-0.000 282 508 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 508(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
-0.000 282 508(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:

|-0.000 282 508| = 0.000 282 508


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: 0.000 282 508.

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) 0.000 282 508 × 2 = 0 + 0.000 565 016;
  • 2) 0.000 565 016 × 2 = 0 + 0.001 130 032;
  • 3) 0.001 130 032 × 2 = 0 + 0.002 260 064;
  • 4) 0.002 260 064 × 2 = 0 + 0.004 520 128;
  • 5) 0.004 520 128 × 2 = 0 + 0.009 040 256;
  • 6) 0.009 040 256 × 2 = 0 + 0.018 080 512;
  • 7) 0.018 080 512 × 2 = 0 + 0.036 161 024;
  • 8) 0.036 161 024 × 2 = 0 + 0.072 322 048;
  • 9) 0.072 322 048 × 2 = 0 + 0.144 644 096;
  • 10) 0.144 644 096 × 2 = 0 + 0.289 288 192;
  • 11) 0.289 288 192 × 2 = 0 + 0.578 576 384;
  • 12) 0.578 576 384 × 2 = 1 + 0.157 152 768;
  • 13) 0.157 152 768 × 2 = 0 + 0.314 305 536;
  • 14) 0.314 305 536 × 2 = 0 + 0.628 611 072;
  • 15) 0.628 611 072 × 2 = 1 + 0.257 222 144;
  • 16) 0.257 222 144 × 2 = 0 + 0.514 444 288;
  • 17) 0.514 444 288 × 2 = 1 + 0.028 888 576;
  • 18) 0.028 888 576 × 2 = 0 + 0.057 777 152;
  • 19) 0.057 777 152 × 2 = 0 + 0.115 554 304;
  • 20) 0.115 554 304 × 2 = 0 + 0.231 108 608;
  • 21) 0.231 108 608 × 2 = 0 + 0.462 217 216;
  • 22) 0.462 217 216 × 2 = 0 + 0.924 434 432;
  • 23) 0.924 434 432 × 2 = 1 + 0.848 868 864;
  • 24) 0.848 868 864 × 2 = 1 + 0.697 737 728;
  • 25) 0.697 737 728 × 2 = 1 + 0.395 475 456;
  • 26) 0.395 475 456 × 2 = 0 + 0.790 950 912;
  • 27) 0.790 950 912 × 2 = 1 + 0.581 901 824;
  • 28) 0.581 901 824 × 2 = 1 + 0.163 803 648;
  • 29) 0.163 803 648 × 2 = 0 + 0.327 607 296;
  • 30) 0.327 607 296 × 2 = 0 + 0.655 214 592;
  • 31) 0.655 214 592 × 2 = 1 + 0.310 429 184;
  • 32) 0.310 429 184 × 2 = 0 + 0.620 858 368;
  • 33) 0.620 858 368 × 2 = 1 + 0.241 716 736;
  • 34) 0.241 716 736 × 2 = 0 + 0.483 433 472;
  • 35) 0.483 433 472 × 2 = 0 + 0.966 866 944;
  • 36) 0.966 866 944 × 2 = 1 + 0.933 733 888;
  • 37) 0.933 733 888 × 2 = 1 + 0.867 467 776;
  • 38) 0.867 467 776 × 2 = 1 + 0.734 935 552;
  • 39) 0.734 935 552 × 2 = 1 + 0.469 871 104;
  • 40) 0.469 871 104 × 2 = 0 + 0.939 742 208;
  • 41) 0.939 742 208 × 2 = 1 + 0.879 484 416;
  • 42) 0.879 484 416 × 2 = 1 + 0.758 968 832;
  • 43) 0.758 968 832 × 2 = 1 + 0.517 937 664;
  • 44) 0.517 937 664 × 2 = 1 + 0.035 875 328;
  • 45) 0.035 875 328 × 2 = 0 + 0.071 750 656;
  • 46) 0.071 750 656 × 2 = 0 + 0.143 501 312;
  • 47) 0.143 501 312 × 2 = 0 + 0.287 002 624;
  • 48) 0.287 002 624 × 2 = 0 + 0.574 005 248;
  • 49) 0.574 005 248 × 2 = 1 + 0.148 010 496;
  • 50) 0.148 010 496 × 2 = 0 + 0.296 020 992;
  • 51) 0.296 020 992 × 2 = 0 + 0.592 041 984;
  • 52) 0.592 041 984 × 2 = 1 + 0.184 083 968;
  • 53) 0.184 083 968 × 2 = 0 + 0.368 167 936;
  • 54) 0.368 167 936 × 2 = 0 + 0.736 335 872;
  • 55) 0.736 335 872 × 2 = 1 + 0.472 671 744;
  • 56) 0.472 671 744 × 2 = 0 + 0.945 343 488;
  • 57) 0.945 343 488 × 2 = 1 + 0.890 686 976;
  • 58) 0.890 686 976 × 2 = 1 + 0.781 373 952;
  • 59) 0.781 373 952 × 2 = 1 + 0.562 747 904;
  • 60) 0.562 747 904 × 2 = 1 + 0.125 495 808;
  • 61) 0.125 495 808 × 2 = 0 + 0.250 991 616;
  • 62) 0.250 991 616 × 2 = 0 + 0.501 983 232;
  • 63) 0.501 983 232 × 2 = 1 + 0.003 966 464;
  • 64) 0.003 966 464 × 2 = 0 + 0.007 932 928;

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:


0.000 282 508(10) =

0.0000 0000 0001 0010 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 0010(2)

6. Positive number before normalization:

0.000 282 508(10) =

0.0000 0000 0001 0010 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 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:


0.000 282 508(10) =

0.0000 0000 0001 0010 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 0010(2) =

0.0000 0000 0001 0010 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 0010(2) × 20 =

1.0010 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 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 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 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 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 0010 =

0010 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 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 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 0010


Decimal number -0.000 282 508 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0011 1011 0010 1001 1110 1111 0000 1001 0010 1111 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