32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 4 141 974 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 4 141 974(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;
  • 4 141 974 ÷ 2 = 2 070 987 + 0;
  • 2 070 987 ÷ 2 = 1 035 493 + 1;
  • 1 035 493 ÷ 2 = 517 746 + 1;
  • 517 746 ÷ 2 = 258 873 + 0;
  • 258 873 ÷ 2 = 129 436 + 1;
  • 129 436 ÷ 2 = 64 718 + 0;
  • 64 718 ÷ 2 = 32 359 + 0;
  • 32 359 ÷ 2 = 16 179 + 1;
  • 16 179 ÷ 2 = 8 089 + 1;
  • 8 089 ÷ 2 = 4 044 + 1;
  • 4 044 ÷ 2 = 2 022 + 0;
  • 2 022 ÷ 2 = 1 011 + 0;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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.


4 141 974(10) =


11 1111 0011 0011 1001 0110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the left, so that only one non zero digit remains to the left of it:


4 141 974(10) =


11 1111 0011 0011 1001 0110(2) =


11 1111 0011 0011 1001 0110(2) × 20 =


1.1111 1001 1001 1100 1011 0(2) × 221


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): 21


Mantissa (not normalized):
1.1111 1001 1001 1100 1011 0


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


21 + 2(8-1) - 1 =


(21 + 127)(10) =


148(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;
  • 148 ÷ 2 = 74 + 0;
  • 74 ÷ 2 = 37 + 0;
  • 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) =


148(10) =


1001 0100(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 adding the necessary number of zeros to the right.


Mantissa (normalized) =


1. 1 1111 0011 0011 1001 0110 00 =


111 1100 1100 1110 0101 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 0100


Mantissa (23 bits) =
111 1100 1100 1110 0101 1000


The base ten decimal number 4 141 974 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 0100 - 111 1100 1100 1110 0101 1000

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