-0.000 281 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 74(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
-0.000 281 74(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:

|-0.000 281 74| = 0.000 281 74


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: 0.000 281 74.

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) 0.000 281 74 × 2 = 0 + 0.000 563 48;
  • 2) 0.000 563 48 × 2 = 0 + 0.001 126 96;
  • 3) 0.001 126 96 × 2 = 0 + 0.002 253 92;
  • 4) 0.002 253 92 × 2 = 0 + 0.004 507 84;
  • 5) 0.004 507 84 × 2 = 0 + 0.009 015 68;
  • 6) 0.009 015 68 × 2 = 0 + 0.018 031 36;
  • 7) 0.018 031 36 × 2 = 0 + 0.036 062 72;
  • 8) 0.036 062 72 × 2 = 0 + 0.072 125 44;
  • 9) 0.072 125 44 × 2 = 0 + 0.144 250 88;
  • 10) 0.144 250 88 × 2 = 0 + 0.288 501 76;
  • 11) 0.288 501 76 × 2 = 0 + 0.577 003 52;
  • 12) 0.577 003 52 × 2 = 1 + 0.154 007 04;
  • 13) 0.154 007 04 × 2 = 0 + 0.308 014 08;
  • 14) 0.308 014 08 × 2 = 0 + 0.616 028 16;
  • 15) 0.616 028 16 × 2 = 1 + 0.232 056 32;
  • 16) 0.232 056 32 × 2 = 0 + 0.464 112 64;
  • 17) 0.464 112 64 × 2 = 0 + 0.928 225 28;
  • 18) 0.928 225 28 × 2 = 1 + 0.856 450 56;
  • 19) 0.856 450 56 × 2 = 1 + 0.712 901 12;
  • 20) 0.712 901 12 × 2 = 1 + 0.425 802 24;
  • 21) 0.425 802 24 × 2 = 0 + 0.851 604 48;
  • 22) 0.851 604 48 × 2 = 1 + 0.703 208 96;
  • 23) 0.703 208 96 × 2 = 1 + 0.406 417 92;
  • 24) 0.406 417 92 × 2 = 0 + 0.812 835 84;
  • 25) 0.812 835 84 × 2 = 1 + 0.625 671 68;
  • 26) 0.625 671 68 × 2 = 1 + 0.251 343 36;
  • 27) 0.251 343 36 × 2 = 0 + 0.502 686 72;
  • 28) 0.502 686 72 × 2 = 1 + 0.005 373 44;
  • 29) 0.005 373 44 × 2 = 0 + 0.010 746 88;
  • 30) 0.010 746 88 × 2 = 0 + 0.021 493 76;
  • 31) 0.021 493 76 × 2 = 0 + 0.042 987 52;
  • 32) 0.042 987 52 × 2 = 0 + 0.085 975 04;
  • 33) 0.085 975 04 × 2 = 0 + 0.171 950 08;
  • 34) 0.171 950 08 × 2 = 0 + 0.343 900 16;
  • 35) 0.343 900 16 × 2 = 0 + 0.687 800 32;
  • 36) 0.687 800 32 × 2 = 1 + 0.375 600 64;
  • 37) 0.375 600 64 × 2 = 0 + 0.751 201 28;
  • 38) 0.751 201 28 × 2 = 1 + 0.502 402 56;
  • 39) 0.502 402 56 × 2 = 1 + 0.004 805 12;
  • 40) 0.004 805 12 × 2 = 0 + 0.009 610 24;
  • 41) 0.009 610 24 × 2 = 0 + 0.019 220 48;
  • 42) 0.019 220 48 × 2 = 0 + 0.038 440 96;
  • 43) 0.038 440 96 × 2 = 0 + 0.076 881 92;
  • 44) 0.076 881 92 × 2 = 0 + 0.153 763 84;
  • 45) 0.153 763 84 × 2 = 0 + 0.307 527 68;
  • 46) 0.307 527 68 × 2 = 0 + 0.615 055 36;
  • 47) 0.615 055 36 × 2 = 1 + 0.230 110 72;
  • 48) 0.230 110 72 × 2 = 0 + 0.460 221 44;
  • 49) 0.460 221 44 × 2 = 0 + 0.920 442 88;
  • 50) 0.920 442 88 × 2 = 1 + 0.840 885 76;
  • 51) 0.840 885 76 × 2 = 1 + 0.681 771 52;
  • 52) 0.681 771 52 × 2 = 1 + 0.363 543 04;
  • 53) 0.363 543 04 × 2 = 0 + 0.727 086 08;
  • 54) 0.727 086 08 × 2 = 1 + 0.454 172 16;
  • 55) 0.454 172 16 × 2 = 0 + 0.908 344 32;
  • 56) 0.908 344 32 × 2 = 1 + 0.816 688 64;
  • 57) 0.816 688 64 × 2 = 1 + 0.633 377 28;
  • 58) 0.633 377 28 × 2 = 1 + 0.266 754 56;
  • 59) 0.266 754 56 × 2 = 0 + 0.533 509 12;
  • 60) 0.533 509 12 × 2 = 1 + 0.067 018 24;
  • 61) 0.067 018 24 × 2 = 0 + 0.134 036 48;
  • 62) 0.134 036 48 × 2 = 0 + 0.268 072 96;
  • 63) 0.268 072 96 × 2 = 0 + 0.536 145 92;
  • 64) 0.536 145 92 × 2 = 1 + 0.072 291 84;

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:


0.000 281 74(10) =

0.0000 0000 0001 0010 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 0001(2)

6. Positive number before normalization:

0.000 281 74(10) =

0.0000 0000 0001 0010 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 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:


0.000 281 74(10) =

0.0000 0000 0001 0010 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 0001(2) =

0.0000 0000 0001 0010 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 0001(2) × 20 =

1.0010 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 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 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 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 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 0001 =

0010 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 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 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 0001


Decimal number -0.000 281 74 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 0110 1101 0000 0001 0110 0000 0010 0111 0101 1101 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