-0.000 282 005 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 19(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
-0.000 282 005 19(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:

|-0.000 282 005 19| = 0.000 282 005 19


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: 0.000 282 005 19.

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) 0.000 282 005 19 × 2 = 0 + 0.000 564 010 38;
  • 2) 0.000 564 010 38 × 2 = 0 + 0.001 128 020 76;
  • 3) 0.001 128 020 76 × 2 = 0 + 0.002 256 041 52;
  • 4) 0.002 256 041 52 × 2 = 0 + 0.004 512 083 04;
  • 5) 0.004 512 083 04 × 2 = 0 + 0.009 024 166 08;
  • 6) 0.009 024 166 08 × 2 = 0 + 0.018 048 332 16;
  • 7) 0.018 048 332 16 × 2 = 0 + 0.036 096 664 32;
  • 8) 0.036 096 664 32 × 2 = 0 + 0.072 193 328 64;
  • 9) 0.072 193 328 64 × 2 = 0 + 0.144 386 657 28;
  • 10) 0.144 386 657 28 × 2 = 0 + 0.288 773 314 56;
  • 11) 0.288 773 314 56 × 2 = 0 + 0.577 546 629 12;
  • 12) 0.577 546 629 12 × 2 = 1 + 0.155 093 258 24;
  • 13) 0.155 093 258 24 × 2 = 0 + 0.310 186 516 48;
  • 14) 0.310 186 516 48 × 2 = 0 + 0.620 373 032 96;
  • 15) 0.620 373 032 96 × 2 = 1 + 0.240 746 065 92;
  • 16) 0.240 746 065 92 × 2 = 0 + 0.481 492 131 84;
  • 17) 0.481 492 131 84 × 2 = 0 + 0.962 984 263 68;
  • 18) 0.962 984 263 68 × 2 = 1 + 0.925 968 527 36;
  • 19) 0.925 968 527 36 × 2 = 1 + 0.851 937 054 72;
  • 20) 0.851 937 054 72 × 2 = 1 + 0.703 874 109 44;
  • 21) 0.703 874 109 44 × 2 = 1 + 0.407 748 218 88;
  • 22) 0.407 748 218 88 × 2 = 0 + 0.815 496 437 76;
  • 23) 0.815 496 437 76 × 2 = 1 + 0.630 992 875 52;
  • 24) 0.630 992 875 52 × 2 = 1 + 0.261 985 751 04;
  • 25) 0.261 985 751 04 × 2 = 0 + 0.523 971 502 08;
  • 26) 0.523 971 502 08 × 2 = 1 + 0.047 943 004 16;
  • 27) 0.047 943 004 16 × 2 = 0 + 0.095 886 008 32;
  • 28) 0.095 886 008 32 × 2 = 0 + 0.191 772 016 64;
  • 29) 0.191 772 016 64 × 2 = 0 + 0.383 544 033 28;
  • 30) 0.383 544 033 28 × 2 = 0 + 0.767 088 066 56;
  • 31) 0.767 088 066 56 × 2 = 1 + 0.534 176 133 12;
  • 32) 0.534 176 133 12 × 2 = 1 + 0.068 352 266 24;
  • 33) 0.068 352 266 24 × 2 = 0 + 0.136 704 532 48;
  • 34) 0.136 704 532 48 × 2 = 0 + 0.273 409 064 96;
  • 35) 0.273 409 064 96 × 2 = 0 + 0.546 818 129 92;
  • 36) 0.546 818 129 92 × 2 = 1 + 0.093 636 259 84;
  • 37) 0.093 636 259 84 × 2 = 0 + 0.187 272 519 68;
  • 38) 0.187 272 519 68 × 2 = 0 + 0.374 545 039 36;
  • 39) 0.374 545 039 36 × 2 = 0 + 0.749 090 078 72;
  • 40) 0.749 090 078 72 × 2 = 1 + 0.498 180 157 44;
  • 41) 0.498 180 157 44 × 2 = 0 + 0.996 360 314 88;
  • 42) 0.996 360 314 88 × 2 = 1 + 0.992 720 629 76;
  • 43) 0.992 720 629 76 × 2 = 1 + 0.985 441 259 52;
  • 44) 0.985 441 259 52 × 2 = 1 + 0.970 882 519 04;
  • 45) 0.970 882 519 04 × 2 = 1 + 0.941 765 038 08;
  • 46) 0.941 765 038 08 × 2 = 1 + 0.883 530 076 16;
  • 47) 0.883 530 076 16 × 2 = 1 + 0.767 060 152 32;
  • 48) 0.767 060 152 32 × 2 = 1 + 0.534 120 304 64;
  • 49) 0.534 120 304 64 × 2 = 1 + 0.068 240 609 28;
  • 50) 0.068 240 609 28 × 2 = 0 + 0.136 481 218 56;
  • 51) 0.136 481 218 56 × 2 = 0 + 0.272 962 437 12;
  • 52) 0.272 962 437 12 × 2 = 0 + 0.545 924 874 24;
  • 53) 0.545 924 874 24 × 2 = 1 + 0.091 849 748 48;
  • 54) 0.091 849 748 48 × 2 = 0 + 0.183 699 496 96;
  • 55) 0.183 699 496 96 × 2 = 0 + 0.367 398 993 92;
  • 56) 0.367 398 993 92 × 2 = 0 + 0.734 797 987 84;
  • 57) 0.734 797 987 84 × 2 = 1 + 0.469 595 975 68;
  • 58) 0.469 595 975 68 × 2 = 0 + 0.939 191 951 36;
  • 59) 0.939 191 951 36 × 2 = 1 + 0.878 383 902 72;
  • 60) 0.878 383 902 72 × 2 = 1 + 0.756 767 805 44;
  • 61) 0.756 767 805 44 × 2 = 1 + 0.513 535 610 88;
  • 62) 0.513 535 610 88 × 2 = 1 + 0.027 071 221 76;
  • 63) 0.027 071 221 76 × 2 = 0 + 0.054 142 443 52;
  • 64) 0.054 142 443 52 × 2 = 0 + 0.108 284 887 04;

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:


0.000 282 005 19(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100(2)

6. Positive number before normalization:

0.000 282 005 19(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100(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:


0.000 282 005 19(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100(2) × 20 =

1.0010 0111 1011 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100(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 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100


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 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100 =

0010 0111 1011 0100 0011 0001 0001 0111 1111 1000 1000 1011 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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0011 0001 0001 0111 1111 1000 1000 1011 1100


Decimal number -0.000 282 005 19 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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