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I was cracking my head over the following proof (by induction) from Spivak's calculus.

Givens: $ \binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k} $ and $ n \ge k $

Task: Proof by induction that $ \binom{n}{k} $ is always a natural number.

My approach was:

(1) For the base case I take $ \binom{1}{k} $.

(2) I assume that $ k = 1 $ because $ n \ge k $ giving $\binom{1}{1}=1 $.

(3) I assume that $ \binom{n}{k} \in\mathbb N $ holds and prove $ \binom{n+1}{k}\in\mathbb N $.

(4) I apply the known formula $ \binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k} $.

(5) I know from (3) that $\binom{n}{k}\in\mathbb N $ holds.

(6) I do not know how to conclude that $ \binom{n}{k-1} \in\mathbb N $

(7) I do not know how to conclude that the sum of two natural numbers is a natural number so that even if I knew that $ \binom{n}{k-1} \in\mathbb N $ and take the assumption that $\binom{n}{k}\in\mathbb N$ I cannot conclude that $ \binom{n+1}{k}\in\mathbb N $.

Please help solving this. Please also check my exposition. Any comments on every aspect of my incomplete and possibly faulty proof are appreciated. Thanks in advance

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As you said, we will induct on $n$. But the claim will be, for a given $n$, that $\binom{n}{k}$ is an integer for any $k$. The base case will be $n=1$, and indeed $\binom{1}{0}=1,$ $\binom{1}{1}=1,$ and $\binom{1}{k}=0$ for $k\not=0,1$. Now assume the claim holds for some $n$. Then for any $k$ we have $$ \binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}. $$ Now our induction hypothesis assumes that both terms on the right hand side are integers, so their sum is an integer.

The trick is making the claim about all $k$, instead of a specific $k$.

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