32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 12.840 805 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 12.840 805(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. First, convert to binary (in base 2) the integer part: 12.
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;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


12(10) =


1100(2)


3. Convert to binary (base 2) the fractional part: 0.840 805.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.840 805 × 2 = 1 + 0.681 61;
  • 2) 0.681 61 × 2 = 1 + 0.363 22;
  • 3) 0.363 22 × 2 = 0 + 0.726 44;
  • 4) 0.726 44 × 2 = 1 + 0.452 88;
  • 5) 0.452 88 × 2 = 0 + 0.905 76;
  • 6) 0.905 76 × 2 = 1 + 0.811 52;
  • 7) 0.811 52 × 2 = 1 + 0.623 04;
  • 8) 0.623 04 × 2 = 1 + 0.246 08;
  • 9) 0.246 08 × 2 = 0 + 0.492 16;
  • 10) 0.492 16 × 2 = 0 + 0.984 32;
  • 11) 0.984 32 × 2 = 1 + 0.968 64;
  • 12) 0.968 64 × 2 = 1 + 0.937 28;
  • 13) 0.937 28 × 2 = 1 + 0.874 56;
  • 14) 0.874 56 × 2 = 1 + 0.749 12;
  • 15) 0.749 12 × 2 = 1 + 0.498 24;
  • 16) 0.498 24 × 2 = 0 + 0.996 48;
  • 17) 0.996 48 × 2 = 1 + 0.992 96;
  • 18) 0.992 96 × 2 = 1 + 0.985 92;
  • 19) 0.985 92 × 2 = 1 + 0.971 84;
  • 20) 0.971 84 × 2 = 1 + 0.943 68;
  • 21) 0.943 68 × 2 = 1 + 0.887 36;
  • 22) 0.887 36 × 2 = 1 + 0.774 72;
  • 23) 0.774 72 × 2 = 1 + 0.549 44;
  • 24) 0.549 44 × 2 = 1 + 0.098 88;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.840 805(10) =


0.1101 0111 0011 1110 1111 1111(2)


5. Positive number before normalization:

12.840 805(10) =


1100.1101 0111 0011 1110 1111 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


12.840 805(10) =


1100.1101 0111 0011 1110 1111 1111(2) =


1100.1101 0111 0011 1110 1111 1111(2) × 20 =


1.1001 1010 1110 0111 1101 1111 111(2) × 23


7. 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): 3


Mantissa (not normalized):
1.1001 1010 1110 0111 1101 1111 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


3 + 2(8-1) - 1 =


(3 + 127)(10) =


130(10)


9. 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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


130(10) =


1000 0010(2)


11. 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 1101 0111 0011 1110 1111 1111 =


100 1101 0111 0011 1110 1111


12. 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) =
1000 0010


Mantissa (23 bits) =
100 1101 0111 0011 1110 1111


The base ten decimal number 12.840 805 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0010 - 100 1101 0111 0011 1110 1111

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