-0.000 278 86 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 278 86(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
-0.000 278 86(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:

|-0.000 278 86| = 0.000 278 86


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: 0.000 278 86.

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) 0.000 278 86 × 2 = 0 + 0.000 557 72;
  • 2) 0.000 557 72 × 2 = 0 + 0.001 115 44;
  • 3) 0.001 115 44 × 2 = 0 + 0.002 230 88;
  • 4) 0.002 230 88 × 2 = 0 + 0.004 461 76;
  • 5) 0.004 461 76 × 2 = 0 + 0.008 923 52;
  • 6) 0.008 923 52 × 2 = 0 + 0.017 847 04;
  • 7) 0.017 847 04 × 2 = 0 + 0.035 694 08;
  • 8) 0.035 694 08 × 2 = 0 + 0.071 388 16;
  • 9) 0.071 388 16 × 2 = 0 + 0.142 776 32;
  • 10) 0.142 776 32 × 2 = 0 + 0.285 552 64;
  • 11) 0.285 552 64 × 2 = 0 + 0.571 105 28;
  • 12) 0.571 105 28 × 2 = 1 + 0.142 210 56;
  • 13) 0.142 210 56 × 2 = 0 + 0.284 421 12;
  • 14) 0.284 421 12 × 2 = 0 + 0.568 842 24;
  • 15) 0.568 842 24 × 2 = 1 + 0.137 684 48;
  • 16) 0.137 684 48 × 2 = 0 + 0.275 368 96;
  • 17) 0.275 368 96 × 2 = 0 + 0.550 737 92;
  • 18) 0.550 737 92 × 2 = 1 + 0.101 475 84;
  • 19) 0.101 475 84 × 2 = 0 + 0.202 951 68;
  • 20) 0.202 951 68 × 2 = 0 + 0.405 903 36;
  • 21) 0.405 903 36 × 2 = 0 + 0.811 806 72;
  • 22) 0.811 806 72 × 2 = 1 + 0.623 613 44;
  • 23) 0.623 613 44 × 2 = 1 + 0.247 226 88;
  • 24) 0.247 226 88 × 2 = 0 + 0.494 453 76;
  • 25) 0.494 453 76 × 2 = 0 + 0.988 907 52;
  • 26) 0.988 907 52 × 2 = 1 + 0.977 815 04;
  • 27) 0.977 815 04 × 2 = 1 + 0.955 630 08;
  • 28) 0.955 630 08 × 2 = 1 + 0.911 260 16;
  • 29) 0.911 260 16 × 2 = 1 + 0.822 520 32;
  • 30) 0.822 520 32 × 2 = 1 + 0.645 040 64;
  • 31) 0.645 040 64 × 2 = 1 + 0.290 081 28;
  • 32) 0.290 081 28 × 2 = 0 + 0.580 162 56;
  • 33) 0.580 162 56 × 2 = 1 + 0.160 325 12;
  • 34) 0.160 325 12 × 2 = 0 + 0.320 650 24;
  • 35) 0.320 650 24 × 2 = 0 + 0.641 300 48;
  • 36) 0.641 300 48 × 2 = 1 + 0.282 600 96;
  • 37) 0.282 600 96 × 2 = 0 + 0.565 201 92;
  • 38) 0.565 201 92 × 2 = 1 + 0.130 403 84;
  • 39) 0.130 403 84 × 2 = 0 + 0.260 807 68;
  • 40) 0.260 807 68 × 2 = 0 + 0.521 615 36;
  • 41) 0.521 615 36 × 2 = 1 + 0.043 230 72;
  • 42) 0.043 230 72 × 2 = 0 + 0.086 461 44;
  • 43) 0.086 461 44 × 2 = 0 + 0.172 922 88;
  • 44) 0.172 922 88 × 2 = 0 + 0.345 845 76;
  • 45) 0.345 845 76 × 2 = 0 + 0.691 691 52;
  • 46) 0.691 691 52 × 2 = 1 + 0.383 383 04;
  • 47) 0.383 383 04 × 2 = 0 + 0.766 766 08;
  • 48) 0.766 766 08 × 2 = 1 + 0.533 532 16;
  • 49) 0.533 532 16 × 2 = 1 + 0.067 064 32;
  • 50) 0.067 064 32 × 2 = 0 + 0.134 128 64;
  • 51) 0.134 128 64 × 2 = 0 + 0.268 257 28;
  • 52) 0.268 257 28 × 2 = 0 + 0.536 514 56;
  • 53) 0.536 514 56 × 2 = 1 + 0.073 029 12;
  • 54) 0.073 029 12 × 2 = 0 + 0.146 058 24;
  • 55) 0.146 058 24 × 2 = 0 + 0.292 116 48;
  • 56) 0.292 116 48 × 2 = 0 + 0.584 232 96;
  • 57) 0.584 232 96 × 2 = 1 + 0.168 465 92;
  • 58) 0.168 465 92 × 2 = 0 + 0.336 931 84;
  • 59) 0.336 931 84 × 2 = 0 + 0.673 863 68;
  • 60) 0.673 863 68 × 2 = 1 + 0.347 727 36;
  • 61) 0.347 727 36 × 2 = 0 + 0.695 454 72;
  • 62) 0.695 454 72 × 2 = 1 + 0.390 909 44;
  • 63) 0.390 909 44 × 2 = 0 + 0.781 818 88;
  • 64) 0.781 818 88 × 2 = 1 + 0.563 637 76;

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:


0.000 278 86(10) =

0.0000 0000 0001 0010 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 0101(2)

6. Positive number before normalization:

0.000 278 86(10) =

0.0000 0000 0001 0010 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 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:


0.000 278 86(10) =

0.0000 0000 0001 0010 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 0101(2) =

0.0000 0000 0001 0010 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 0101(2) × 20 =

1.0010 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 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 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 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 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 0101 =

0010 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 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 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 0101


Decimal number -0.000 278 86 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0100 0110 0111 1110 1001 0100 1000 0101 1000 1000 1001 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