0.000 000 235 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 235(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
0.000 000 235(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: 0.000 000 235.

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) 0.000 000 235 × 2 = 0 + 0.000 000 47;
  • 2) 0.000 000 47 × 2 = 0 + 0.000 000 94;
  • 3) 0.000 000 94 × 2 = 0 + 0.000 001 88;
  • 4) 0.000 001 88 × 2 = 0 + 0.000 003 76;
  • 5) 0.000 003 76 × 2 = 0 + 0.000 007 52;
  • 6) 0.000 007 52 × 2 = 0 + 0.000 015 04;
  • 7) 0.000 015 04 × 2 = 0 + 0.000 030 08;
  • 8) 0.000 030 08 × 2 = 0 + 0.000 060 16;
  • 9) 0.000 060 16 × 2 = 0 + 0.000 120 32;
  • 10) 0.000 120 32 × 2 = 0 + 0.000 240 64;
  • 11) 0.000 240 64 × 2 = 0 + 0.000 481 28;
  • 12) 0.000 481 28 × 2 = 0 + 0.000 962 56;
  • 13) 0.000 962 56 × 2 = 0 + 0.001 925 12;
  • 14) 0.001 925 12 × 2 = 0 + 0.003 850 24;
  • 15) 0.003 850 24 × 2 = 0 + 0.007 700 48;
  • 16) 0.007 700 48 × 2 = 0 + 0.015 400 96;
  • 17) 0.015 400 96 × 2 = 0 + 0.030 801 92;
  • 18) 0.030 801 92 × 2 = 0 + 0.061 603 84;
  • 19) 0.061 603 84 × 2 = 0 + 0.123 207 68;
  • 20) 0.123 207 68 × 2 = 0 + 0.246 415 36;
  • 21) 0.246 415 36 × 2 = 0 + 0.492 830 72;
  • 22) 0.492 830 72 × 2 = 0 + 0.985 661 44;
  • 23) 0.985 661 44 × 2 = 1 + 0.971 322 88;
  • 24) 0.971 322 88 × 2 = 1 + 0.942 645 76;
  • 25) 0.942 645 76 × 2 = 1 + 0.885 291 52;
  • 26) 0.885 291 52 × 2 = 1 + 0.770 583 04;
  • 27) 0.770 583 04 × 2 = 1 + 0.541 166 08;
  • 28) 0.541 166 08 × 2 = 1 + 0.082 332 16;
  • 29) 0.082 332 16 × 2 = 0 + 0.164 664 32;
  • 30) 0.164 664 32 × 2 = 0 + 0.329 328 64;
  • 31) 0.329 328 64 × 2 = 0 + 0.658 657 28;
  • 32) 0.658 657 28 × 2 = 1 + 0.317 314 56;
  • 33) 0.317 314 56 × 2 = 0 + 0.634 629 12;
  • 34) 0.634 629 12 × 2 = 1 + 0.269 258 24;
  • 35) 0.269 258 24 × 2 = 0 + 0.538 516 48;
  • 36) 0.538 516 48 × 2 = 1 + 0.077 032 96;
  • 37) 0.077 032 96 × 2 = 0 + 0.154 065 92;
  • 38) 0.154 065 92 × 2 = 0 + 0.308 131 84;
  • 39) 0.308 131 84 × 2 = 0 + 0.616 263 68;
  • 40) 0.616 263 68 × 2 = 1 + 0.232 527 36;
  • 41) 0.232 527 36 × 2 = 0 + 0.465 054 72;
  • 42) 0.465 054 72 × 2 = 0 + 0.930 109 44;
  • 43) 0.930 109 44 × 2 = 1 + 0.860 218 88;
  • 44) 0.860 218 88 × 2 = 1 + 0.720 437 76;
  • 45) 0.720 437 76 × 2 = 1 + 0.440 875 52;
  • 46) 0.440 875 52 × 2 = 0 + 0.881 751 04;

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:


0.000 000 235(10) =

0.0000 0000 0000 0000 0000 0011 1111 0001 0101 0001 0011 10(2)

5. Positive number before normalization:

0.000 000 235(10) =

0.0000 0000 0000 0000 0000 0011 1111 0001 0101 0001 0011 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 235(10) =

0.0000 0000 0000 0000 0000 0011 1111 0001 0101 0001 0011 10(2) =

0.0000 0000 0000 0000 0000 0011 1111 0001 0101 0001 0011 10(2) × 20 =

1.1111 1000 1010 1000 1001 110(2) × 2-23


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): -23

Mantissa (not normalized):
1.1111 1000 1010 1000 1001 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 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) =

104(10) =

0110 1000(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. 111 1100 0101 0100 0100 1110 =

111 1100 0101 0100 0100 1110


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 1000

Mantissa (23 bits) =
111 1100 0101 0100 0100 1110


Decimal number 0.000 000 235 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 111 1100 0101 0100 0100 1110

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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