The context of this work is Spivak's Calculus book, wherein Chapter 2, Problem 3(e)(ii), where you are asked to prove that:
$$\sum_{j=0}^{n}(-1)^j{n\choose j}={n \choose 0} - {n \choose 1} + \cdots \pm {n \choose n} = 0$$
Although this has been answered here on this site, I have come up with my own proof based on induction, which is in the spirit of this book (no combinatorics is assumed, rather a limited set of properties proven in Chapter 1 and an introduction to induction in Chapter 2).
My proof is below, but please remember that the only thing one is supposed to know about the "n choose k" notation at this point of the book, is given in the beginning of Problem 3 and is just this:
If $0 \le k \le n$, the "binomial coefficient" $\binom{n}{k}$ is defined by: $\binom{n}{k}=\frac{n!}{k!(n-k)!}, k \ne 0,n$ and ${n \choose 0} = {n \choose n} = 1$.
Specifically, nothing is discussed about negative values for $k$. That is where my proof runs into trouble as I manage to get $k=-1$ along the way.
Here is the proof which uses something proven in part (a) of the question which is:
$$ {n+1 \choose k} = {n \choose k-1}+{n \choose k} \tag a $$
First of all it seems that this does NOT hold for $n=0$ as $\sum_{j=0}^{0}(-1)^j{0\choose j}=(-1)^0{0 \choose 0}=1$.
It does however hold for $n=1$ as $\sum_{j=0}^{1}(-1)^j{1\choose j}=(-1)^0{1 \choose 0} - (-1)^1{1 \choose 1} = 1 - 1 = 0$.
Therefore let us assume that the equation holds for some $n\ge 1$ and thus $\sum_{j=0}^{n}(-1)^j{n\choose j}=0$
Then for $n+1$ we have:
$$ \begin{align} \sum_{j=0}^{n+1}(-1)^j{n+1\choose j} = && \text{extract last term of sum} \\ (-1)^{n+1}{n+1 \choose n+1} + \sum_{j=0}^{n}(-1)^j{n+1\choose j} = && \\ (-1)^{n+1} + \sum_{j=0}^{n}(-1)^j{n+1\choose j} = && \text{by (a)} \\ (-1)^{n+1} + \sum_{j=0}^{n}(-1)^j\left({n\choose j} + {n\choose j-1}\right) = && \text{split sum} \\ (-1)^{n+1} + \sum_{j=0}^{n}(-1)^j{n\choose j} + \sum_{j=0}^{n}(-1)^j{n\choose j-1} = && \text{by induction assumption} \\ (-1)^{n+1} + 0 + \sum_{j=0}^{n}(-1)^j{n\choose j-1} = && \text{extract first term of sum} \\ (-1)^{n+1} + (-1)^0{n \choose 0-1} + \sum_{j=1}^{n}(-1)^j{n\choose j-1} = && \text{let } k=j-1 \\ (-1)^{n+1} + {n \choose -1} + \sum_{k=0}^{n-1}(-1)^{k+1}{n\choose k} = && \\ (-1)^{n+1} + {n \choose -1} + (-1)\sum_{k=0}^{n-1}(-1)^{k}{n\choose k} = && \text{extend sum to }k=n \\ (-1)^{n+1} + {n \choose -1} + (-1)\left(\sum_{k=0}^{n}\left((-1)^{k}{n\choose k}\right) -(-1)^{n}{n\choose n}\right) = && \text{by induction assumption} \\ (-1)^{n+1} + {n \choose -1} + (-1)\left(0 -(-1)^{n}1\right) = && \\ (-1)^{n+1} + {n \choose -1} + (-1)(-(-1)^{n}) = && \\ (-1)^{n+1} + {n \choose -1} - (-1)^{n+1} = && \\ {n \choose -1} && \\ \end{align} $$
At this point I am tempted to say that choosing a negative number of items from a set of $n$ items is not possible and thus ${n \choose -1} = 0$ and the proof is complete. That does not feel correct, as it feels to me like it contradicts the spirit of this book.
My feeling is that what is expected in this book is something that uses induction and just the definition of "n choose k" provided at the top which is restricted to non-negative values of k.
I would really appreciate any suggestions on some other angle of attack along these lines, or even better: on how to "fix" my proof to avoid $k=-1$.
