-0.000 281 83 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 83(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 83(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 83| = 0.000 281 83


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 83.

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 83 × 2 = 0 + 0.000 563 66;
  • 2) 0.000 563 66 × 2 = 0 + 0.001 127 32;
  • 3) 0.001 127 32 × 2 = 0 + 0.002 254 64;
  • 4) 0.002 254 64 × 2 = 0 + 0.004 509 28;
  • 5) 0.004 509 28 × 2 = 0 + 0.009 018 56;
  • 6) 0.009 018 56 × 2 = 0 + 0.018 037 12;
  • 7) 0.018 037 12 × 2 = 0 + 0.036 074 24;
  • 8) 0.036 074 24 × 2 = 0 + 0.072 148 48;
  • 9) 0.072 148 48 × 2 = 0 + 0.144 296 96;
  • 10) 0.144 296 96 × 2 = 0 + 0.288 593 92;
  • 11) 0.288 593 92 × 2 = 0 + 0.577 187 84;
  • 12) 0.577 187 84 × 2 = 1 + 0.154 375 68;
  • 13) 0.154 375 68 × 2 = 0 + 0.308 751 36;
  • 14) 0.308 751 36 × 2 = 0 + 0.617 502 72;
  • 15) 0.617 502 72 × 2 = 1 + 0.235 005 44;
  • 16) 0.235 005 44 × 2 = 0 + 0.470 010 88;
  • 17) 0.470 010 88 × 2 = 0 + 0.940 021 76;
  • 18) 0.940 021 76 × 2 = 1 + 0.880 043 52;
  • 19) 0.880 043 52 × 2 = 1 + 0.760 087 04;
  • 20) 0.760 087 04 × 2 = 1 + 0.520 174 08;
  • 21) 0.520 174 08 × 2 = 1 + 0.040 348 16;
  • 22) 0.040 348 16 × 2 = 0 + 0.080 696 32;
  • 23) 0.080 696 32 × 2 = 0 + 0.161 392 64;
  • 24) 0.161 392 64 × 2 = 0 + 0.322 785 28;
  • 25) 0.322 785 28 × 2 = 0 + 0.645 570 56;
  • 26) 0.645 570 56 × 2 = 1 + 0.291 141 12;
  • 27) 0.291 141 12 × 2 = 0 + 0.582 282 24;
  • 28) 0.582 282 24 × 2 = 1 + 0.164 564 48;
  • 29) 0.164 564 48 × 2 = 0 + 0.329 128 96;
  • 30) 0.329 128 96 × 2 = 0 + 0.658 257 92;
  • 31) 0.658 257 92 × 2 = 1 + 0.316 515 84;
  • 32) 0.316 515 84 × 2 = 0 + 0.633 031 68;
  • 33) 0.633 031 68 × 2 = 1 + 0.266 063 36;
  • 34) 0.266 063 36 × 2 = 0 + 0.532 126 72;
  • 35) 0.532 126 72 × 2 = 1 + 0.064 253 44;
  • 36) 0.064 253 44 × 2 = 0 + 0.128 506 88;
  • 37) 0.128 506 88 × 2 = 0 + 0.257 013 76;
  • 38) 0.257 013 76 × 2 = 0 + 0.514 027 52;
  • 39) 0.514 027 52 × 2 = 1 + 0.028 055 04;
  • 40) 0.028 055 04 × 2 = 0 + 0.056 110 08;
  • 41) 0.056 110 08 × 2 = 0 + 0.112 220 16;
  • 42) 0.112 220 16 × 2 = 0 + 0.224 440 32;
  • 43) 0.224 440 32 × 2 = 0 + 0.448 880 64;
  • 44) 0.448 880 64 × 2 = 0 + 0.897 761 28;
  • 45) 0.897 761 28 × 2 = 1 + 0.795 522 56;
  • 46) 0.795 522 56 × 2 = 1 + 0.591 045 12;
  • 47) 0.591 045 12 × 2 = 1 + 0.182 090 24;
  • 48) 0.182 090 24 × 2 = 0 + 0.364 180 48;
  • 49) 0.364 180 48 × 2 = 0 + 0.728 360 96;
  • 50) 0.728 360 96 × 2 = 1 + 0.456 721 92;
  • 51) 0.456 721 92 × 2 = 0 + 0.913 443 84;
  • 52) 0.913 443 84 × 2 = 1 + 0.826 887 68;
  • 53) 0.826 887 68 × 2 = 1 + 0.653 775 36;
  • 54) 0.653 775 36 × 2 = 1 + 0.307 550 72;
  • 55) 0.307 550 72 × 2 = 0 + 0.615 101 44;
  • 56) 0.615 101 44 × 2 = 1 + 0.230 202 88;
  • 57) 0.230 202 88 × 2 = 0 + 0.460 405 76;
  • 58) 0.460 405 76 × 2 = 0 + 0.920 811 52;
  • 59) 0.920 811 52 × 2 = 1 + 0.841 623 04;
  • 60) 0.841 623 04 × 2 = 1 + 0.683 246 08;
  • 61) 0.683 246 08 × 2 = 1 + 0.366 492 16;
  • 62) 0.366 492 16 × 2 = 0 + 0.732 984 32;
  • 63) 0.732 984 32 × 2 = 1 + 0.465 968 64;
  • 64) 0.465 968 64 × 2 = 0 + 0.931 937 28;

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 83(10) =

0.0000 0000 0001 0010 0111 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010(2)

6. Positive number before normalization:

0.000 281 83(10) =

0.0000 0000 0001 0010 0111 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010(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 83(10) =

0.0000 0000 0001 0010 0111 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010(2) =

0.0000 0000 0001 0010 0111 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010(2) × 20 =

1.0010 0111 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010(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 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010


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 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010 =

0010 0111 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010


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 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010


Decimal number -0.000 281 83 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1000 0101 0010 1010 0010 0000 1110 0101 1101 0011 1010

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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