1 = negative, 0 = positive.

0

0101 1111

100 1000 0000 0000 0000 0001

100 1000 0000 0000 0000 0001_{(2)} =

#### 1 × 2^{-1} + 0 × 2^{-2} + 0 × 2^{-3} + 1 × 2^{-4} + 0 × 2^{-5} + 0 × 2^{-6} + 0 × 2^{-7} + 0 × 2^{-8} + 0 × 2^{-9} + 0 × 2^{-10} + 0 × 2^{-11} + 0 × 2^{-12} + 0 × 2^{-13} + 0 × 2^{-14} + 0 × 2^{-15} + 0 × 2^{-16} + 0 × 2^{-17} + 0 × 2^{-18} + 0 × 2^{-19} + 0 × 2^{-20} + 0 × 2^{-21} + 0 × 2^{-22} + 1 × 2^{-23} =

#### 0.5 + 0 + 0 + 0.062 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 119 209 289 550 781 25 =

#### 0.5 + 0.062 5 + 0.000 000 119 209 289 550 781 25 =

#### 0.562 500 119 209 289 550 781 25_{(10)}

- 1. Identify the three elements that make up the binary representation of the number:

First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.

The next 8 bits contain the exponent.

The last 23 bits contain the mantissa. - 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10).
- 3. Adjust the exponent, subtract the excess bits, 2
^{(8 - 1)}- 1 = 127, that is due to the 8 bit excess/bias notation. - 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10).
- 5. Put all the numbers into expression to calculate the single precision floating point decimal value:

(-1)^{Sign}× (1 + Mantissa) × 2^{(Exponent adjusted)}

- 1. Identify the elements that make up the binary representation of the number:

First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.

The next 8 bits contain the exponent: 1000 0001

The last 23 bits contain the mantissa: 100 0001 0000 0010 0000 0000 - 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10): 1000 0001
_{(2)}=

1 × 2^{7}+ 0 × 2^{6}+ 0 × 2^{5}+ 0 × 2^{4}+ 0 × 2^{3}+ 0 × 2^{2}+ 0 × 2^{1}+ 1 × 2^{0}=

128 + 0 + 0 + 0 + 0 + 0 + 0 + 1 =

128 + 1 =

129_{(10)} - 3. Adjust the exponent, subtract the excess bits, 2
^{(8 - 1)}- 1 = 127, that is due to the 8 bit excess/bias notation:

Exponent adjusted = 129 - 127 = 2 - 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10): 100 0001 0000 0010 0000 0000
_{(2)}=

1 × 2^{-1}+ 0 × 2^{-2}+ 0 × 2^{-3}+ 0 × 2^{-4}+ 0 × 2^{-5}+ 0 × 2^{-6}+ 1 × 2^{-7}+ 0 × 2^{-8}+ 0 × 2^{-9}+ 0 × 2^{-10}+ 0 × 2^{-11}+ 0 × 2^{-12}+ 0 × 2^{-13}+ 1 × 2^{-14}+ 0 × 2^{-15}+ 0 × 2^{-16}+ 0 × 2^{-17}+ 0 × 2^{-18}+ 0 × 2^{-19}+ 0 × 2^{-20}+ 0 × 2^{-21}+ 0 × 2^{-22}+ 0 × 2^{-23}=

0.5 + 0 + 0 + 0 + 0 + 0 + 0.007 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =

0.5 + 0.007 812 5 + 0.000 061 035 156 25 =

0.507 873 535 156 25_{(10)} - 5. Put all the numbers into expression to calculate the single precision floating point decimal value: (-1)
^{Sign}× (1 + Mantissa) × 2^{(Exponent adjusted)}=

(-1)^{1}× (1 + 0.507 873 535 156 25) × 2^{2}=

-1.507 873 535 156 25 × 2^{2}=

-6.031 494 140 625 #### 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 converted from 32 bit single precision IEEE 754 binary floating point representation to decimal number (float) in decimal system (in base 10) = -6.031 494 140 625

_{(10)}

### 1.1. Unsigned integer -> Unsigned binary

### 1.2. Signed integer -> Signed binary

### 1.3. Signed integer -> Signed binary one's complement

### 1.4. Signed integer -> Signed binary two's complement

### 2.1. Decimal -> 32bit single precision IEEE 754 binary floating point

### 2.2. Decimal -> 64bit double precision IEEE 754 binary floating point

### 3.1. Unsigned binary -> Unsigned integer

### 3.2. Signed binary -> Signed integer

### 3.3. Signed binary one's complement -> Signed integer

### 3.4. Signed binary two's complement -> Signed integer