32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 167 899 528 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 167 899 528(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;
  • 167 899 528 ÷ 2 = 83 949 764 + 0;
  • 83 949 764 ÷ 2 = 41 974 882 + 0;
  • 41 974 882 ÷ 2 = 20 987 441 + 0;
  • 20 987 441 ÷ 2 = 10 493 720 + 1;
  • 10 493 720 ÷ 2 = 5 246 860 + 0;
  • 5 246 860 ÷ 2 = 2 623 430 + 0;
  • 2 623 430 ÷ 2 = 1 311 715 + 0;
  • 1 311 715 ÷ 2 = 655 857 + 1;
  • 655 857 ÷ 2 = 327 928 + 1;
  • 327 928 ÷ 2 = 163 964 + 0;
  • 163 964 ÷ 2 = 81 982 + 0;
  • 81 982 ÷ 2 = 40 991 + 0;
  • 40 991 ÷ 2 = 20 495 + 1;
  • 20 495 ÷ 2 = 10 247 + 1;
  • 10 247 ÷ 2 = 5 123 + 1;
  • 5 123 ÷ 2 = 2 561 + 1;
  • 2 561 ÷ 2 = 1 280 + 1;
  • 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.


167 899 528(10) =


1010 0000 0001 1111 0001 1000 1000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the left, so that only one non zero digit remains to the left of it:


167 899 528(10) =


1010 0000 0001 1111 0001 1000 1000(2) =


1010 0000 0001 1111 0001 1000 1000(2) × 20 =


1.0100 0000 0011 1110 0011 0001 000(2) × 227


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): 27


Mantissa (not normalized):
1.0100 0000 0011 1110 0011 0001 000


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


27 + 2(8-1) - 1 =


(27 + 127)(10) =


154(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;
  • 154 ÷ 2 = 77 + 0;
  • 77 ÷ 2 = 38 + 1;
  • 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) =


154(10) =


1001 1010(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. 010 0000 0001 1111 0001 1000 1000 =


010 0000 0001 1111 0001 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 1010


Mantissa (23 bits) =
010 0000 0001 1111 0001 1000


The base ten decimal number 167 899 528 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 1010 - 010 0000 0001 1111 0001 1000

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