32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 88.852 31 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 88.852 31(10) converted and written in 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: 88.
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;
  • 88 ÷ 2 = 44 + 0;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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.


88(10) =


101 1000(2)


3. Convert to binary (base 2) the fractional part: 0.852 31.

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) 0.852 31 × 2 = 1 + 0.704 62;
  • 2) 0.704 62 × 2 = 1 + 0.409 24;
  • 3) 0.409 24 × 2 = 0 + 0.818 48;
  • 4) 0.818 48 × 2 = 1 + 0.636 96;
  • 5) 0.636 96 × 2 = 1 + 0.273 92;
  • 6) 0.273 92 × 2 = 0 + 0.547 84;
  • 7) 0.547 84 × 2 = 1 + 0.095 68;
  • 8) 0.095 68 × 2 = 0 + 0.191 36;
  • 9) 0.191 36 × 2 = 0 + 0.382 72;
  • 10) 0.382 72 × 2 = 0 + 0.765 44;
  • 11) 0.765 44 × 2 = 1 + 0.530 88;
  • 12) 0.530 88 × 2 = 1 + 0.061 76;
  • 13) 0.061 76 × 2 = 0 + 0.123 52;
  • 14) 0.123 52 × 2 = 0 + 0.247 04;
  • 15) 0.247 04 × 2 = 0 + 0.494 08;
  • 16) 0.494 08 × 2 = 0 + 0.988 16;
  • 17) 0.988 16 × 2 = 1 + 0.976 32;
  • 18) 0.976 32 × 2 = 1 + 0.952 64;
  • 19) 0.952 64 × 2 = 1 + 0.905 28;
  • 20) 0.905 28 × 2 = 1 + 0.810 56;
  • 21) 0.810 56 × 2 = 1 + 0.621 12;
  • 22) 0.621 12 × 2 = 1 + 0.242 24;
  • 23) 0.242 24 × 2 = 0 + 0.484 48;
  • 24) 0.484 48 × 2 = 0 + 0.968 96;

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...)


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:


0.852 31(10) =


0.1101 1010 0011 0000 1111 1100(2)


5. Positive number before normalization:

88.852 31(10) =


101 1000.1101 1010 0011 0000 1111 1100(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:


88.852 31(10) =


101 1000.1101 1010 0011 0000 1111 1100(2) =


101 1000.1101 1010 0011 0000 1111 1100(2) × 20 =


1.0110 0011 0110 1000 1100 0011 1111 00(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):
1.0110 0011 0110 1000 1100 0011 1111 00


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. 011 0001 1011 0100 0110 0001 111 1100 =


011 0001 1011 0100 0110 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) =
1000 0101


Mantissa (23 bits) =
011 0001 1011 0100 0110 0001


The base ten decimal number 88.852 31 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0101 - 011 0001 1011 0100 0110 0001

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