32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -11 491 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -11 491(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. Start with the positive version of the number:

|-11 491| = 11 491

2. 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;
  • 11 491 ÷ 2 = 5 745 + 1;
  • 5 745 ÷ 2 = 2 872 + 1;
  • 2 872 ÷ 2 = 1 436 + 0;
  • 1 436 ÷ 2 = 718 + 0;
  • 718 ÷ 2 = 359 + 0;
  • 359 ÷ 2 = 179 + 1;
  • 179 ÷ 2 = 89 + 1;
  • 89 ÷ 2 = 44 + 1;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


11 491(10) =


10 1100 1110 0011(2)


4. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the left, so that only one non zero digit remains to the left of it:


11 491(10) =


10 1100 1110 0011(2) =


10 1100 1110 0011(2) × 20 =


1.0110 0111 0001 1(2) × 213


5. 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 1 (a negative number)


Exponent (unadjusted): 13


Mantissa (not normalized):
1.0110 0111 0001 1


6. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


13 + 2(8-1) - 1 =


(13 + 127)(10) =


140(10)


7. 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;
  • 140 ÷ 2 = 70 + 0;
  • 70 ÷ 2 = 35 + 0;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


140(10) =


1000 1100(2)


9. 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. 0 1100 1110 0011 00 0000 0000 =


011 0011 1000 1100 0000 0000


10. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (8 bits) =
1000 1100


Mantissa (23 bits) =
011 0011 1000 1100 0000 0000


The base ten decimal number -11 491 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1000 1100 - 011 0011 1000 1100 0000 0000

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