-0.000 282 805 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 805(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
-0.000 282 805(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:

|-0.000 282 805| = 0.000 282 805


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: 0.000 282 805.

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) 0.000 282 805 × 2 = 0 + 0.000 565 61;
  • 2) 0.000 565 61 × 2 = 0 + 0.001 131 22;
  • 3) 0.001 131 22 × 2 = 0 + 0.002 262 44;
  • 4) 0.002 262 44 × 2 = 0 + 0.004 524 88;
  • 5) 0.004 524 88 × 2 = 0 + 0.009 049 76;
  • 6) 0.009 049 76 × 2 = 0 + 0.018 099 52;
  • 7) 0.018 099 52 × 2 = 0 + 0.036 199 04;
  • 8) 0.036 199 04 × 2 = 0 + 0.072 398 08;
  • 9) 0.072 398 08 × 2 = 0 + 0.144 796 16;
  • 10) 0.144 796 16 × 2 = 0 + 0.289 592 32;
  • 11) 0.289 592 32 × 2 = 0 + 0.579 184 64;
  • 12) 0.579 184 64 × 2 = 1 + 0.158 369 28;
  • 13) 0.158 369 28 × 2 = 0 + 0.316 738 56;
  • 14) 0.316 738 56 × 2 = 0 + 0.633 477 12;
  • 15) 0.633 477 12 × 2 = 1 + 0.266 954 24;
  • 16) 0.266 954 24 × 2 = 0 + 0.533 908 48;
  • 17) 0.533 908 48 × 2 = 1 + 0.067 816 96;
  • 18) 0.067 816 96 × 2 = 0 + 0.135 633 92;
  • 19) 0.135 633 92 × 2 = 0 + 0.271 267 84;
  • 20) 0.271 267 84 × 2 = 0 + 0.542 535 68;
  • 21) 0.542 535 68 × 2 = 1 + 0.085 071 36;
  • 22) 0.085 071 36 × 2 = 0 + 0.170 142 72;
  • 23) 0.170 142 72 × 2 = 0 + 0.340 285 44;
  • 24) 0.340 285 44 × 2 = 0 + 0.680 570 88;
  • 25) 0.680 570 88 × 2 = 1 + 0.361 141 76;
  • 26) 0.361 141 76 × 2 = 0 + 0.722 283 52;
  • 27) 0.722 283 52 × 2 = 1 + 0.444 567 04;
  • 28) 0.444 567 04 × 2 = 0 + 0.889 134 08;
  • 29) 0.889 134 08 × 2 = 1 + 0.778 268 16;
  • 30) 0.778 268 16 × 2 = 1 + 0.556 536 32;
  • 31) 0.556 536 32 × 2 = 1 + 0.113 072 64;
  • 32) 0.113 072 64 × 2 = 0 + 0.226 145 28;
  • 33) 0.226 145 28 × 2 = 0 + 0.452 290 56;
  • 34) 0.452 290 56 × 2 = 0 + 0.904 581 12;
  • 35) 0.904 581 12 × 2 = 1 + 0.809 162 24;
  • 36) 0.809 162 24 × 2 = 1 + 0.618 324 48;
  • 37) 0.618 324 48 × 2 = 1 + 0.236 648 96;
  • 38) 0.236 648 96 × 2 = 0 + 0.473 297 92;
  • 39) 0.473 297 92 × 2 = 0 + 0.946 595 84;
  • 40) 0.946 595 84 × 2 = 1 + 0.893 191 68;
  • 41) 0.893 191 68 × 2 = 1 + 0.786 383 36;
  • 42) 0.786 383 36 × 2 = 1 + 0.572 766 72;
  • 43) 0.572 766 72 × 2 = 1 + 0.145 533 44;
  • 44) 0.145 533 44 × 2 = 0 + 0.291 066 88;
  • 45) 0.291 066 88 × 2 = 0 + 0.582 133 76;
  • 46) 0.582 133 76 × 2 = 1 + 0.164 267 52;
  • 47) 0.164 267 52 × 2 = 0 + 0.328 535 04;
  • 48) 0.328 535 04 × 2 = 0 + 0.657 070 08;
  • 49) 0.657 070 08 × 2 = 1 + 0.314 140 16;
  • 50) 0.314 140 16 × 2 = 0 + 0.628 280 32;
  • 51) 0.628 280 32 × 2 = 1 + 0.256 560 64;
  • 52) 0.256 560 64 × 2 = 0 + 0.513 121 28;
  • 53) 0.513 121 28 × 2 = 1 + 0.026 242 56;
  • 54) 0.026 242 56 × 2 = 0 + 0.052 485 12;
  • 55) 0.052 485 12 × 2 = 0 + 0.104 970 24;
  • 56) 0.104 970 24 × 2 = 0 + 0.209 940 48;
  • 57) 0.209 940 48 × 2 = 0 + 0.419 880 96;
  • 58) 0.419 880 96 × 2 = 0 + 0.839 761 92;
  • 59) 0.839 761 92 × 2 = 1 + 0.679 523 84;
  • 60) 0.679 523 84 × 2 = 1 + 0.359 047 68;
  • 61) 0.359 047 68 × 2 = 0 + 0.718 095 36;
  • 62) 0.718 095 36 × 2 = 1 + 0.436 190 72;
  • 63) 0.436 190 72 × 2 = 0 + 0.872 381 44;
  • 64) 0.872 381 44 × 2 = 1 + 0.744 762 88;

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:


0.000 282 805(10) =

0.0000 0000 0001 0010 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101(2)

6. Positive number before normalization:

0.000 282 805(10) =

0.0000 0000 0001 0010 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101(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:


0.000 282 805(10) =

0.0000 0000 0001 0010 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101(2) =

0.0000 0000 0001 0010 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101(2) × 20 =

1.0010 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101(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 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101


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 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101 =

0010 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101


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 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101


Decimal number -0.000 282 805 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 1000 1010 1110 0011 1001 1110 0100 1010 1000 0011 0101

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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