32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 11 372 246 189 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 11 372 246 189(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;
  • 11 372 246 189 ÷ 2 = 5 686 123 094 + 1;
  • 5 686 123 094 ÷ 2 = 2 843 061 547 + 0;
  • 2 843 061 547 ÷ 2 = 1 421 530 773 + 1;
  • 1 421 530 773 ÷ 2 = 710 765 386 + 1;
  • 710 765 386 ÷ 2 = 355 382 693 + 0;
  • 355 382 693 ÷ 2 = 177 691 346 + 1;
  • 177 691 346 ÷ 2 = 88 845 673 + 0;
  • 88 845 673 ÷ 2 = 44 422 836 + 1;
  • 44 422 836 ÷ 2 = 22 211 418 + 0;
  • 22 211 418 ÷ 2 = 11 105 709 + 0;
  • 11 105 709 ÷ 2 = 5 552 854 + 1;
  • 5 552 854 ÷ 2 = 2 776 427 + 0;
  • 2 776 427 ÷ 2 = 1 388 213 + 1;
  • 1 388 213 ÷ 2 = 694 106 + 1;
  • 694 106 ÷ 2 = 347 053 + 0;
  • 347 053 ÷ 2 = 173 526 + 1;
  • 173 526 ÷ 2 = 86 763 + 0;
  • 86 763 ÷ 2 = 43 381 + 1;
  • 43 381 ÷ 2 = 21 690 + 1;
  • 21 690 ÷ 2 = 10 845 + 0;
  • 10 845 ÷ 2 = 5 422 + 1;
  • 5 422 ÷ 2 = 2 711 + 0;
  • 2 711 ÷ 2 = 1 355 + 1;
  • 1 355 ÷ 2 = 677 + 1;
  • 677 ÷ 2 = 338 + 1;
  • 338 ÷ 2 = 169 + 0;
  • 169 ÷ 2 = 84 + 1;
  • 84 ÷ 2 = 42 + 0;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 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.


11 372 246 189(10) =


10 1010 0101 1101 0110 1011 0100 1010 1101(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 33 positions to the left, so that only one non zero digit remains to the left of it:


11 372 246 189(10) =


10 1010 0101 1101 0110 1011 0100 1010 1101(2) =


10 1010 0101 1101 0110 1011 0100 1010 1101(2) × 20 =


1.0101 0010 1110 1011 0101 1010 0101 0110 1(2) × 233


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): 33


Mantissa (not normalized):
1.0101 0010 1110 1011 0101 1010 0101 0110 1


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


33 + 2(8-1) - 1 =


(33 + 127)(10) =


160(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;
  • 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;

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) =


160(10) =


1010 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 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 1001 0111 0101 1010 1101 00 1010 1101 =


010 1001 0111 0101 1010 1101


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) =
1010 0000


Mantissa (23 bits) =
010 1001 0111 0101 1010 1101


The base ten decimal number 11 372 246 189 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1010 0000 - 010 1001 0111 0101 1010 1101

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