-0.000 282 005 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 06(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 06(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 06| = 0.000 282 005 06


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 06.

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 06 × 2 = 0 + 0.000 564 010 12;
  • 2) 0.000 564 010 12 × 2 = 0 + 0.001 128 020 24;
  • 3) 0.001 128 020 24 × 2 = 0 + 0.002 256 040 48;
  • 4) 0.002 256 040 48 × 2 = 0 + 0.004 512 080 96;
  • 5) 0.004 512 080 96 × 2 = 0 + 0.009 024 161 92;
  • 6) 0.009 024 161 92 × 2 = 0 + 0.018 048 323 84;
  • 7) 0.018 048 323 84 × 2 = 0 + 0.036 096 647 68;
  • 8) 0.036 096 647 68 × 2 = 0 + 0.072 193 295 36;
  • 9) 0.072 193 295 36 × 2 = 0 + 0.144 386 590 72;
  • 10) 0.144 386 590 72 × 2 = 0 + 0.288 773 181 44;
  • 11) 0.288 773 181 44 × 2 = 0 + 0.577 546 362 88;
  • 12) 0.577 546 362 88 × 2 = 1 + 0.155 092 725 76;
  • 13) 0.155 092 725 76 × 2 = 0 + 0.310 185 451 52;
  • 14) 0.310 185 451 52 × 2 = 0 + 0.620 370 903 04;
  • 15) 0.620 370 903 04 × 2 = 1 + 0.240 741 806 08;
  • 16) 0.240 741 806 08 × 2 = 0 + 0.481 483 612 16;
  • 17) 0.481 483 612 16 × 2 = 0 + 0.962 967 224 32;
  • 18) 0.962 967 224 32 × 2 = 1 + 0.925 934 448 64;
  • 19) 0.925 934 448 64 × 2 = 1 + 0.851 868 897 28;
  • 20) 0.851 868 897 28 × 2 = 1 + 0.703 737 794 56;
  • 21) 0.703 737 794 56 × 2 = 1 + 0.407 475 589 12;
  • 22) 0.407 475 589 12 × 2 = 0 + 0.814 951 178 24;
  • 23) 0.814 951 178 24 × 2 = 1 + 0.629 902 356 48;
  • 24) 0.629 902 356 48 × 2 = 1 + 0.259 804 712 96;
  • 25) 0.259 804 712 96 × 2 = 0 + 0.519 609 425 92;
  • 26) 0.519 609 425 92 × 2 = 1 + 0.039 218 851 84;
  • 27) 0.039 218 851 84 × 2 = 0 + 0.078 437 703 68;
  • 28) 0.078 437 703 68 × 2 = 0 + 0.156 875 407 36;
  • 29) 0.156 875 407 36 × 2 = 0 + 0.313 750 814 72;
  • 30) 0.313 750 814 72 × 2 = 0 + 0.627 501 629 44;
  • 31) 0.627 501 629 44 × 2 = 1 + 0.255 003 258 88;
  • 32) 0.255 003 258 88 × 2 = 0 + 0.510 006 517 76;
  • 33) 0.510 006 517 76 × 2 = 1 + 0.020 013 035 52;
  • 34) 0.020 013 035 52 × 2 = 0 + 0.040 026 071 04;
  • 35) 0.040 026 071 04 × 2 = 0 + 0.080 052 142 08;
  • 36) 0.080 052 142 08 × 2 = 0 + 0.160 104 284 16;
  • 37) 0.160 104 284 16 × 2 = 0 + 0.320 208 568 32;
  • 38) 0.320 208 568 32 × 2 = 0 + 0.640 417 136 64;
  • 39) 0.640 417 136 64 × 2 = 1 + 0.280 834 273 28;
  • 40) 0.280 834 273 28 × 2 = 0 + 0.561 668 546 56;
  • 41) 0.561 668 546 56 × 2 = 1 + 0.123 337 093 12;
  • 42) 0.123 337 093 12 × 2 = 0 + 0.246 674 186 24;
  • 43) 0.246 674 186 24 × 2 = 0 + 0.493 348 372 48;
  • 44) 0.493 348 372 48 × 2 = 0 + 0.986 696 744 96;
  • 45) 0.986 696 744 96 × 2 = 1 + 0.973 393 489 92;
  • 46) 0.973 393 489 92 × 2 = 1 + 0.946 786 979 84;
  • 47) 0.946 786 979 84 × 2 = 1 + 0.893 573 959 68;
  • 48) 0.893 573 959 68 × 2 = 1 + 0.787 147 919 36;
  • 49) 0.787 147 919 36 × 2 = 1 + 0.574 295 838 72;
  • 50) 0.574 295 838 72 × 2 = 1 + 0.148 591 677 44;
  • 51) 0.148 591 677 44 × 2 = 0 + 0.297 183 354 88;
  • 52) 0.297 183 354 88 × 2 = 0 + 0.594 366 709 76;
  • 53) 0.594 366 709 76 × 2 = 1 + 0.188 733 419 52;
  • 54) 0.188 733 419 52 × 2 = 0 + 0.377 466 839 04;
  • 55) 0.377 466 839 04 × 2 = 0 + 0.754 933 678 08;
  • 56) 0.754 933 678 08 × 2 = 1 + 0.509 867 356 16;
  • 57) 0.509 867 356 16 × 2 = 1 + 0.019 734 712 32;
  • 58) 0.019 734 712 32 × 2 = 0 + 0.039 469 424 64;
  • 59) 0.039 469 424 64 × 2 = 0 + 0.078 938 849 28;
  • 60) 0.078 938 849 28 × 2 = 0 + 0.157 877 698 56;
  • 61) 0.157 877 698 56 × 2 = 0 + 0.315 755 397 12;
  • 62) 0.315 755 397 12 × 2 = 0 + 0.631 510 794 24;
  • 63) 0.631 510 794 24 × 2 = 1 + 0.263 021 588 48;
  • 64) 0.263 021 588 48 × 2 = 0 + 0.526 043 176 96;

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 06(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1000 0010 1000 1111 1100 1001 1000 0010(2)

6. Positive number before normalization:

0.000 282 005 06(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1000 0010 1000 1111 1100 1001 1000 0010(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 06(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1000 0010 1000 1111 1100 1001 1000 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1000 0010 1000 1111 1100 1001 1000 0010(2) × 20 =

1.0010 0111 1011 0100 0010 1000 0010 1000 1111 1100 1001 1000 0010(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 0010 1000 0010 1000 1111 1100 1001 1000 0010


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 0010 1000 0010 1000 1111 1100 1001 1000 0010 =

0010 0111 1011 0100 0010 1000 0010 1000 1111 1100 1001 1000 0010


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 0010 1000 0010 1000 1111 1100 1001 1000 0010


Decimal number -0.000 282 005 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0010 1000 0010 1000 1111 1100 1001 1000 0010

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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