Chain of implications shows equivalence of several conditions

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Pick two conditions, say, $k$ and $l$ where $k<l$. Then $k\implies k+1\implies...\implies l-1\implies l$, so $k\implies l$. Further, $l\implies l+1\implies ...\implies n-1 \implies n\implies 1\implies...\implies k$, so $l \implies k$. Thus, $k\iff l$ for any two conditions $k$ and $l$.

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Updated on August 15, 2022

Comments

  • wwww
    wwww 2 months

    In mathematical articles, theorems frequently have the following form:

    The following (conditions) are equivalent:

    1. (first condition)

    2. (second condition)

    3. (third condition)

    ...

    where several (usually more than two) conditions are listed.

    Can you convince me that proving that showing the chain of implications 1 → 2 → 3 → ... → n → 1 (where n is the last condition) suffices to prove the equivalence of the n conditions?

    • Git Gud
      Git Gud over 6 years
      How convincing a proof do you need?
    • Git Gud
      Git Gud over 6 years
      Take $n+1$ statements in the given condition and assume it works for $n$ statements. Use the induction hypothesis on the first $n$ statements. You already know that the $n^{\text{th}}$ statement implies the $(n+1)^{\text{th}}$ one, so you're only missing the other direction. You also know that the last statement implies the first. Now, due to the induction hypothesis the first implies the next-to-last one, completing the induction step.
  • Git Gud
    Git Gud over 6 years
    I really don't think it's possible that someone who would ask this question would find this convincing.