Range of Integers Representable by $n$-bit Two’s Complement

In an $n$-bit two’s complement representation, the most significant bit (MSB) acts as the sign bit, and there are $2^n$ distinct bit patterns total. The set of representable signed integers is:

• Minimum: $-2^{n-1}$
• Maximum: $2^{n-1}-1$

So the representable range is $-2^{n-1} \;\le\; x \;\le\; 2^{n-1}-1.$

Key ideas:

• Two’s complement encodes negatives using modular arithmetic on $2^n$.
• MSB (most significant bit) determines whether a bit pattern is interpreted as nonnegative or negative.
• Sign extension preserves the numeric value.

Note: The range formula above is the standard result for two’s complement encodings, widely used in computer architecture and digital logic.

Deriving the range (conceptual proof)

Let bits be $b_{n-1}b_{n-2}\dots b_1b_0$, where $b_{n-1}$ is the MSB. Two’s complement interprets the bit string as a signed value via a weighted sum: $x = -b_{n-1}\cdot 2^{n-1} + \sum_{i=0}^{n-2} b_i\cdot 2^i.$

From this:

• If $b_{n-1}=0$ (nonnegative), then the sum is maximized when all remaining bits are 1: $x_{\max} = \sum_{i=0}^{n-2} 2^i = 2^{n-1}-1.$
• If $b_{n-1}=1$ (negative), the most negative value occurs when all remaining bits are 0: $x_{\min} = -2^{n-1}.$

Therefore: $x \in \left[-2^{n-1},\; 2^{n-1}-1\right].$

Key terms:

• [Modular arithmetic]{def="Working with integers modulo a fixed value such as $2^n$"} explains why bit patterns wrap around.
• Integer encoding clarifies “representation range.”
• Bit pattern.

Examples for intuition

$n=4$ (4-bit two’s complement)

• Minimum: $-2^{3} = -8$
• Maximum: $2^{3}-1 = 7$

So the range is: $-8 \le x \le 7.$

$n=8$ (8-bit two’s complement, common in bytes)

• Minimum: $-2^{7} = -128$
• Maximum: $2^{7}-1 = 127$

So the range is: $-128 \le x \le 127.$