Definition. The two’s complement of an $n$-bit binary number $x$ is given by $\bar{x}+1$, where $\bar{x}$ is constructed from $x$ by inverting the $0$s and $1$s.
Definition. The two’s complement representation of a signed $n$-bit number $a_{n-1}\cdots a_1a_0$ (or simply “two’s complement number”) is given by
\[-2^{n-1} a_{n-1} + 2^{n-2}a_{n-2}+ \cdots 2^1 a_1 + 2^0a_0\]where the first bit is the sign bit.
Theorem. Let $x_c$ denote the two’s complement of an $n$-bit binary number $x$. Then $x + x_c = 2^n$ in decimal.
proof. The binary representating of $2^n$ is
\[1\underbrace{00\cdots 0}_{\text{n}}.\]Write this as
\[1 + \underbrace{11\cdots 1}_{\text{n}}.\]Note that by definition of $\bar{x}$,
\[\underbrace{11\cdots 1}_{\text{n}} - x = \bar{x}.\]So
\[x+x_c = x + (\bar{x}+1) = x + (\underbrace{11\cdots 1}_{\text{n}} - x) + 1 = \underbrace{11\cdots 1}_{\text{n}} + 1 = 1\underbrace{00\cdots 0}_{\text{n}},\]which is $2^n$ in decimal.
Corollary. The two’s complement representation of the negative of a binary number $x$ is given by $-x = \bar{x} + 1$. That is, the two’s complement representation of the negative of a binary number can be constructed by inverting the $0$s and $1$s and adding $1$.
proof. By the definition of $\bar{x}$, $x + \bar{x} = \underbrace{11\cdots 1}_{\text{n}}$, which is $-1$ in two’s complement representation because
\[-2^{n-1} + 2^{n-2} + \cdots + 2 + 1 = -2^{n-1} + \sum_{i=0}^{n-2} 2^i = -2^{n-1} + 1\cdot (2^{n-1}-1) = -1.\]So in two’s complement representation, $-x = \bar{x}+1$.