-0.000 281 34 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 34(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 34(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 34| = 0.000 281 34


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 34.

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 34 × 2 = 0 + 0.000 562 68;
  • 2) 0.000 562 68 × 2 = 0 + 0.001 125 36;
  • 3) 0.001 125 36 × 2 = 0 + 0.002 250 72;
  • 4) 0.002 250 72 × 2 = 0 + 0.004 501 44;
  • 5) 0.004 501 44 × 2 = 0 + 0.009 002 88;
  • 6) 0.009 002 88 × 2 = 0 + 0.018 005 76;
  • 7) 0.018 005 76 × 2 = 0 + 0.036 011 52;
  • 8) 0.036 011 52 × 2 = 0 + 0.072 023 04;
  • 9) 0.072 023 04 × 2 = 0 + 0.144 046 08;
  • 10) 0.144 046 08 × 2 = 0 + 0.288 092 16;
  • 11) 0.288 092 16 × 2 = 0 + 0.576 184 32;
  • 12) 0.576 184 32 × 2 = 1 + 0.152 368 64;
  • 13) 0.152 368 64 × 2 = 0 + 0.304 737 28;
  • 14) 0.304 737 28 × 2 = 0 + 0.609 474 56;
  • 15) 0.609 474 56 × 2 = 1 + 0.218 949 12;
  • 16) 0.218 949 12 × 2 = 0 + 0.437 898 24;
  • 17) 0.437 898 24 × 2 = 0 + 0.875 796 48;
  • 18) 0.875 796 48 × 2 = 1 + 0.751 592 96;
  • 19) 0.751 592 96 × 2 = 1 + 0.503 185 92;
  • 20) 0.503 185 92 × 2 = 1 + 0.006 371 84;
  • 21) 0.006 371 84 × 2 = 0 + 0.012 743 68;
  • 22) 0.012 743 68 × 2 = 0 + 0.025 487 36;
  • 23) 0.025 487 36 × 2 = 0 + 0.050 974 72;
  • 24) 0.050 974 72 × 2 = 0 + 0.101 949 44;
  • 25) 0.101 949 44 × 2 = 0 + 0.203 898 88;
  • 26) 0.203 898 88 × 2 = 0 + 0.407 797 76;
  • 27) 0.407 797 76 × 2 = 0 + 0.815 595 52;
  • 28) 0.815 595 52 × 2 = 1 + 0.631 191 04;
  • 29) 0.631 191 04 × 2 = 1 + 0.262 382 08;
  • 30) 0.262 382 08 × 2 = 0 + 0.524 764 16;
  • 31) 0.524 764 16 × 2 = 1 + 0.049 528 32;
  • 32) 0.049 528 32 × 2 = 0 + 0.099 056 64;
  • 33) 0.099 056 64 × 2 = 0 + 0.198 113 28;
  • 34) 0.198 113 28 × 2 = 0 + 0.396 226 56;
  • 35) 0.396 226 56 × 2 = 0 + 0.792 453 12;
  • 36) 0.792 453 12 × 2 = 1 + 0.584 906 24;
  • 37) 0.584 906 24 × 2 = 1 + 0.169 812 48;
  • 38) 0.169 812 48 × 2 = 0 + 0.339 624 96;
  • 39) 0.339 624 96 × 2 = 0 + 0.679 249 92;
  • 40) 0.679 249 92 × 2 = 1 + 0.358 499 84;
  • 41) 0.358 499 84 × 2 = 0 + 0.716 999 68;
  • 42) 0.716 999 68 × 2 = 1 + 0.433 999 36;
  • 43) 0.433 999 36 × 2 = 0 + 0.867 998 72;
  • 44) 0.867 998 72 × 2 = 1 + 0.735 997 44;
  • 45) 0.735 997 44 × 2 = 1 + 0.471 994 88;
  • 46) 0.471 994 88 × 2 = 0 + 0.943 989 76;
  • 47) 0.943 989 76 × 2 = 1 + 0.887 979 52;
  • 48) 0.887 979 52 × 2 = 1 + 0.775 959 04;
  • 49) 0.775 959 04 × 2 = 1 + 0.551 918 08;
  • 50) 0.551 918 08 × 2 = 1 + 0.103 836 16;
  • 51) 0.103 836 16 × 2 = 0 + 0.207 672 32;
  • 52) 0.207 672 32 × 2 = 0 + 0.415 344 64;
  • 53) 0.415 344 64 × 2 = 0 + 0.830 689 28;
  • 54) 0.830 689 28 × 2 = 1 + 0.661 378 56;
  • 55) 0.661 378 56 × 2 = 1 + 0.322 757 12;
  • 56) 0.322 757 12 × 2 = 0 + 0.645 514 24;
  • 57) 0.645 514 24 × 2 = 1 + 0.291 028 48;
  • 58) 0.291 028 48 × 2 = 0 + 0.582 056 96;
  • 59) 0.582 056 96 × 2 = 1 + 0.164 113 92;
  • 60) 0.164 113 92 × 2 = 0 + 0.328 227 84;
  • 61) 0.328 227 84 × 2 = 0 + 0.656 455 68;
  • 62) 0.656 455 68 × 2 = 1 + 0.312 911 36;
  • 63) 0.312 911 36 × 2 = 0 + 0.625 822 72;
  • 64) 0.625 822 72 × 2 = 1 + 0.251 645 44;

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 34(10) =

0.0000 0000 0001 0010 0111 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101(2)

6. Positive number before normalization:

0.000 281 34(10) =

0.0000 0000 0001 0010 0111 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101(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 34(10) =

0.0000 0000 0001 0010 0111 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101(2) =

0.0000 0000 0001 0010 0111 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101(2) × 20 =

1.0010 0111 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101(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 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101


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 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101 =

0010 0111 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101


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 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101


Decimal number -0.000 281 34 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 0000 0001 1010 0001 1001 0101 1011 1100 0110 1010 0101

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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