# Thread: Infimum Proof

1. ## 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, ...

2. ## Re: Infimum Proof

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

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

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

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

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

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

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

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