Question: If we can generate a maximum of 4 Boolean functions using n Boolean variables, what is the minimum value of n?

Formula: Number of Boolean functions of $n$ variables is:

\[ 2^{2^n} \]

Condition: We are told the total functions must be ≤ 4:

\[ 2^{2^n} \leq 4 \]

✅ Try values of $n$:

• $n = 0$: $2^{2^0} = 2^1 = 2$ ✅
• $n = 1$: $2^{2^1} = 2^2 = 4$ ✅
• $n = 2$: $2^{2^2} = 2^4 = 16$ ❌

Minimum $n$ for which number of Boolean functions ≤ 4 is:

\[ \boxed{1} \]

✅ Final Answer: $\boxed{1}$

✅ Given:

The floating-point binary number is \( +1001.11_2 \).

We need to convert it into an 8-bit fraction and a 6-bit exponent format.

✅ Step 1: Normalize the Binary Number

We start by normalizing the binary number into scientific notation of the form:

\( 1.xxxx \times 2^n \)

Converting \( 1001.11_2 \) into scientific notation gives:

\( 1001.11_2 = 1.00111_2 \times 2^3 \)

The exponent is \( 3 \) (because the binary point is shifted 3 places to the left).

✅ Step 2: Convert the Exponent to Binary

The exponent is \( 3 \) in decimal. To represent this in binary using 6 bits, we get:

\( \text{Exponent} = 000100_2 \)

✅ Step 3: Convert the Fraction to 8 Bits

The fractional part of the normalized binary number is \( 00111 \). We need to extend it to 8 bits:

\( \text{Fraction} = 01001110_2 \)

✅ Final Answer:

The floating-point binary number \( +1001.11_2 \) in 8-bit fraction and 6-bit exponent format is:

Exponent: \( 000100_2 \), Fraction: \( 01001110_2 \)

10-bit 2's Complement Representation of –35

Format: 10-bit signed integer using 2's complement representation.

Step-by-Step:

1. First, convert 35 to 10-bit binary: 0000100011
2. Find 1's complement: 1111011100
3. Add 1 (to get 2's complement): 1111011101

✅ Final Answer: 1111011101

 –35 in 10-bit 2's complement: 1111011101