<s>[INST] Prove decimal number pattern for its n-bit 2's compliment If we let $B =  b_{n−1}b_{n−2} · · · b_1b_0$ be an $n$-bit $2$’s complement integer.
How can we show that the decimal value of $B$ is $−b_{n−1}\cdot2^{n−1} + b_{n−2} \cdot 2^{n−2} + b_{n−3}\cdot 2^{n−3} + \cdots + b_1\cdot 2 + b_0$.
 [/INST] Hint: let $\;A=-B\;$ then, by the definition of $2$'s complement, $A$ is obtained from $B$ by flipping all bits and adding $1\,$:
$$
A = (1-b_{n-1})\cdot 2^{n-1} + (1-b_{n-2})\cdot 2^{n-2} + \cdots + (1-b_1) \cdot 2 + (1-b_0) + 1
$$
Let $B\,'$ be the given expression $B\,' = −b_{n−1}\cdot2^{n−1} + b_{n−2} \cdot 2^{n−2} + \cdots + b_1\cdot 2 + b_0\,$. Then:
$$
\require{cancel}
\begin{align}
A + B\,' & = - 2 b_{n-1}\cdot 2^{n-1} + (2^{n-1}+2^{n-2}+\cdots+2+1) + 1 = \\
 & = -b_{n-1} \cdot 2^{n} + (2^n - \cancel{1}) + \cancel{1} \\
 & \equiv \;0 \pmod{2^n}
\end{align}
$$
Since $2$'s complement arithmetic is $\bmod 2^n\;$ it follows that $A+B\,'=0$ $\iff$ $B\,'=-A=B$.
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