-0.000 282 005 27 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 27(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 27(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 27| = 0.000 282 005 27


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 27.

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 27 × 2 = 0 + 0.000 564 010 54;
  • 2) 0.000 564 010 54 × 2 = 0 + 0.001 128 021 08;
  • 3) 0.001 128 021 08 × 2 = 0 + 0.002 256 042 16;
  • 4) 0.002 256 042 16 × 2 = 0 + 0.004 512 084 32;
  • 5) 0.004 512 084 32 × 2 = 0 + 0.009 024 168 64;
  • 6) 0.009 024 168 64 × 2 = 0 + 0.018 048 337 28;
  • 7) 0.018 048 337 28 × 2 = 0 + 0.036 096 674 56;
  • 8) 0.036 096 674 56 × 2 = 0 + 0.072 193 349 12;
  • 9) 0.072 193 349 12 × 2 = 0 + 0.144 386 698 24;
  • 10) 0.144 386 698 24 × 2 = 0 + 0.288 773 396 48;
  • 11) 0.288 773 396 48 × 2 = 0 + 0.577 546 792 96;
  • 12) 0.577 546 792 96 × 2 = 1 + 0.155 093 585 92;
  • 13) 0.155 093 585 92 × 2 = 0 + 0.310 187 171 84;
  • 14) 0.310 187 171 84 × 2 = 0 + 0.620 374 343 68;
  • 15) 0.620 374 343 68 × 2 = 1 + 0.240 748 687 36;
  • 16) 0.240 748 687 36 × 2 = 0 + 0.481 497 374 72;
  • 17) 0.481 497 374 72 × 2 = 0 + 0.962 994 749 44;
  • 18) 0.962 994 749 44 × 2 = 1 + 0.925 989 498 88;
  • 19) 0.925 989 498 88 × 2 = 1 + 0.851 978 997 76;
  • 20) 0.851 978 997 76 × 2 = 1 + 0.703 957 995 52;
  • 21) 0.703 957 995 52 × 2 = 1 + 0.407 915 991 04;
  • 22) 0.407 915 991 04 × 2 = 0 + 0.815 831 982 08;
  • 23) 0.815 831 982 08 × 2 = 1 + 0.631 663 964 16;
  • 24) 0.631 663 964 16 × 2 = 1 + 0.263 327 928 32;
  • 25) 0.263 327 928 32 × 2 = 0 + 0.526 655 856 64;
  • 26) 0.526 655 856 64 × 2 = 1 + 0.053 311 713 28;
  • 27) 0.053 311 713 28 × 2 = 0 + 0.106 623 426 56;
  • 28) 0.106 623 426 56 × 2 = 0 + 0.213 246 853 12;
  • 29) 0.213 246 853 12 × 2 = 0 + 0.426 493 706 24;
  • 30) 0.426 493 706 24 × 2 = 0 + 0.852 987 412 48;
  • 31) 0.852 987 412 48 × 2 = 1 + 0.705 974 824 96;
  • 32) 0.705 974 824 96 × 2 = 1 + 0.411 949 649 92;
  • 33) 0.411 949 649 92 × 2 = 0 + 0.823 899 299 84;
  • 34) 0.823 899 299 84 × 2 = 1 + 0.647 798 599 68;
  • 35) 0.647 798 599 68 × 2 = 1 + 0.295 597 199 36;
  • 36) 0.295 597 199 36 × 2 = 0 + 0.591 194 398 72;
  • 37) 0.591 194 398 72 × 2 = 1 + 0.182 388 797 44;
  • 38) 0.182 388 797 44 × 2 = 0 + 0.364 777 594 88;
  • 39) 0.364 777 594 88 × 2 = 0 + 0.729 555 189 76;
  • 40) 0.729 555 189 76 × 2 = 1 + 0.459 110 379 52;
  • 41) 0.459 110 379 52 × 2 = 0 + 0.918 220 759 04;
  • 42) 0.918 220 759 04 × 2 = 1 + 0.836 441 518 08;
  • 43) 0.836 441 518 08 × 2 = 1 + 0.672 883 036 16;
  • 44) 0.672 883 036 16 × 2 = 1 + 0.345 766 072 32;
  • 45) 0.345 766 072 32 × 2 = 0 + 0.691 532 144 64;
  • 46) 0.691 532 144 64 × 2 = 1 + 0.383 064 289 28;
  • 47) 0.383 064 289 28 × 2 = 0 + 0.766 128 578 56;
  • 48) 0.766 128 578 56 × 2 = 1 + 0.532 257 157 12;
  • 49) 0.532 257 157 12 × 2 = 1 + 0.064 514 314 24;
  • 50) 0.064 514 314 24 × 2 = 0 + 0.129 028 628 48;
  • 51) 0.129 028 628 48 × 2 = 0 + 0.258 057 256 96;
  • 52) 0.258 057 256 96 × 2 = 0 + 0.516 114 513 92;
  • 53) 0.516 114 513 92 × 2 = 1 + 0.032 229 027 84;
  • 54) 0.032 229 027 84 × 2 = 0 + 0.064 458 055 68;
  • 55) 0.064 458 055 68 × 2 = 0 + 0.128 916 111 36;
  • 56) 0.128 916 111 36 × 2 = 0 + 0.257 832 222 72;
  • 57) 0.257 832 222 72 × 2 = 0 + 0.515 664 445 44;
  • 58) 0.515 664 445 44 × 2 = 1 + 0.031 328 890 88;
  • 59) 0.031 328 890 88 × 2 = 0 + 0.062 657 781 76;
  • 60) 0.062 657 781 76 × 2 = 0 + 0.125 315 563 52;
  • 61) 0.125 315 563 52 × 2 = 0 + 0.250 631 127 04;
  • 62) 0.250 631 127 04 × 2 = 0 + 0.501 262 254 08;
  • 63) 0.501 262 254 08 × 2 = 1 + 0.002 524 508 16;
  • 64) 0.002 524 508 16 × 2 = 0 + 0.005 049 016 32;

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 27(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0110 1001 0111 0101 1000 1000 0100 0010(2)

6. Positive number before normalization:

0.000 282 005 27(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0110 1001 0111 0101 1000 1000 0100 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 27(10) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0110 1001 0111 0101 1000 1000 0100 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0011 0110 1001 0111 0101 1000 1000 0100 0010(2) × 20 =

1.0010 0111 1011 0100 0011 0110 1001 0111 0101 1000 1000 0100 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 0011 0110 1001 0111 0101 1000 1000 0100 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 0011 0110 1001 0111 0101 1000 1000 0100 0010 =

0010 0111 1011 0100 0011 0110 1001 0111 0101 1000 1000 0100 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 0011 0110 1001 0111 0101 1000 1000 0100 0010


Decimal number -0.000 282 005 27 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0011 0110 1001 0111 0101 1000 1000 0100 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