32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 16 777 215 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 16 777 215(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;
  • 16 777 215 ÷ 2 = 8 388 607 + 1;
  • 8 388 607 ÷ 2 = 4 194 303 + 1;
  • 4 194 303 ÷ 2 = 2 097 151 + 1;
  • 2 097 151 ÷ 2 = 1 048 575 + 1;
  • 1 048 575 ÷ 2 = 524 287 + 1;
  • 524 287 ÷ 2 = 262 143 + 1;
  • 262 143 ÷ 2 = 131 071 + 1;
  • 131 071 ÷ 2 = 65 535 + 1;
  • 65 535 ÷ 2 = 32 767 + 1;
  • 32 767 ÷ 2 = 16 383 + 1;
  • 16 383 ÷ 2 = 8 191 + 1;
  • 8 191 ÷ 2 = 4 095 + 1;
  • 4 095 ÷ 2 = 2 047 + 1;
  • 2 047 ÷ 2 = 1 023 + 1;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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.


16 777 215(10) =


1111 1111 1111 1111 1111 1111(2)


3. Normalize the binary representation of the number.

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


16 777 215(10) =


1111 1111 1111 1111 1111 1111(2) =


1111 1111 1111 1111 1111 1111(2) × 20 =


1.1111 1111 1111 1111 1111 111(2) × 223


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): 23


Mantissa (not normalized):
1.1111 1111 1111 1111 1111 111


5. 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) =


150(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;
  • 150 ÷ 2 = 75 + 0;
  • 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) =


150(10) =


1001 0110(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, only if necessary (not the case here).


Mantissa (normalized) =


1. 111 1111 1111 1111 1111 1111 =


111 1111 1111 1111 1111 1111


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 0110


Mantissa (23 bits) =
111 1111 1111 1111 1111 1111


The base ten decimal number 16 777 215 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 0110 - 111 1111 1111 1111 1111 1111

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation