-0.000 281 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 37(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 37(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 37| = 0.000 281 37


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 37.

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 37 × 2 = 0 + 0.000 562 74;
  • 2) 0.000 562 74 × 2 = 0 + 0.001 125 48;
  • 3) 0.001 125 48 × 2 = 0 + 0.002 250 96;
  • 4) 0.002 250 96 × 2 = 0 + 0.004 501 92;
  • 5) 0.004 501 92 × 2 = 0 + 0.009 003 84;
  • 6) 0.009 003 84 × 2 = 0 + 0.018 007 68;
  • 7) 0.018 007 68 × 2 = 0 + 0.036 015 36;
  • 8) 0.036 015 36 × 2 = 0 + 0.072 030 72;
  • 9) 0.072 030 72 × 2 = 0 + 0.144 061 44;
  • 10) 0.144 061 44 × 2 = 0 + 0.288 122 88;
  • 11) 0.288 122 88 × 2 = 0 + 0.576 245 76;
  • 12) 0.576 245 76 × 2 = 1 + 0.152 491 52;
  • 13) 0.152 491 52 × 2 = 0 + 0.304 983 04;
  • 14) 0.304 983 04 × 2 = 0 + 0.609 966 08;
  • 15) 0.609 966 08 × 2 = 1 + 0.219 932 16;
  • 16) 0.219 932 16 × 2 = 0 + 0.439 864 32;
  • 17) 0.439 864 32 × 2 = 0 + 0.879 728 64;
  • 18) 0.879 728 64 × 2 = 1 + 0.759 457 28;
  • 19) 0.759 457 28 × 2 = 1 + 0.518 914 56;
  • 20) 0.518 914 56 × 2 = 1 + 0.037 829 12;
  • 21) 0.037 829 12 × 2 = 0 + 0.075 658 24;
  • 22) 0.075 658 24 × 2 = 0 + 0.151 316 48;
  • 23) 0.151 316 48 × 2 = 0 + 0.302 632 96;
  • 24) 0.302 632 96 × 2 = 0 + 0.605 265 92;
  • 25) 0.605 265 92 × 2 = 1 + 0.210 531 84;
  • 26) 0.210 531 84 × 2 = 0 + 0.421 063 68;
  • 27) 0.421 063 68 × 2 = 0 + 0.842 127 36;
  • 28) 0.842 127 36 × 2 = 1 + 0.684 254 72;
  • 29) 0.684 254 72 × 2 = 1 + 0.368 509 44;
  • 30) 0.368 509 44 × 2 = 0 + 0.737 018 88;
  • 31) 0.737 018 88 × 2 = 1 + 0.474 037 76;
  • 32) 0.474 037 76 × 2 = 0 + 0.948 075 52;
  • 33) 0.948 075 52 × 2 = 1 + 0.896 151 04;
  • 34) 0.896 151 04 × 2 = 1 + 0.792 302 08;
  • 35) 0.792 302 08 × 2 = 1 + 0.584 604 16;
  • 36) 0.584 604 16 × 2 = 1 + 0.169 208 32;
  • 37) 0.169 208 32 × 2 = 0 + 0.338 416 64;
  • 38) 0.338 416 64 × 2 = 0 + 0.676 833 28;
  • 39) 0.676 833 28 × 2 = 1 + 0.353 666 56;
  • 40) 0.353 666 56 × 2 = 0 + 0.707 333 12;
  • 41) 0.707 333 12 × 2 = 1 + 0.414 666 24;
  • 42) 0.414 666 24 × 2 = 0 + 0.829 332 48;
  • 43) 0.829 332 48 × 2 = 1 + 0.658 664 96;
  • 44) 0.658 664 96 × 2 = 1 + 0.317 329 92;
  • 45) 0.317 329 92 × 2 = 0 + 0.634 659 84;
  • 46) 0.634 659 84 × 2 = 1 + 0.269 319 68;
  • 47) 0.269 319 68 × 2 = 0 + 0.538 639 36;
  • 48) 0.538 639 36 × 2 = 1 + 0.077 278 72;
  • 49) 0.077 278 72 × 2 = 0 + 0.154 557 44;
  • 50) 0.154 557 44 × 2 = 0 + 0.309 114 88;
  • 51) 0.309 114 88 × 2 = 0 + 0.618 229 76;
  • 52) 0.618 229 76 × 2 = 1 + 0.236 459 52;
  • 53) 0.236 459 52 × 2 = 0 + 0.472 919 04;
  • 54) 0.472 919 04 × 2 = 0 + 0.945 838 08;
  • 55) 0.945 838 08 × 2 = 1 + 0.891 676 16;
  • 56) 0.891 676 16 × 2 = 1 + 0.783 352 32;
  • 57) 0.783 352 32 × 2 = 1 + 0.566 704 64;
  • 58) 0.566 704 64 × 2 = 1 + 0.133 409 28;
  • 59) 0.133 409 28 × 2 = 0 + 0.266 818 56;
  • 60) 0.266 818 56 × 2 = 0 + 0.533 637 12;
  • 61) 0.533 637 12 × 2 = 1 + 0.067 274 24;
  • 62) 0.067 274 24 × 2 = 0 + 0.134 548 48;
  • 63) 0.134 548 48 × 2 = 0 + 0.269 096 96;
  • 64) 0.269 096 96 × 2 = 0 + 0.538 193 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:


0.000 281 37(10) =

0.0000 0000 0001 0010 0111 0000 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000(2)

6. Positive number before normalization:

0.000 281 37(10) =

0.0000 0000 0001 0010 0111 0000 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000(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 37(10) =

0.0000 0000 0001 0010 0111 0000 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000(2) =

0.0000 0000 0001 0010 0111 0000 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000(2) × 20 =

1.0010 0111 0000 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000(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 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000


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 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000 =

0010 0111 0000 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000


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 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000


Decimal number -0.000 281 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 0000 1001 1010 1111 0010 1011 0101 0001 0011 1100 1000

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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