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

Problem of 1’s Complement Representation

1. Two Zeros Exist
• Positive Zero: 0000
• Negative Zero: 1111
? This creates ambiguity, because logically both are zero but they have different bit patterns.

2. End-Around Carry Needed
When adding numbers, if a carry comes out of the MSB, it must be added back to the LSB.
(+5) = 0101
(-5) = 1010   (1’s complement of 0101)
----------------
Add: 1111 → End-around carry = 1
Final: 0000 (after adding carry)
? Result is zero, but note that two different zeros are possible.

3. Hardware Complexity
Extra logic is required so that +0 and -0 are treated the same, making design slower and costlier.

✅ Conclusion

That’s why modern systems use 2’s complement. It has only one zero and simplifies arithmetic operations.

IEEE 754 Single Precision Floating Point

Total Size: 32 bits

• Sign bit: 1 bit
• Exponent: 8 bits (biased, excess-127 notation)
• Mantissa (Fraction): 23 bits

? Exponent Representation

The exponent is stored in 8-bit biased form with a bias of 127.
Stored exponent = Actual exponent + 127

Example:
Actual exponent = 3
Stored exponent = 3 + 127 = 130 = 10000010

✅ Conclusion

In IEEE single precision, the exponent is represented in 8-bit biased (excess-127) notation.

2's Complement to Hexadecimal Conversion

Given the 2's complement binary number: (011010)
Find its equivalent hexadecimal representation.

Base (Radix) Problem Solution

Question: Find the base r such that

(312)_r ÷ (20)_r = (13.1)_r

Expansion in base 10

• (312)_r = 3r² + 1r + 2
• (20)_r = 2r
• (13.1)_r = r + 3 + 1/r

Equation

(3r² + r + 2) / (2r) = r + 3 + 1/r

Multiply both sides by 2r

3r² + r + 2 = 2r² + 6r + 2

Simplification

r² - 5r = 0
⇒ r(r - 5) = 0

✅ Final Answer

Since base must be > 1, the valid solution is:
Base (radix) = 5