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Prove the multiplicative property of products by induction

by RoRi

Prove the multiplicative property of the product, i.e.,

    \[ \prod_{k=1}^n (a_k b_k) = \left( \prod_{k=1}^n a_k \right) \left( \prod_{k=1}^n b_k \right). \]

Proof. The proof is by induction. For n=1, we have

    \[ \prod_{k=1}^1 a_k b_k = a_1 b_1, \qquad \text{and} \qquad \left(\prod_{k=1}^1 a_k \right) \left(\prod_{k=1}^1 b_k \right) = a_1 b_1. \]

Thus, the multiplicative property holds for the case n =1. Assume then that it holds for some n=m \in \mathbb{Z}_{>0}. Then,

    \begin{align*} &&\prod_{k=1}^m a_k b_k &= \left( \prod_{k=1}^m a_k \right) \left( \prod_{k=1}^m b_k \right) \\ \implies && \prod_{k=1}^{m+1} a_k b_k &= \prod_{k=1}^m a_k b_k \cdot (a_{m+1} b_{m+1}) & (\text{Def. of prod}) \\ \implies &&&= \left(\prod_{k=1}^m a_k \right) a_{m+1} \cdot \left( \prod_{k=1}^m b_k \right) b_{m+1} & (\text{Ind. Hyp.})\\ \implies &&&= \left( \prod_{k=1}^{m+1} a_k \right) \left( \prod_{k=1}^{m+1} b_k \right) & (\text{Def. of prod}). \end{align*}

Thus, the m+1 case is true; therefore, the property holds for all n \in \mathbb{Z}_{>0}. \qquad \blacksquare

One comment

  1. Rishabh Goswami says:

    A very good initiative. It help’s the novice learner’s to understand the problems and develop heuristics. Keep up the Good Work. (A+)

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