104.433 333 333 333 333 333 333 361 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 104.433 333 333 333 333 333 333 361 1(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
104.433 333 333 333 333 333 333 361 1(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: 104.
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;
  • 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;

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.

104(10) =

110 1000(2)


3. Convert to binary (base 2) the fractional part: 0.433 333 333 333 333 333 333 361 1.

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) 0.433 333 333 333 333 333 333 361 1 × 2 = 0 + 0.866 666 666 666 666 666 666 722 2;
  • 2) 0.866 666 666 666 666 666 666 722 2 × 2 = 1 + 0.733 333 333 333 333 333 333 444 4;
  • 3) 0.733 333 333 333 333 333 333 444 4 × 2 = 1 + 0.466 666 666 666 666 666 666 888 8;
  • 4) 0.466 666 666 666 666 666 666 888 8 × 2 = 0 + 0.933 333 333 333 333 333 333 777 6;
  • 5) 0.933 333 333 333 333 333 333 777 6 × 2 = 1 + 0.866 666 666 666 666 666 667 555 2;
  • 6) 0.866 666 666 666 666 666 667 555 2 × 2 = 1 + 0.733 333 333 333 333 333 335 110 4;
  • 7) 0.733 333 333 333 333 333 335 110 4 × 2 = 1 + 0.466 666 666 666 666 666 670 220 8;
  • 8) 0.466 666 666 666 666 666 670 220 8 × 2 = 0 + 0.933 333 333 333 333 333 340 441 6;
  • 9) 0.933 333 333 333 333 333 340 441 6 × 2 = 1 + 0.866 666 666 666 666 666 680 883 2;
  • 10) 0.866 666 666 666 666 666 680 883 2 × 2 = 1 + 0.733 333 333 333 333 333 361 766 4;
  • 11) 0.733 333 333 333 333 333 361 766 4 × 2 = 1 + 0.466 666 666 666 666 666 723 532 8;
  • 12) 0.466 666 666 666 666 666 723 532 8 × 2 = 0 + 0.933 333 333 333 333 333 447 065 6;
  • 13) 0.933 333 333 333 333 333 447 065 6 × 2 = 1 + 0.866 666 666 666 666 666 894 131 2;
  • 14) 0.866 666 666 666 666 666 894 131 2 × 2 = 1 + 0.733 333 333 333 333 333 788 262 4;
  • 15) 0.733 333 333 333 333 333 788 262 4 × 2 = 1 + 0.466 666 666 666 666 667 576 524 8;
  • 16) 0.466 666 666 666 666 667 576 524 8 × 2 = 0 + 0.933 333 333 333 333 335 153 049 6;
  • 17) 0.933 333 333 333 333 335 153 049 6 × 2 = 1 + 0.866 666 666 666 666 670 306 099 2;
  • 18) 0.866 666 666 666 666 670 306 099 2 × 2 = 1 + 0.733 333 333 333 333 340 612 198 4;
  • 19) 0.733 333 333 333 333 340 612 198 4 × 2 = 1 + 0.466 666 666 666 666 681 224 396 8;
  • 20) 0.466 666 666 666 666 681 224 396 8 × 2 = 0 + 0.933 333 333 333 333 362 448 793 6;
  • 21) 0.933 333 333 333 333 362 448 793 6 × 2 = 1 + 0.866 666 666 666 666 724 897 587 2;
  • 22) 0.866 666 666 666 666 724 897 587 2 × 2 = 1 + 0.733 333 333 333 333 449 795 174 4;
  • 23) 0.733 333 333 333 333 449 795 174 4 × 2 = 1 + 0.466 666 666 666 666 899 590 348 8;
  • 24) 0.466 666 666 666 666 899 590 348 8 × 2 = 0 + 0.933 333 333 333 333 799 180 697 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:


0.433 333 333 333 333 333 333 361 1(10) =

0.0110 1110 1110 1110 1110 1110(2)

5. Positive number before normalization:

104.433 333 333 333 333 333 333 361 1(10) =

110 1000.0110 1110 1110 1110 1110 1110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


104.433 333 333 333 333 333 333 361 1(10) =

110 1000.0110 1110 1110 1110 1110 1110(2) =

110 1000.0110 1110 1110 1110 1110 1110(2) × 20 =

1.1010 0001 1011 1011 1011 1011 1011 10(2) × 26


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): 6

Mantissa (not normalized):
1.1010 0001 1011 1011 1011 1011 1011 10


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

6 + 2(8-1) - 1 =

(6 + 127)(10) =

133(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;
  • 133 ÷ 2 = 66 + 1;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

133(10) =

1000 0101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 101 0000 1101 1101 1101 1101 110 1110 =

101 0000 1101 1101 1101 1101


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) =
1000 0101

Mantissa (23 bits) =
101 0000 1101 1101 1101 1101


Decimal number 104.433 333 333 333 333 333 333 361 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0101 - 101 0000 1101 1101 1101 1101

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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