-0.000 282 373 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 373(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
-0.000 282 373(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:

|-0.000 282 373| = 0.000 282 373


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: 0.000 282 373.

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) 0.000 282 373 × 2 = 0 + 0.000 564 746;
  • 2) 0.000 564 746 × 2 = 0 + 0.001 129 492;
  • 3) 0.001 129 492 × 2 = 0 + 0.002 258 984;
  • 4) 0.002 258 984 × 2 = 0 + 0.004 517 968;
  • 5) 0.004 517 968 × 2 = 0 + 0.009 035 936;
  • 6) 0.009 035 936 × 2 = 0 + 0.018 071 872;
  • 7) 0.018 071 872 × 2 = 0 + 0.036 143 744;
  • 8) 0.036 143 744 × 2 = 0 + 0.072 287 488;
  • 9) 0.072 287 488 × 2 = 0 + 0.144 574 976;
  • 10) 0.144 574 976 × 2 = 0 + 0.289 149 952;
  • 11) 0.289 149 952 × 2 = 0 + 0.578 299 904;
  • 12) 0.578 299 904 × 2 = 1 + 0.156 599 808;
  • 13) 0.156 599 808 × 2 = 0 + 0.313 199 616;
  • 14) 0.313 199 616 × 2 = 0 + 0.626 399 232;
  • 15) 0.626 399 232 × 2 = 1 + 0.252 798 464;
  • 16) 0.252 798 464 × 2 = 0 + 0.505 596 928;
  • 17) 0.505 596 928 × 2 = 1 + 0.011 193 856;
  • 18) 0.011 193 856 × 2 = 0 + 0.022 387 712;
  • 19) 0.022 387 712 × 2 = 0 + 0.044 775 424;
  • 20) 0.044 775 424 × 2 = 0 + 0.089 550 848;
  • 21) 0.089 550 848 × 2 = 0 + 0.179 101 696;
  • 22) 0.179 101 696 × 2 = 0 + 0.358 203 392;
  • 23) 0.358 203 392 × 2 = 0 + 0.716 406 784;
  • 24) 0.716 406 784 × 2 = 1 + 0.432 813 568;
  • 25) 0.432 813 568 × 2 = 0 + 0.865 627 136;
  • 26) 0.865 627 136 × 2 = 1 + 0.731 254 272;
  • 27) 0.731 254 272 × 2 = 1 + 0.462 508 544;
  • 28) 0.462 508 544 × 2 = 0 + 0.925 017 088;
  • 29) 0.925 017 088 × 2 = 1 + 0.850 034 176;
  • 30) 0.850 034 176 × 2 = 1 + 0.700 068 352;
  • 31) 0.700 068 352 × 2 = 1 + 0.400 136 704;
  • 32) 0.400 136 704 × 2 = 0 + 0.800 273 408;
  • 33) 0.800 273 408 × 2 = 1 + 0.600 546 816;
  • 34) 0.600 546 816 × 2 = 1 + 0.201 093 632;
  • 35) 0.201 093 632 × 2 = 0 + 0.402 187 264;
  • 36) 0.402 187 264 × 2 = 0 + 0.804 374 528;
  • 37) 0.804 374 528 × 2 = 1 + 0.608 749 056;
  • 38) 0.608 749 056 × 2 = 1 + 0.217 498 112;
  • 39) 0.217 498 112 × 2 = 0 + 0.434 996 224;
  • 40) 0.434 996 224 × 2 = 0 + 0.869 992 448;
  • 41) 0.869 992 448 × 2 = 1 + 0.739 984 896;
  • 42) 0.739 984 896 × 2 = 1 + 0.479 969 792;
  • 43) 0.479 969 792 × 2 = 0 + 0.959 939 584;
  • 44) 0.959 939 584 × 2 = 1 + 0.919 879 168;
  • 45) 0.919 879 168 × 2 = 1 + 0.839 758 336;
  • 46) 0.839 758 336 × 2 = 1 + 0.679 516 672;
  • 47) 0.679 516 672 × 2 = 1 + 0.359 033 344;
  • 48) 0.359 033 344 × 2 = 0 + 0.718 066 688;
  • 49) 0.718 066 688 × 2 = 1 + 0.436 133 376;
  • 50) 0.436 133 376 × 2 = 0 + 0.872 266 752;
  • 51) 0.872 266 752 × 2 = 1 + 0.744 533 504;
  • 52) 0.744 533 504 × 2 = 1 + 0.489 067 008;
  • 53) 0.489 067 008 × 2 = 0 + 0.978 134 016;
  • 54) 0.978 134 016 × 2 = 1 + 0.956 268 032;
  • 55) 0.956 268 032 × 2 = 1 + 0.912 536 064;
  • 56) 0.912 536 064 × 2 = 1 + 0.825 072 128;
  • 57) 0.825 072 128 × 2 = 1 + 0.650 144 256;
  • 58) 0.650 144 256 × 2 = 1 + 0.300 288 512;
  • 59) 0.300 288 512 × 2 = 0 + 0.600 577 024;
  • 60) 0.600 577 024 × 2 = 1 + 0.201 154 048;
  • 61) 0.201 154 048 × 2 = 0 + 0.402 308 096;
  • 62) 0.402 308 096 × 2 = 0 + 0.804 616 192;
  • 63) 0.804 616 192 × 2 = 1 + 0.609 232 384;
  • 64) 0.609 232 384 × 2 = 1 + 0.218 464 768;

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:


0.000 282 373(10) =

0.0000 0000 0001 0010 1000 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011(2)

6. Positive number before normalization:

0.000 282 373(10) =

0.0000 0000 0001 0010 1000 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011(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:


0.000 282 373(10) =

0.0000 0000 0001 0010 1000 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011(2) =

0.0000 0000 0001 0010 1000 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011(2) × 20 =

1.0010 1000 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011(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 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011


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 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011 =

0010 1000 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011


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 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011


Decimal number -0.000 282 373 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0001 0110 1110 1100 1100 1101 1110 1011 0111 1101 0011

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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