32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 50 361 649 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 50 361 649(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;
  • 50 361 649 ÷ 2 = 25 180 824 + 1;
  • 25 180 824 ÷ 2 = 12 590 412 + 0;
  • 12 590 412 ÷ 2 = 6 295 206 + 0;
  • 6 295 206 ÷ 2 = 3 147 603 + 0;
  • 3 147 603 ÷ 2 = 1 573 801 + 1;
  • 1 573 801 ÷ 2 = 786 900 + 1;
  • 786 900 ÷ 2 = 393 450 + 0;
  • 393 450 ÷ 2 = 196 725 + 0;
  • 196 725 ÷ 2 = 98 362 + 1;
  • 98 362 ÷ 2 = 49 181 + 0;
  • 49 181 ÷ 2 = 24 590 + 1;
  • 24 590 ÷ 2 = 12 295 + 0;
  • 12 295 ÷ 2 = 6 147 + 1;
  • 6 147 ÷ 2 = 3 073 + 1;
  • 3 073 ÷ 2 = 1 536 + 1;
  • 1 536 ÷ 2 = 768 + 0;
  • 768 ÷ 2 = 384 + 0;
  • 384 ÷ 2 = 192 + 0;
  • 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.


50 361 649(10) =


11 0000 0000 0111 0101 0011 0001(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the left, so that only one non zero digit remains to the left of it:


50 361 649(10) =


11 0000 0000 0111 0101 0011 0001(2) =


11 0000 0000 0111 0101 0011 0001(2) × 20 =


1.1000 0000 0011 1010 1001 1000 1(2) × 225


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): 25


Mantissa (not normalized):
1.1000 0000 0011 1010 1001 1000 1


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


25 + 2(8-1) - 1 =


(25 + 127)(10) =


152(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;
  • 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;

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) =


152(10) =


1001 1000(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 0001 1101 0100 1100 01 =


100 0000 0001 1101 0100 1100


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 1000


Mantissa (23 bits) =
100 0000 0001 1101 0100 1100


The base ten decimal number 50 361 649 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 1000 - 100 0000 0001 1101 0100 1100

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