Determine the hexadecimal IEEE FPS pattern that represents the decimal value 21.375.

To convert the decimal value **21.375** into its IEEE 754 single-precision floating-point (32-bit) representation, we follow these steps:

### Step 1: Convert 21.375 to binary form
First, split the decimal number into its integer and fractional parts.

- The integer part is **21**.
- The fractional part is **0.375**.

#### 1.1 Convert the integer part (21) to binary:
Divide 21 by 2 repeatedly:

\[
21 \div 2 = 10 \text{ remainder } 1
\]
\[
10 \div 2 = 5 \text{ remainder } 0
\]
\[
5 \div 2 = 2 \text{ remainder } 1
\]
\[
2 \div 2 = 1 \text{ remainder } 0
\]
\[
1 \div 2 = 0 \text{ remainder } 1
\]

So, the binary representation of **21** is \(10101_2\).

#### 1.2 Convert the fractional part (0.375) to binary:
Multiply the fractional part by 2 repeatedly:

\[
0.375 \times 2 = 0.75 \quad \text{(integer part = 0)}
\]
\[
0.75 \times 2 = 1.5 \quad \text{(integer part = 1)}
\]
\[
0.5 \times 2 = 1.0 \quad \text{(integer part = 1)}
\]

So, the binary representation of **0.375** is \(0.011_2\).

#### 1.3 Combine the binary representations:
The binary form of **21.375** is:

\[
21.375_{10} = 10101.011_2
\]

### Step 2: Normalize the binary number
To normalize the number, move the binary point so that there is one non-zero digit to the left of the point.

\[
10101.011_2 = 1.0101011 \times 2^4
\]

Now, the number is in the form \(1.xxxx \times 2^E\), where \(E\) is the exponent.

### Step 3: Determine the exponent and the bias
The exponent in this case is \(4\), but in IEEE 754 format, the exponent is stored with a bias of **127**. So:

\[
\text{biased exponent} = 4 + 127 = 131
\]

In binary, **131** is:

\[
131_{10} = 10000011_2
\]

### Step 4: Determine the mantissa
The mantissa (fraction) is the binary digits to the right of the binary point in the normalized form. From the normalized number \(1.0101011_2\), the mantissa is:

\[
01010110000000000000000
\]

(The mantissa is 23 bits long, so we pad with zeros.)

### Step 5: Put it all together in IEEE 754 format
The IEEE 754 single-precision floating-point format is:

- 1 bit for the sign (0 for positive numbers)
- 8 bits for the exponent
- 23 bits for the mantissa

So, for **21.375**:

- **Sign bit** = 0
- **Exponent** = \(10000011_2\)
- **Mantissa** = \(01010110000000000000000_2\)

Thus, the full 32-bit IEEE 754 representation is:

\[
0 \, 10000011 \, 01010110000000000000000
\]

### Step 6: Convert to hexadecimal
Now, group the bits into 4-bit sections:

\[
0100 \ 0001 \ 1010 \ 1011 \ 0000 \ 0000 \ 0000 \ 0000
\]

Convert each group to hexadecimal:

\[
0100 = 4, \ 0001 = 1, \ 1010 = A, \ 1011 = B, \ 0000 = 0, \ 0000 = 0
\]

Thus, the hexadecimal representation is:

\[
\boxed{41AB0000}
\]

This is the IEEE 754 hexadecimal representation of **21.375**.

---

Would you like further details on any of these steps? Here are some related questions:

1. How does IEEE 754 handle negative numbers?
2. What is the difference between single-precision and double-precision floating-point formats?
3. How can rounding errors occur in floating-point arithmetic?
4. What is the bias in the IEEE 754 floating-point representation, and how is it chosen?
5. How do you convert a hexadecimal floating-point representation back to decimal?

**Tip**: When converting decimal numbers to binary, it helps to split the number into integer and fractional parts and handle them separately.

Learn how to convert the decimal number 21.375 into its IEEE 754 single-precision floating-point hexadecimal representation. This step-by-step guide covers binary conversion, normalization, and hexadecimal conversion, following the IEEE 754 standard. Suitable for advanced high school or undergraduate students.

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