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Let * be an associative binary operation on a set 'A' with identity element e. Let 'B' be a subset of 'A' that is closed under *. Let b1, b2, b3, ... bn ∈ B. Prove that b1 * b2 * b3... bn ∈ B.

We know that b1, b2, b3 and so on are elements of B under an associative binary operation, but how would one prove this with induction?

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We can prove this by induction on $n$.

Base case: $b_1 \in B$ by definition.

Inductive case: we must show that, for all $k$, if $b_1 * \ldots * b_k \in B$ then $b_1 * \ldots * b_k * b_{k+1} \in B$.

Notice that, since $*$ is associative, $b_1 * \ldots * b_k * b_{k+1} = (b_1 * \ldots * b_k) * b_{k+1}$. We know by our inductive hypothesis that $b_1 * \ldots * b_k \in B$, and we know that $B$ is closed under $*$. Therefore, $(b_1 * \ldots * b_k) * b_{k+1} \in B$ and we are done.

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  • $\begingroup$ Thank you for your answer, it made sense to me! $\endgroup$ – Earthbound27 Apr 8 '16 at 18:10

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