# 32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 9 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

## Number 9_{(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**; - 9 ÷ 2 = 4 +
**1**; - 4 ÷ 2 = 2 +
**0**; - 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.

#### 9_{(10) } =

#### 1001_{(2) }

### 3. 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:

#### 9_{(10) }=

#### 1001_{(2) }=

#### 1001_{(2) } × 2^{0 }=

#### 1.001_{(2) } × 2^{3 }

### 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): 3

#### Mantissa (not normalized):

1.001

### 5. 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) }

### 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**; - 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**;

### 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) =

#### 130_{(10) } =

#### 1000 0010_{(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 adding the necessary number of zeros to the right.

#### Mantissa (normalized) =

#### 1. 001 0000 0000 0000 0000 0000 =

#### 001 0000 0000 0000 0000 0000

### 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) =

1000 0010

#### Mantissa (23 bits) =

001 0000 0000 0000 0000 0000

## The base ten decimal number 9 converted and written in 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0010 - 001 0000 0000 0000 0000 0000

### More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation: