-0.000 287 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 287 1(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
-0.000 287 1(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:

|-0.000 287 1| = 0.000 287 1


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: 0.000 287 1.

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) 0.000 287 1 × 2 = 0 + 0.000 574 2;
  • 2) 0.000 574 2 × 2 = 0 + 0.001 148 4;
  • 3) 0.001 148 4 × 2 = 0 + 0.002 296 8;
  • 4) 0.002 296 8 × 2 = 0 + 0.004 593 6;
  • 5) 0.004 593 6 × 2 = 0 + 0.009 187 2;
  • 6) 0.009 187 2 × 2 = 0 + 0.018 374 4;
  • 7) 0.018 374 4 × 2 = 0 + 0.036 748 8;
  • 8) 0.036 748 8 × 2 = 0 + 0.073 497 6;
  • 9) 0.073 497 6 × 2 = 0 + 0.146 995 2;
  • 10) 0.146 995 2 × 2 = 0 + 0.293 990 4;
  • 11) 0.293 990 4 × 2 = 0 + 0.587 980 8;
  • 12) 0.587 980 8 × 2 = 1 + 0.175 961 6;
  • 13) 0.175 961 6 × 2 = 0 + 0.351 923 2;
  • 14) 0.351 923 2 × 2 = 0 + 0.703 846 4;
  • 15) 0.703 846 4 × 2 = 1 + 0.407 692 8;
  • 16) 0.407 692 8 × 2 = 0 + 0.815 385 6;
  • 17) 0.815 385 6 × 2 = 1 + 0.630 771 2;
  • 18) 0.630 771 2 × 2 = 1 + 0.261 542 4;
  • 19) 0.261 542 4 × 2 = 0 + 0.523 084 8;
  • 20) 0.523 084 8 × 2 = 1 + 0.046 169 6;
  • 21) 0.046 169 6 × 2 = 0 + 0.092 339 2;
  • 22) 0.092 339 2 × 2 = 0 + 0.184 678 4;
  • 23) 0.184 678 4 × 2 = 0 + 0.369 356 8;
  • 24) 0.369 356 8 × 2 = 0 + 0.738 713 6;
  • 25) 0.738 713 6 × 2 = 1 + 0.477 427 2;
  • 26) 0.477 427 2 × 2 = 0 + 0.954 854 4;
  • 27) 0.954 854 4 × 2 = 1 + 0.909 708 8;
  • 28) 0.909 708 8 × 2 = 1 + 0.819 417 6;
  • 29) 0.819 417 6 × 2 = 1 + 0.638 835 2;
  • 30) 0.638 835 2 × 2 = 1 + 0.277 670 4;
  • 31) 0.277 670 4 × 2 = 0 + 0.555 340 8;
  • 32) 0.555 340 8 × 2 = 1 + 0.110 681 6;
  • 33) 0.110 681 6 × 2 = 0 + 0.221 363 2;
  • 34) 0.221 363 2 × 2 = 0 + 0.442 726 4;
  • 35) 0.442 726 4 × 2 = 0 + 0.885 452 8;
  • 36) 0.885 452 8 × 2 = 1 + 0.770 905 6;
  • 37) 0.770 905 6 × 2 = 1 + 0.541 811 2;
  • 38) 0.541 811 2 × 2 = 1 + 0.083 622 4;
  • 39) 0.083 622 4 × 2 = 0 + 0.167 244 8;
  • 40) 0.167 244 8 × 2 = 0 + 0.334 489 6;
  • 41) 0.334 489 6 × 2 = 0 + 0.668 979 2;
  • 42) 0.668 979 2 × 2 = 1 + 0.337 958 4;
  • 43) 0.337 958 4 × 2 = 0 + 0.675 916 8;
  • 44) 0.675 916 8 × 2 = 1 + 0.351 833 6;
  • 45) 0.351 833 6 × 2 = 0 + 0.703 667 2;
  • 46) 0.703 667 2 × 2 = 1 + 0.407 334 4;
  • 47) 0.407 334 4 × 2 = 0 + 0.814 668 8;
  • 48) 0.814 668 8 × 2 = 1 + 0.629 337 6;
  • 49) 0.629 337 6 × 2 = 1 + 0.258 675 2;
  • 50) 0.258 675 2 × 2 = 0 + 0.517 350 4;
  • 51) 0.517 350 4 × 2 = 1 + 0.034 700 8;
  • 52) 0.034 700 8 × 2 = 0 + 0.069 401 6;
  • 53) 0.069 401 6 × 2 = 0 + 0.138 803 2;
  • 54) 0.138 803 2 × 2 = 0 + 0.277 606 4;
  • 55) 0.277 606 4 × 2 = 0 + 0.555 212 8;
  • 56) 0.555 212 8 × 2 = 1 + 0.110 425 6;
  • 57) 0.110 425 6 × 2 = 0 + 0.220 851 2;
  • 58) 0.220 851 2 × 2 = 0 + 0.441 702 4;
  • 59) 0.441 702 4 × 2 = 0 + 0.883 404 8;
  • 60) 0.883 404 8 × 2 = 1 + 0.766 809 6;
  • 61) 0.766 809 6 × 2 = 1 + 0.533 619 2;
  • 62) 0.533 619 2 × 2 = 1 + 0.067 238 4;
  • 63) 0.067 238 4 × 2 = 0 + 0.134 476 8;
  • 64) 0.134 476 8 × 2 = 0 + 0.268 953 6;

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:


0.000 287 1(10) =

0.0000 0000 0001 0010 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 1100(2)

6. Positive number before normalization:

0.000 287 1(10) =

0.0000 0000 0001 0010 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 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:


0.000 287 1(10) =

0.0000 0000 0001 0010 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 1100(2) =

0.0000 0000 0001 0010 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 1100(2) × 20 =

1.0010 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 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 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 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 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 1100 =

0010 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 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 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 1100


Decimal number -0.000 287 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1101 0000 1011 1101 0001 1100 0101 0101 1010 0001 0001 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