-0.000 282 006 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 72(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
-0.000 282 006 72(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:

|-0.000 282 006 72| = 0.000 282 006 72


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: 0.000 282 006 72.

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) 0.000 282 006 72 × 2 = 0 + 0.000 564 013 44;
  • 2) 0.000 564 013 44 × 2 = 0 + 0.001 128 026 88;
  • 3) 0.001 128 026 88 × 2 = 0 + 0.002 256 053 76;
  • 4) 0.002 256 053 76 × 2 = 0 + 0.004 512 107 52;
  • 5) 0.004 512 107 52 × 2 = 0 + 0.009 024 215 04;
  • 6) 0.009 024 215 04 × 2 = 0 + 0.018 048 430 08;
  • 7) 0.018 048 430 08 × 2 = 0 + 0.036 096 860 16;
  • 8) 0.036 096 860 16 × 2 = 0 + 0.072 193 720 32;
  • 9) 0.072 193 720 32 × 2 = 0 + 0.144 387 440 64;
  • 10) 0.144 387 440 64 × 2 = 0 + 0.288 774 881 28;
  • 11) 0.288 774 881 28 × 2 = 0 + 0.577 549 762 56;
  • 12) 0.577 549 762 56 × 2 = 1 + 0.155 099 525 12;
  • 13) 0.155 099 525 12 × 2 = 0 + 0.310 199 050 24;
  • 14) 0.310 199 050 24 × 2 = 0 + 0.620 398 100 48;
  • 15) 0.620 398 100 48 × 2 = 1 + 0.240 796 200 96;
  • 16) 0.240 796 200 96 × 2 = 0 + 0.481 592 401 92;
  • 17) 0.481 592 401 92 × 2 = 0 + 0.963 184 803 84;
  • 18) 0.963 184 803 84 × 2 = 1 + 0.926 369 607 68;
  • 19) 0.926 369 607 68 × 2 = 1 + 0.852 739 215 36;
  • 20) 0.852 739 215 36 × 2 = 1 + 0.705 478 430 72;
  • 21) 0.705 478 430 72 × 2 = 1 + 0.410 956 861 44;
  • 22) 0.410 956 861 44 × 2 = 0 + 0.821 913 722 88;
  • 23) 0.821 913 722 88 × 2 = 1 + 0.643 827 445 76;
  • 24) 0.643 827 445 76 × 2 = 1 + 0.287 654 891 52;
  • 25) 0.287 654 891 52 × 2 = 0 + 0.575 309 783 04;
  • 26) 0.575 309 783 04 × 2 = 1 + 0.150 619 566 08;
  • 27) 0.150 619 566 08 × 2 = 0 + 0.301 239 132 16;
  • 28) 0.301 239 132 16 × 2 = 0 + 0.602 478 264 32;
  • 29) 0.602 478 264 32 × 2 = 1 + 0.204 956 528 64;
  • 30) 0.204 956 528 64 × 2 = 0 + 0.409 913 057 28;
  • 31) 0.409 913 057 28 × 2 = 0 + 0.819 826 114 56;
  • 32) 0.819 826 114 56 × 2 = 1 + 0.639 652 229 12;
  • 33) 0.639 652 229 12 × 2 = 1 + 0.279 304 458 24;
  • 34) 0.279 304 458 24 × 2 = 0 + 0.558 608 916 48;
  • 35) 0.558 608 916 48 × 2 = 1 + 0.117 217 832 96;
  • 36) 0.117 217 832 96 × 2 = 0 + 0.234 435 665 92;
  • 37) 0.234 435 665 92 × 2 = 0 + 0.468 871 331 84;
  • 38) 0.468 871 331 84 × 2 = 0 + 0.937 742 663 68;
  • 39) 0.937 742 663 68 × 2 = 1 + 0.875 485 327 36;
  • 40) 0.875 485 327 36 × 2 = 1 + 0.750 970 654 72;
  • 41) 0.750 970 654 72 × 2 = 1 + 0.501 941 309 44;
  • 42) 0.501 941 309 44 × 2 = 1 + 0.003 882 618 88;
  • 43) 0.003 882 618 88 × 2 = 0 + 0.007 765 237 76;
  • 44) 0.007 765 237 76 × 2 = 0 + 0.015 530 475 52;
  • 45) 0.015 530 475 52 × 2 = 0 + 0.031 060 951 04;
  • 46) 0.031 060 951 04 × 2 = 0 + 0.062 121 902 08;
  • 47) 0.062 121 902 08 × 2 = 0 + 0.124 243 804 16;
  • 48) 0.124 243 804 16 × 2 = 0 + 0.248 487 608 32;
  • 49) 0.248 487 608 32 × 2 = 0 + 0.496 975 216 64;
  • 50) 0.496 975 216 64 × 2 = 0 + 0.993 950 433 28;
  • 51) 0.993 950 433 28 × 2 = 1 + 0.987 900 866 56;
  • 52) 0.987 900 866 56 × 2 = 1 + 0.975 801 733 12;
  • 53) 0.975 801 733 12 × 2 = 1 + 0.951 603 466 24;
  • 54) 0.951 603 466 24 × 2 = 1 + 0.903 206 932 48;
  • 55) 0.903 206 932 48 × 2 = 1 + 0.806 413 864 96;
  • 56) 0.806 413 864 96 × 2 = 1 + 0.612 827 729 92;
  • 57) 0.612 827 729 92 × 2 = 1 + 0.225 655 459 84;
  • 58) 0.225 655 459 84 × 2 = 0 + 0.451 310 919 68;
  • 59) 0.451 310 919 68 × 2 = 0 + 0.902 621 839 36;
  • 60) 0.902 621 839 36 × 2 = 1 + 0.805 243 678 72;
  • 61) 0.805 243 678 72 × 2 = 1 + 0.610 487 357 44;
  • 62) 0.610 487 357 44 × 2 = 1 + 0.220 974 714 88;
  • 63) 0.220 974 714 88 × 2 = 0 + 0.441 949 429 76;
  • 64) 0.441 949 429 76 × 2 = 0 + 0.883 898 859 52;

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:


0.000 282 006 72(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 1100(2)

6. Positive number before normalization:

0.000 282 006 72(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 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:


0.000 282 006 72(10) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 1100(2) × 20 =

1.0010 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 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 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 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 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 1100 =

0010 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 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 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 1100


Decimal number -0.000 282 006 72 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1001 1010 0011 1100 0000 0011 1111 1001 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