0.000 000 305 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 305(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 305(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 305.

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) 0.000 000 305 × 2 = 0 + 0.000 000 61;
  • 2) 0.000 000 61 × 2 = 0 + 0.000 001 22;
  • 3) 0.000 001 22 × 2 = 0 + 0.000 002 44;
  • 4) 0.000 002 44 × 2 = 0 + 0.000 004 88;
  • 5) 0.000 004 88 × 2 = 0 + 0.000 009 76;
  • 6) 0.000 009 76 × 2 = 0 + 0.000 019 52;
  • 7) 0.000 019 52 × 2 = 0 + 0.000 039 04;
  • 8) 0.000 039 04 × 2 = 0 + 0.000 078 08;
  • 9) 0.000 078 08 × 2 = 0 + 0.000 156 16;
  • 10) 0.000 156 16 × 2 = 0 + 0.000 312 32;
  • 11) 0.000 312 32 × 2 = 0 + 0.000 624 64;
  • 12) 0.000 624 64 × 2 = 0 + 0.001 249 28;
  • 13) 0.001 249 28 × 2 = 0 + 0.002 498 56;
  • 14) 0.002 498 56 × 2 = 0 + 0.004 997 12;
  • 15) 0.004 997 12 × 2 = 0 + 0.009 994 24;
  • 16) 0.009 994 24 × 2 = 0 + 0.019 988 48;
  • 17) 0.019 988 48 × 2 = 0 + 0.039 976 96;
  • 18) 0.039 976 96 × 2 = 0 + 0.079 953 92;
  • 19) 0.079 953 92 × 2 = 0 + 0.159 907 84;
  • 20) 0.159 907 84 × 2 = 0 + 0.319 815 68;
  • 21) 0.319 815 68 × 2 = 0 + 0.639 631 36;
  • 22) 0.639 631 36 × 2 = 1 + 0.279 262 72;
  • 23) 0.279 262 72 × 2 = 0 + 0.558 525 44;
  • 24) 0.558 525 44 × 2 = 1 + 0.117 050 88;
  • 25) 0.117 050 88 × 2 = 0 + 0.234 101 76;
  • 26) 0.234 101 76 × 2 = 0 + 0.468 203 52;
  • 27) 0.468 203 52 × 2 = 0 + 0.936 407 04;
  • 28) 0.936 407 04 × 2 = 1 + 0.872 814 08;
  • 29) 0.872 814 08 × 2 = 1 + 0.745 628 16;
  • 30) 0.745 628 16 × 2 = 1 + 0.491 256 32;
  • 31) 0.491 256 32 × 2 = 0 + 0.982 512 64;
  • 32) 0.982 512 64 × 2 = 1 + 0.965 025 28;
  • 33) 0.965 025 28 × 2 = 1 + 0.930 050 56;
  • 34) 0.930 050 56 × 2 = 1 + 0.860 101 12;
  • 35) 0.860 101 12 × 2 = 1 + 0.720 202 24;
  • 36) 0.720 202 24 × 2 = 1 + 0.440 404 48;
  • 37) 0.440 404 48 × 2 = 0 + 0.880 808 96;
  • 38) 0.880 808 96 × 2 = 1 + 0.761 617 92;
  • 39) 0.761 617 92 × 2 = 1 + 0.523 235 84;
  • 40) 0.523 235 84 × 2 = 1 + 0.046 471 68;
  • 41) 0.046 471 68 × 2 = 0 + 0.092 943 36;
  • 42) 0.092 943 36 × 2 = 0 + 0.185 886 72;
  • 43) 0.185 886 72 × 2 = 0 + 0.371 773 44;
  • 44) 0.371 773 44 × 2 = 0 + 0.743 546 88;
  • 45) 0.743 546 88 × 2 = 1 + 0.487 093 76;

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).


4. 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:


0.000 000 305(10) =

0.0000 0000 0000 0000 0000 0101 0001 1101 1111 0111 0000 1(2)

5. Positive number before normalization:

0.000 000 305(10) =

0.0000 0000 0000 0000 0000 0101 0001 1101 1111 0111 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 305(10) =

0.0000 0000 0000 0000 0000 0101 0001 1101 1111 0111 0000 1(2) =

0.0000 0000 0000 0000 0000 0101 0001 1101 1111 0111 0000 1(2) × 20 =

1.0100 0111 0111 1101 1100 001(2) × 2-22


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -22

Mantissa (not normalized):
1.0100 0111 0111 1101 1100 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

105(10) =

0110 1001(2)


11. 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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 010 0011 1011 1110 1110 0001 =

010 0011 1011 1110 1110 0001


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 1001

Mantissa (23 bits) =
010 0011 1011 1110 1110 0001


Decimal number 0.000 000 305 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 010 0011 1011 1110 1110 0001

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111