Question

convert the following real numbers to binary (five binary places)  a.9.26

Answer 1

To convert the decimal number 9.26 to binary with five binary places, you will need to separate it into its integer and fractional parts.

### Step 1: Convert the Integer Part (9)

1. Divide 9 by 2:
- 9 ÷ 2 = 4 remainder 1
2. Divide 4 by 2:
- 4 ÷ 2 = 2 remainder 0
3. Divide 2 by 2:
- 2 ÷ 2 = 1 remainder 0
4. Divide 1 by 2:
- 1 ÷ 2 = 0 remainder 1

Now read the remainders from bottom to top:
Thus, \(9\) in binary is \(1001\).

### Step 2: Convert the Fractional Part (0.26)

To convert the fractional part, we multiply by 2 and take the integer part:

1. \(0.26 \times 2 = 0.52\) → integer part is 0
2. \(0.52 \times 2 = 1.04\) → integer part is 1
3. \(0.04 \times 2 = 0.08\) → integer part is 0
4. \(0.08 \times 2 = 0.16\) → integer part is 0
5. \(0.16 \times 2 = 0.32\) → integer part is 0
6. \(0.32 \times 2 = 0.64\) → integer part is 0
7. \(0.64 \times 2 = 1.28\) → integer part is 1

We take the first five bits after the decimal point:
Thus, \(0.26\) in binary is approximately \(0.01000\).

### Final Result

Combining both the integer and fractional parts, we have:

\[
9.26 \text{ in binary} \approx 1001.01000
\]

So, the binary representation of 9.26 (with five binary places) is:

\[
\textbf{1001.01000}
\]