 0001 0010 0111 1011 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111(2) × 20 =

1.0010 0111 1011 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111(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 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111


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 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111 =

0010 0111 1011 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111


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 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111


Decimal number -0.000 282 009 04 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0011 1001 1010 1001 1110 0011 0001 1100 0111

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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