32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 1 246 799 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 1 246 799(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;
  • 1 246 799 ÷ 2 = 623 399 + 1;
  • 623 399 ÷ 2 = 311 699 + 1;
  • 311 699 ÷ 2 = 155 849 + 1;
  • 155 849 ÷ 2 = 77 924 + 1;
  • 77 924 ÷ 2 = 38 962 + 0;
  • 38 962 ÷ 2 = 19 481 + 0;
  • 19 481 ÷ 2 = 9 740 + 1;
  • 9 740 ÷ 2 = 4 870 + 0;
  • 4 870 ÷ 2 = 2 435 + 0;
  • 2 435 ÷ 2 = 1 217 + 1;
  • 1 217 ÷ 2 = 608 + 1;
  • 608 ÷ 2 = 304 + 0;
  • 304 ÷ 2 = 152 + 0;
  • 152 ÷ 2 = 76 + 0;
  • 76 ÷ 2 = 38 + 0;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.


1 246 799(10) =


1 0011 0000 0110 0100 1111(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the left, so that only one non zero digit remains to the left of it:


1 246 799(10) =


1 0011 0000 0110 0100 1111(2) =


1 0011 0000 0110 0100 1111(2) × 20 =


1.0011 0000 0110 0100 1111(2) × 220


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): 20


Mantissa (not normalized):
1.0011 0000 0110 0100 1111


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


20 + 2(8-1) - 1 =


(20 + 127)(10) =


147(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;
  • 147 ÷ 2 = 73 + 1;
  • 73 ÷ 2 = 36 + 1;
  • 36 ÷ 2 = 18 + 0;
  • 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) =


147(10) =


1001 0011(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. 0011 0000 0110 0100 1111 000 =


001 1000 0011 0010 0111 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 0011


Mantissa (23 bits) =
001 1000 0011 0010 0111 1000


The base ten decimal number 1 246 799 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 0011 - 001 1000 0011 0010 0111 1000

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