32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 163 840 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 163 840(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;
  • 163 840 ÷ 2 = 81 920 + 0;
  • 81 920 ÷ 2 = 40 960 + 0;
  • 40 960 ÷ 2 = 20 480 + 0;
  • 20 480 ÷ 2 = 10 240 + 0;
  • 10 240 ÷ 2 = 5 120 + 0;
  • 5 120 ÷ 2 = 2 560 + 0;
  • 2 560 ÷ 2 = 1 280 + 0;
  • 1 280 ÷ 2 = 640 + 0;
  • 640 ÷ 2 = 320 + 0;
  • 320 ÷ 2 = 160 + 0;
  • 160 ÷ 2 = 80 + 0;
  • 80 ÷ 2 = 40 + 0;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 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.


163 840(10) =


10 1000 0000 0000 0000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the left, so that only one non zero digit remains to the left of it:


163 840(10) =


10 1000 0000 0000 0000(2) =


10 1000 0000 0000 0000(2) × 20 =


1.0100 0000 0000 0000 0(2) × 217


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): 17


Mantissa (not normalized):
1.0100 0000 0000 0000 0


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


17 + 2(8-1) - 1 =


(17 + 127)(10) =


144(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;
  • 144 ÷ 2 = 72 + 0;
  • 72 ÷ 2 = 36 + 0;
  • 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) =


144(10) =


1001 0000(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. 0 1000 0000 0000 0000 00 0000 =


010 0000 0000 0000 0000 0000


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 0000


Mantissa (23 bits) =
010 0000 0000 0000 0000 0000


The base ten decimal number 163 840 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 0000 - 010 0000 0000 0000 0000 0000

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