-0.000 282 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 62(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 62(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 62| = 0.000 282 62


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 62.

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 62 × 2 = 0 + 0.000 565 24;
  • 2) 0.000 565 24 × 2 = 0 + 0.001 130 48;
  • 3) 0.001 130 48 × 2 = 0 + 0.002 260 96;
  • 4) 0.002 260 96 × 2 = 0 + 0.004 521 92;
  • 5) 0.004 521 92 × 2 = 0 + 0.009 043 84;
  • 6) 0.009 043 84 × 2 = 0 + 0.018 087 68;
  • 7) 0.018 087 68 × 2 = 0 + 0.036 175 36;
  • 8) 0.036 175 36 × 2 = 0 + 0.072 350 72;
  • 9) 0.072 350 72 × 2 = 0 + 0.144 701 44;
  • 10) 0.144 701 44 × 2 = 0 + 0.289 402 88;
  • 11) 0.289 402 88 × 2 = 0 + 0.578 805 76;
  • 12) 0.578 805 76 × 2 = 1 + 0.157 611 52;
  • 13) 0.157 611 52 × 2 = 0 + 0.315 223 04;
  • 14) 0.315 223 04 × 2 = 0 + 0.630 446 08;
  • 15) 0.630 446 08 × 2 = 1 + 0.260 892 16;
  • 16) 0.260 892 16 × 2 = 0 + 0.521 784 32;
  • 17) 0.521 784 32 × 2 = 1 + 0.043 568 64;
  • 18) 0.043 568 64 × 2 = 0 + 0.087 137 28;
  • 19) 0.087 137 28 × 2 = 0 + 0.174 274 56;
  • 20) 0.174 274 56 × 2 = 0 + 0.348 549 12;
  • 21) 0.348 549 12 × 2 = 0 + 0.697 098 24;
  • 22) 0.697 098 24 × 2 = 1 + 0.394 196 48;
  • 23) 0.394 196 48 × 2 = 0 + 0.788 392 96;
  • 24) 0.788 392 96 × 2 = 1 + 0.576 785 92;
  • 25) 0.576 785 92 × 2 = 1 + 0.153 571 84;
  • 26) 0.153 571 84 × 2 = 0 + 0.307 143 68;
  • 27) 0.307 143 68 × 2 = 0 + 0.614 287 36;
  • 28) 0.614 287 36 × 2 = 1 + 0.228 574 72;
  • 29) 0.228 574 72 × 2 = 0 + 0.457 149 44;
  • 30) 0.457 149 44 × 2 = 0 + 0.914 298 88;
  • 31) 0.914 298 88 × 2 = 1 + 0.828 597 76;
  • 32) 0.828 597 76 × 2 = 1 + 0.657 195 52;
  • 33) 0.657 195 52 × 2 = 1 + 0.314 391 04;
  • 34) 0.314 391 04 × 2 = 0 + 0.628 782 08;
  • 35) 0.628 782 08 × 2 = 1 + 0.257 564 16;
  • 36) 0.257 564 16 × 2 = 0 + 0.515 128 32;
  • 37) 0.515 128 32 × 2 = 1 + 0.030 256 64;
  • 38) 0.030 256 64 × 2 = 0 + 0.060 513 28;
  • 39) 0.060 513 28 × 2 = 0 + 0.121 026 56;
  • 40) 0.121 026 56 × 2 = 0 + 0.242 053 12;
  • 41) 0.242 053 12 × 2 = 0 + 0.484 106 24;
  • 42) 0.484 106 24 × 2 = 0 + 0.968 212 48;
  • 43) 0.968 212 48 × 2 = 1 + 0.936 424 96;
  • 44) 0.936 424 96 × 2 = 1 + 0.872 849 92;
  • 45) 0.872 849 92 × 2 = 1 + 0.745 699 84;
  • 46) 0.745 699 84 × 2 = 1 + 0.491 399 68;
  • 47) 0.491 399 68 × 2 = 0 + 0.982 799 36;
  • 48) 0.982 799 36 × 2 = 1 + 0.965 598 72;
  • 49) 0.965 598 72 × 2 = 1 + 0.931 197 44;
  • 50) 0.931 197 44 × 2 = 1 + 0.862 394 88;
  • 51) 0.862 394 88 × 2 = 1 + 0.724 789 76;
  • 52) 0.724 789 76 × 2 = 1 + 0.449 579 52;
  • 53) 0.449 579 52 × 2 = 0 + 0.899 159 04;
  • 54) 0.899 159 04 × 2 = 1 + 0.798 318 08;
  • 55) 0.798 318 08 × 2 = 1 + 0.596 636 16;
  • 56) 0.596 636 16 × 2 = 1 + 0.193 272 32;
  • 57) 0.193 272 32 × 2 = 0 + 0.386 544 64;
  • 58) 0.386 544 64 × 2 = 0 + 0.773 089 28;
  • 59) 0.773 089 28 × 2 = 1 + 0.546 178 56;
  • 60) 0.546 178 56 × 2 = 1 + 0.092 357 12;
  • 61) 0.092 357 12 × 2 = 0 + 0.184 714 24;
  • 62) 0.184 714 24 × 2 = 0 + 0.369 428 48;
  • 63) 0.369 428 48 × 2 = 0 + 0.738 856 96;
  • 64) 0.738 856 96 × 2 = 1 + 0.477 713 92;

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 62(10) =

0.0000 0000 0001 0010 1000 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001(2)

6. Positive number before normalization:

0.000 282 62(10) =

0.0000 0000 0001 0010 1000 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001(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 62(10) =

0.0000 0000 0001 0010 1000 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001(2) =

0.0000 0000 0001 0010 1000 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001(2) × 20 =

1.0010 1000 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001(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 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001


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 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001 =

0010 1000 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001


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 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001


Decimal number -0.000 282 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0101 1001 0011 1010 1000 0011 1101 1111 0111 0011 0001

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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