32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 1 010 101 100 110 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 1 010 101 100 110(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;
  • 1 010 101 100 110 ÷ 2 = 505 050 550 055 + 0;
  • 505 050 550 055 ÷ 2 = 252 525 275 027 + 1;
  • 252 525 275 027 ÷ 2 = 126 262 637 513 + 1;
  • 126 262 637 513 ÷ 2 = 63 131 318 756 + 1;
  • 63 131 318 756 ÷ 2 = 31 565 659 378 + 0;
  • 31 565 659 378 ÷ 2 = 15 782 829 689 + 0;
  • 15 782 829 689 ÷ 2 = 7 891 414 844 + 1;
  • 7 891 414 844 ÷ 2 = 3 945 707 422 + 0;
  • 3 945 707 422 ÷ 2 = 1 972 853 711 + 0;
  • 1 972 853 711 ÷ 2 = 986 426 855 + 1;
  • 986 426 855 ÷ 2 = 493 213 427 + 1;
  • 493 213 427 ÷ 2 = 246 606 713 + 1;
  • 246 606 713 ÷ 2 = 123 303 356 + 1;
  • 123 303 356 ÷ 2 = 61 651 678 + 0;
  • 61 651 678 ÷ 2 = 30 825 839 + 0;
  • 30 825 839 ÷ 2 = 15 412 919 + 1;
  • 15 412 919 ÷ 2 = 7 706 459 + 1;
  • 7 706 459 ÷ 2 = 3 853 229 + 1;
  • 3 853 229 ÷ 2 = 1 926 614 + 1;
  • 1 926 614 ÷ 2 = 963 307 + 0;
  • 963 307 ÷ 2 = 481 653 + 1;
  • 481 653 ÷ 2 = 240 826 + 1;
  • 240 826 ÷ 2 = 120 413 + 0;
  • 120 413 ÷ 2 = 60 206 + 1;
  • 60 206 ÷ 2 = 30 103 + 0;
  • 30 103 ÷ 2 = 15 051 + 1;
  • 15 051 ÷ 2 = 7 525 + 1;
  • 7 525 ÷ 2 = 3 762 + 1;
  • 3 762 ÷ 2 = 1 881 + 0;
  • 1 881 ÷ 2 = 940 + 1;
  • 940 ÷ 2 = 470 + 0;
  • 470 ÷ 2 = 235 + 0;
  • 235 ÷ 2 = 117 + 1;
  • 117 ÷ 2 = 58 + 1;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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.


1 010 101 100 110(10) =


1110 1011 0010 1110 1011 0111 1001 1110 0100 1110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 39 positions to the left, so that only one non zero digit remains to the left of it:


1 010 101 100 110(10) =


1110 1011 0010 1110 1011 0111 1001 1110 0100 1110(2) =


1110 1011 0010 1110 1011 0111 1001 1110 0100 1110(2) × 20 =


1.1101 0110 0101 1101 0110 1111 0011 1100 1001 110(2) × 239


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): 39


Mantissa (not normalized):
1.1101 0110 0101 1101 0110 1111 0011 1100 1001 110


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


39 + 2(8-1) - 1 =


(39 + 127)(10) =


166(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;
  • 166 ÷ 2 = 83 + 0;
  • 83 ÷ 2 = 41 + 1;
  • 41 ÷ 2 = 20 + 1;
  • 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) =


166(10) =


1010 0110(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. 110 1011 0010 1110 1011 0111 1001 1110 0100 1110 =


110 1011 0010 1110 1011 0111


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 0110


Mantissa (23 bits) =
110 1011 0010 1110 1011 0111


The base ten decimal number 1 010 101 100 110 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1010 0110 - 110 1011 0010 1110 1011 0111

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation