-0.000 281 28 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 28(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
-0.000 281 28(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:

|-0.000 281 28| = 0.000 281 28


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: 0.000 281 28.

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) 0.000 281 28 × 2 = 0 + 0.000 562 56;
  • 2) 0.000 562 56 × 2 = 0 + 0.001 125 12;
  • 3) 0.001 125 12 × 2 = 0 + 0.002 250 24;
  • 4) 0.002 250 24 × 2 = 0 + 0.004 500 48;
  • 5) 0.004 500 48 × 2 = 0 + 0.009 000 96;
  • 6) 0.009 000 96 × 2 = 0 + 0.018 001 92;
  • 7) 0.018 001 92 × 2 = 0 + 0.036 003 84;
  • 8) 0.036 003 84 × 2 = 0 + 0.072 007 68;
  • 9) 0.072 007 68 × 2 = 0 + 0.144 015 36;
  • 10) 0.144 015 36 × 2 = 0 + 0.288 030 72;
  • 11) 0.288 030 72 × 2 = 0 + 0.576 061 44;
  • 12) 0.576 061 44 × 2 = 1 + 0.152 122 88;
  • 13) 0.152 122 88 × 2 = 0 + 0.304 245 76;
  • 14) 0.304 245 76 × 2 = 0 + 0.608 491 52;
  • 15) 0.608 491 52 × 2 = 1 + 0.216 983 04;
  • 16) 0.216 983 04 × 2 = 0 + 0.433 966 08;
  • 17) 0.433 966 08 × 2 = 0 + 0.867 932 16;
  • 18) 0.867 932 16 × 2 = 1 + 0.735 864 32;
  • 19) 0.735 864 32 × 2 = 1 + 0.471 728 64;
  • 20) 0.471 728 64 × 2 = 0 + 0.943 457 28;
  • 21) 0.943 457 28 × 2 = 1 + 0.886 914 56;
  • 22) 0.886 914 56 × 2 = 1 + 0.773 829 12;
  • 23) 0.773 829 12 × 2 = 1 + 0.547 658 24;
  • 24) 0.547 658 24 × 2 = 1 + 0.095 316 48;
  • 25) 0.095 316 48 × 2 = 0 + 0.190 632 96;
  • 26) 0.190 632 96 × 2 = 0 + 0.381 265 92;
  • 27) 0.381 265 92 × 2 = 0 + 0.762 531 84;
  • 28) 0.762 531 84 × 2 = 1 + 0.525 063 68;
  • 29) 0.525 063 68 × 2 = 1 + 0.050 127 36;
  • 30) 0.050 127 36 × 2 = 0 + 0.100 254 72;
  • 31) 0.100 254 72 × 2 = 0 + 0.200 509 44;
  • 32) 0.200 509 44 × 2 = 0 + 0.401 018 88;
  • 33) 0.401 018 88 × 2 = 0 + 0.802 037 76;
  • 34) 0.802 037 76 × 2 = 1 + 0.604 075 52;
  • 35) 0.604 075 52 × 2 = 1 + 0.208 151 04;
  • 36) 0.208 151 04 × 2 = 0 + 0.416 302 08;
  • 37) 0.416 302 08 × 2 = 0 + 0.832 604 16;
  • 38) 0.832 604 16 × 2 = 1 + 0.665 208 32;
  • 39) 0.665 208 32 × 2 = 1 + 0.330 416 64;
  • 40) 0.330 416 64 × 2 = 0 + 0.660 833 28;
  • 41) 0.660 833 28 × 2 = 1 + 0.321 666 56;
  • 42) 0.321 666 56 × 2 = 0 + 0.643 333 12;
  • 43) 0.643 333 12 × 2 = 1 + 0.286 666 24;
  • 44) 0.286 666 24 × 2 = 0 + 0.573 332 48;
  • 45) 0.573 332 48 × 2 = 1 + 0.146 664 96;
  • 46) 0.146 664 96 × 2 = 0 + 0.293 329 92;
  • 47) 0.293 329 92 × 2 = 0 + 0.586 659 84;
  • 48) 0.586 659 84 × 2 = 1 + 0.173 319 68;
  • 49) 0.173 319 68 × 2 = 0 + 0.346 639 36;
  • 50) 0.346 639 36 × 2 = 0 + 0.693 278 72;
  • 51) 0.693 278 72 × 2 = 1 + 0.386 557 44;
  • 52) 0.386 557 44 × 2 = 0 + 0.773 114 88;
  • 53) 0.773 114 88 × 2 = 1 + 0.546 229 76;
  • 54) 0.546 229 76 × 2 = 1 + 0.092 459 52;
  • 55) 0.092 459 52 × 2 = 0 + 0.184 919 04;
  • 56) 0.184 919 04 × 2 = 0 + 0.369 838 08;
  • 57) 0.369 838 08 × 2 = 0 + 0.739 676 16;
  • 58) 0.739 676 16 × 2 = 1 + 0.479 352 32;
  • 59) 0.479 352 32 × 2 = 0 + 0.958 704 64;
  • 60) 0.958 704 64 × 2 = 1 + 0.917 409 28;
  • 61) 0.917 409 28 × 2 = 1 + 0.834 818 56;
  • 62) 0.834 818 56 × 2 = 1 + 0.669 637 12;
  • 63) 0.669 637 12 × 2 = 1 + 0.339 274 24;
  • 64) 0.339 274 24 × 2 = 0 + 0.678 548 48;

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:


0.000 281 28(10) =

0.0000 0000 0001 0010 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 1110(2)

6. Positive number before normalization:

0.000 281 28(10) =

0.0000 0000 0001 0010 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 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:


0.000 281 28(10) =

0.0000 0000 0001 0010 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 1110(2) =

0.0000 0000 0001 0010 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 1110(2) × 20 =

1.0010 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 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 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 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 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 1110 =

0010 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 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 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 1110


Decimal number -0.000 281 28 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0110 1111 0001 1000 0110 0110 1010 1001 0010 1100 0101 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