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In college, we've seen some general formulas to compute the decimal value a of a binary representation $a_{N-1} \dots a_2 a_1 a_0$ in two's complement, $N$ being the number of bits. For $a \ge 0$ we've seen $a= \sum_{i=0}^{N-2} a_i 2^i$, which seems correct to me if I work it out. But for $a < 0$ our book says $a=-2^{N-1}+\sum_{i=0}^{N-2} a_i 2^i$ and this is where I go wrong.

Can someone explain with an example why this is correct or which formula is right?

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Let $a_{N-1} \cdots a_0$ be the two's complement representation of the number $a$. The more general formula is: $$a = -a_{N-1}2^{N-1} + \sum_{i=0}^{N-2} a_i 2^i$$ (I am sure you can figure out how your formulas can be derived from this one.)

Why does this hold? For $a \ge 0$, $a_{N - 1} = 0$ and then the value of $a$ is simply the value of the binary representation $a_{N-2} \cdots a_0$. For $a < 0$, on the other hand, let $b = -a \ge 0$ have binary representation $b_{N-1} \cdots b_0$. Then $$\sum_{i=0}^{N-1} a_i 2^i = \sum_{i=0}^{N-1} (1-b_i) 2^i + 1 = \sum_{i=0}^{N-1} 2^i - \sum_{i=0}^{N-1} b_i 2^i + 1 = 2^N - b$$ since the two's complement is defined as the complement of the representation $b_{N-1} \cdots b_0$, plus one, and $\sum_{i=0}^{N-1}2^i = 2^N - 1$. Since $a$ is negative, $a_{N-1} = 1$ and then: $$a = -b = -2^N + \sum_{i=0}^{N-1} a_i 2^i = -2^N + 2^{N-1} + \sum_{i=0}^{N-2} a_i 2^i = -2^{N-1} + \sum_{i=0}^{N-2} a_i 2^i $$


I have refrained from giving an explicit example as you had asked; I believe explaining why the formula works in the first place is far more instructive.

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  • $\begingroup$ Thanks a lot! This makes it very clear. I also see know there's a mistake in the general formule for a. Instead of a(N-1), our book said a(N) and the professor noted there may be a mistake in it, but didn't say what it was. You explained it very well! $\endgroup$ – L3k Dec 6 '18 at 12:43
  • $\begingroup$ @L3k If you believe this a satisfactory answer to your question, don't forget to click the tick mark to the left and mark it as an "accepted" answer ;) $\endgroup$ – dkaeae Dec 6 '18 at 14:01
  • $\begingroup$ I've only just now started to us this, so thanks a lot for the tip :) @dkaeae $\endgroup$ – L3k Dec 29 '18 at 11:03

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