32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 101 011 105 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 101 011 105(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;
  • 101 011 105 ÷ 2 = 50 505 552 + 1;
  • 50 505 552 ÷ 2 = 25 252 776 + 0;
  • 25 252 776 ÷ 2 = 12 626 388 + 0;
  • 12 626 388 ÷ 2 = 6 313 194 + 0;
  • 6 313 194 ÷ 2 = 3 156 597 + 0;
  • 3 156 597 ÷ 2 = 1 578 298 + 1;
  • 1 578 298 ÷ 2 = 789 149 + 0;
  • 789 149 ÷ 2 = 394 574 + 1;
  • 394 574 ÷ 2 = 197 287 + 0;
  • 197 287 ÷ 2 = 98 643 + 1;
  • 98 643 ÷ 2 = 49 321 + 1;
  • 49 321 ÷ 2 = 24 660 + 1;
  • 24 660 ÷ 2 = 12 330 + 0;
  • 12 330 ÷ 2 = 6 165 + 0;
  • 6 165 ÷ 2 = 3 082 + 1;
  • 3 082 ÷ 2 = 1 541 + 0;
  • 1 541 ÷ 2 = 770 + 1;
  • 770 ÷ 2 = 385 + 0;
  • 385 ÷ 2 = 192 + 1;
  • 192 ÷ 2 = 96 + 0;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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.


101 011 105(10) =


110 0000 0101 0100 1110 1010 0001(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the left, so that only one non zero digit remains to the left of it:


101 011 105(10) =


110 0000 0101 0100 1110 1010 0001(2) =


110 0000 0101 0100 1110 1010 0001(2) × 20 =


1.1000 0001 0101 0011 1010 1000 01(2) × 226


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): 26


Mantissa (not normalized):
1.1000 0001 0101 0011 1010 1000 01


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


26 + 2(8-1) - 1 =


(26 + 127)(10) =


153(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;
  • 153 ÷ 2 = 76 + 1;
  • 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;

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) =


153(10) =


1001 1001(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 removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 100 0000 1010 1001 1101 0100 001 =


100 0000 1010 1001 1101 0100


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 1001


Mantissa (23 bits) =
100 0000 1010 1001 1101 0100


The base ten decimal number 101 011 105 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 1001 - 100 0000 1010 1001 1101 0100

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