0.000 005 05 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 005 05(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
0.000 005 05(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: 0.000 005 05.

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) 0.000 005 05 × 2 = 0 + 0.000 010 1;
  • 2) 0.000 010 1 × 2 = 0 + 0.000 020 2;
  • 3) 0.000 020 2 × 2 = 0 + 0.000 040 4;
  • 4) 0.000 040 4 × 2 = 0 + 0.000 080 8;
  • 5) 0.000 080 8 × 2 = 0 + 0.000 161 6;
  • 6) 0.000 161 6 × 2 = 0 + 0.000 323 2;
  • 7) 0.000 323 2 × 2 = 0 + 0.000 646 4;
  • 8) 0.000 646 4 × 2 = 0 + 0.001 292 8;
  • 9) 0.001 292 8 × 2 = 0 + 0.002 585 6;
  • 10) 0.002 585 6 × 2 = 0 + 0.005 171 2;
  • 11) 0.005 171 2 × 2 = 0 + 0.010 342 4;
  • 12) 0.010 342 4 × 2 = 0 + 0.020 684 8;
  • 13) 0.020 684 8 × 2 = 0 + 0.041 369 6;
  • 14) 0.041 369 6 × 2 = 0 + 0.082 739 2;
  • 15) 0.082 739 2 × 2 = 0 + 0.165 478 4;
  • 16) 0.165 478 4 × 2 = 0 + 0.330 956 8;
  • 17) 0.330 956 8 × 2 = 0 + 0.661 913 6;
  • 18) 0.661 913 6 × 2 = 1 + 0.323 827 2;
  • 19) 0.323 827 2 × 2 = 0 + 0.647 654 4;
  • 20) 0.647 654 4 × 2 = 1 + 0.295 308 8;
  • 21) 0.295 308 8 × 2 = 0 + 0.590 617 6;
  • 22) 0.590 617 6 × 2 = 1 + 0.181 235 2;
  • 23) 0.181 235 2 × 2 = 0 + 0.362 470 4;
  • 24) 0.362 470 4 × 2 = 0 + 0.724 940 8;
  • 25) 0.724 940 8 × 2 = 1 + 0.449 881 6;
  • 26) 0.449 881 6 × 2 = 0 + 0.899 763 2;
  • 27) 0.899 763 2 × 2 = 1 + 0.799 526 4;
  • 28) 0.799 526 4 × 2 = 1 + 0.599 052 8;
  • 29) 0.599 052 8 × 2 = 1 + 0.198 105 6;
  • 30) 0.198 105 6 × 2 = 0 + 0.396 211 2;
  • 31) 0.396 211 2 × 2 = 0 + 0.792 422 4;
  • 32) 0.792 422 4 × 2 = 1 + 0.584 844 8;
  • 33) 0.584 844 8 × 2 = 1 + 0.169 689 6;
  • 34) 0.169 689 6 × 2 = 0 + 0.339 379 2;
  • 35) 0.339 379 2 × 2 = 0 + 0.678 758 4;
  • 36) 0.678 758 4 × 2 = 1 + 0.357 516 8;
  • 37) 0.357 516 8 × 2 = 0 + 0.715 033 6;
  • 38) 0.715 033 6 × 2 = 1 + 0.430 067 2;
  • 39) 0.430 067 2 × 2 = 0 + 0.860 134 4;
  • 40) 0.860 134 4 × 2 = 1 + 0.720 268 8;
  • 41) 0.720 268 8 × 2 = 1 + 0.440 537 6;

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:


0.000 005 05(10) =

0.0000 0000 0000 0000 0101 0100 1011 1001 1001 0101 1(2)

5. Positive number before normalization:

0.000 005 05(10) =

0.0000 0000 0000 0000 0101 0100 1011 1001 1001 0101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 005 05(10) =

0.0000 0000 0000 0000 0101 0100 1011 1001 1001 0101 1(2) =

0.0000 0000 0000 0000 0101 0100 1011 1001 1001 0101 1(2) × 20 =

1.0101 0010 1110 0110 0101 011(2) × 2-18


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): -18

Mantissa (not normalized):
1.0101 0010 1110 0110 0101 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(8-1) - 1 =

(-18 + 127)(10) =

109(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;
  • 109 ÷ 2 = 54 + 1;
  • 54 ÷ 2 = 27 + 0;
  • 27 ÷ 2 = 13 + 1;
  • 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) =

109(10) =

0110 1101(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 1001 0111 0011 0010 1011 =

010 1001 0111 0011 0010 1011


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 1101

Mantissa (23 bits) =
010 1001 0111 0011 0010 1011


Decimal number 0.000 005 05 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1101 - 010 1001 0111 0011 0010 1011

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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