-0.000 282 041 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 041(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
-0.000 282 041(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:

|-0.000 282 041| = 0.000 282 041


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: 0.000 282 041.

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) 0.000 282 041 × 2 = 0 + 0.000 564 082;
  • 2) 0.000 564 082 × 2 = 0 + 0.001 128 164;
  • 3) 0.001 128 164 × 2 = 0 + 0.002 256 328;
  • 4) 0.002 256 328 × 2 = 0 + 0.004 512 656;
  • 5) 0.004 512 656 × 2 = 0 + 0.009 025 312;
  • 6) 0.009 025 312 × 2 = 0 + 0.018 050 624;
  • 7) 0.018 050 624 × 2 = 0 + 0.036 101 248;
  • 8) 0.036 101 248 × 2 = 0 + 0.072 202 496;
  • 9) 0.072 202 496 × 2 = 0 + 0.144 404 992;
  • 10) 0.144 404 992 × 2 = 0 + 0.288 809 984;
  • 11) 0.288 809 984 × 2 = 0 + 0.577 619 968;
  • 12) 0.577 619 968 × 2 = 1 + 0.155 239 936;
  • 13) 0.155 239 936 × 2 = 0 + 0.310 479 872;
  • 14) 0.310 479 872 × 2 = 0 + 0.620 959 744;
  • 15) 0.620 959 744 × 2 = 1 + 0.241 919 488;
  • 16) 0.241 919 488 × 2 = 0 + 0.483 838 976;
  • 17) 0.483 838 976 × 2 = 0 + 0.967 677 952;
  • 18) 0.967 677 952 × 2 = 1 + 0.935 355 904;
  • 19) 0.935 355 904 × 2 = 1 + 0.870 711 808;
  • 20) 0.870 711 808 × 2 = 1 + 0.741 423 616;
  • 21) 0.741 423 616 × 2 = 1 + 0.482 847 232;
  • 22) 0.482 847 232 × 2 = 0 + 0.965 694 464;
  • 23) 0.965 694 464 × 2 = 1 + 0.931 388 928;
  • 24) 0.931 388 928 × 2 = 1 + 0.862 777 856;
  • 25) 0.862 777 856 × 2 = 1 + 0.725 555 712;
  • 26) 0.725 555 712 × 2 = 1 + 0.451 111 424;
  • 27) 0.451 111 424 × 2 = 0 + 0.902 222 848;
  • 28) 0.902 222 848 × 2 = 1 + 0.804 445 696;
  • 29) 0.804 445 696 × 2 = 1 + 0.608 891 392;
  • 30) 0.608 891 392 × 2 = 1 + 0.217 782 784;
  • 31) 0.217 782 784 × 2 = 0 + 0.435 565 568;
  • 32) 0.435 565 568 × 2 = 0 + 0.871 131 136;
  • 33) 0.871 131 136 × 2 = 1 + 0.742 262 272;
  • 34) 0.742 262 272 × 2 = 1 + 0.484 524 544;
  • 35) 0.484 524 544 × 2 = 0 + 0.969 049 088;
  • 36) 0.969 049 088 × 2 = 1 + 0.938 098 176;
  • 37) 0.938 098 176 × 2 = 1 + 0.876 196 352;
  • 38) 0.876 196 352 × 2 = 1 + 0.752 392 704;
  • 39) 0.752 392 704 × 2 = 1 + 0.504 785 408;
  • 40) 0.504 785 408 × 2 = 1 + 0.009 570 816;
  • 41) 0.009 570 816 × 2 = 0 + 0.019 141 632;
  • 42) 0.019 141 632 × 2 = 0 + 0.038 283 264;
  • 43) 0.038 283 264 × 2 = 0 + 0.076 566 528;
  • 44) 0.076 566 528 × 2 = 0 + 0.153 133 056;
  • 45) 0.153 133 056 × 2 = 0 + 0.306 266 112;
  • 46) 0.306 266 112 × 2 = 0 + 0.612 532 224;
  • 47) 0.612 532 224 × 2 = 1 + 0.225 064 448;
  • 48) 0.225 064 448 × 2 = 0 + 0.450 128 896;
  • 49) 0.450 128 896 × 2 = 0 + 0.900 257 792;
  • 50) 0.900 257 792 × 2 = 1 + 0.800 515 584;
  • 51) 0.800 515 584 × 2 = 1 + 0.601 031 168;
  • 52) 0.601 031 168 × 2 = 1 + 0.202 062 336;
  • 53) 0.202 062 336 × 2 = 0 + 0.404 124 672;
  • 54) 0.404 124 672 × 2 = 0 + 0.808 249 344;
  • 55) 0.808 249 344 × 2 = 1 + 0.616 498 688;
  • 56) 0.616 498 688 × 2 = 1 + 0.232 997 376;
  • 57) 0.232 997 376 × 2 = 0 + 0.465 994 752;
  • 58) 0.465 994 752 × 2 = 0 + 0.931 989 504;
  • 59) 0.931 989 504 × 2 = 1 + 0.863 979 008;
  • 60) 0.863 979 008 × 2 = 1 + 0.727 958 016;
  • 61) 0.727 958 016 × 2 = 1 + 0.455 916 032;
  • 62) 0.455 916 032 × 2 = 0 + 0.911 832 064;
  • 63) 0.911 832 064 × 2 = 1 + 0.823 664 128;
  • 64) 0.823 664 128 × 2 = 1 + 0.647 328 256;

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:


0.000 282 041(10) =

0.0000 0000 0001 0010 0111 1011 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011(2)

6. Positive number before normalization:

0.000 282 041(10) =

0.0000 0000 0001 0010 0111 1011 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011(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:


0.000 282 041(10) =

0.0000 0000 0001 0010 0111 1011 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011(2) =

0.0000 0000 0001 0010 0111 1011 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011(2) × 20 =

1.0010 0111 1011 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011(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 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011


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 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011 =

0010 0111 1011 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011


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 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011


Decimal number -0.000 282 041 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1101 1100 1101 1111 0000 0010 0111 0011 0011 1011

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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