32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 27 645 233 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 27 645 233(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. 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;
  • 27 645 233 ÷ 2 = 13 822 616 + 1;
  • 13 822 616 ÷ 2 = 6 911 308 + 0;
  • 6 911 308 ÷ 2 = 3 455 654 + 0;
  • 3 455 654 ÷ 2 = 1 727 827 + 0;
  • 1 727 827 ÷ 2 = 863 913 + 1;
  • 863 913 ÷ 2 = 431 956 + 1;
  • 431 956 ÷ 2 = 215 978 + 0;
  • 215 978 ÷ 2 = 107 989 + 0;
  • 107 989 ÷ 2 = 53 994 + 1;
  • 53 994 ÷ 2 = 26 997 + 0;
  • 26 997 ÷ 2 = 13 498 + 1;
  • 13 498 ÷ 2 = 6 749 + 0;
  • 6 749 ÷ 2 = 3 374 + 1;
  • 3 374 ÷ 2 = 1 687 + 0;
  • 1 687 ÷ 2 = 843 + 1;
  • 843 ÷ 2 = 421 + 1;
  • 421 ÷ 2 = 210 + 1;
  • 210 ÷ 2 = 105 + 0;
  • 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;

2. Construct the base 2 representation of the positive number.

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


27 645 233(10) =


1 1010 0101 1101 0101 0011 0001(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the left, so that only one non zero digit remains to the left of it:


27 645 233(10) =


1 1010 0101 1101 0101 0011 0001(2) =


1 1010 0101 1101 0101 0011 0001(2) × 20 =


1.1010 0101 1101 0101 0011 0001(2) × 224


4. 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): 24


Mantissa (not normalized):
1.1010 0101 1101 0101 0011 0001


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


24 + 2(8-1) - 1 =


(24 + 127)(10) =


151(10)


6. 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;
  • 151 ÷ 2 = 75 + 1;
  • 75 ÷ 2 = 37 + 1;
  • 37 ÷ 2 = 18 + 1;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


151(10) =


1001 0111(2)


8. 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 0010 1110 1010 1001 1000 1 =


101 0010 1110 1010 1001 1000


9. 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) =
1001 0111


Mantissa (23 bits) =
101 0010 1110 1010 1001 1000


The base ten decimal number 27 645 233 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 0111 - 101 0010 1110 1010 1001 1000

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