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Math Help - Infimum Proof

  1. #1
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    Infimum Proof

    Without using the Archimedean theorem, show that for each positive real there is a natural number so that 1/n < r.
    I have the feeling its a proof by contradiction, but I'm unsure what to do. I know if I set up a contradiction, I will see that r is a lower bound for the set of 1/1, 1/2, ...
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  2. #2
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    Re: Infimum Proof

    \text{I am going to prove this using contradiction and the supremum definition.}

    \textbf{Proof.} \text{ Supose, to obtain a contradiction, that:}

    \exists \ r \in \mathbb{R}_{++} \ , \ \forall \ n \in \mathbb{N} \ , \ \frac{1}{n} \geq r

    \Rightarrow \frac{1}{r} \geq n \ , \ \forall \ n \in \mathbb{N} \text{ (note that } r > 0 \text{)}

    \Rightarrow \frac{1}{r} = \sup \left \{ n: n \in \mathbb{N} \right \}

    \Rightarrow \exists \ m \in \mathbb{N} \ , \ \frac{1}{r} - 1 < m

    \Rightarrow \frac{1}{r} < (m + 1) \in \mathbb{N}

    \Rightarrow \text{Contradiction!}
    \text{Q.E.D.}
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