Exercise · ZW-ARCHIMEDEAN-PROPERTY-009 · ★★★ Advanced

Which is the correct idea for proving the Archimedean property from the completeness axiom?

Choose one option.

Show answer
  • D. If \(\mathbb N\) were bounded it would have a supremum \(s\); since \(s-1\) is not a bound, some \(n>s-1\), so \(n+1>s\), a contradiction

Explanation

It is a proof by contradiction: \(s-1<s\) prevents \(s-1\) from being an upper bound, and from \(n>s-1\) we get \(n+1>s\) with \(n+1\in\mathbb N\). Saying “it is infinite” or using the limit of \(1/n\) would assume what is to be proved.

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Review the concept: Archimedean property

The property of \(\mathbb R\) by which, however small a positive number \(x\) is, some multiple of it by a natural \(n\in\mathbb N\) exceeds any real \(y\).

\[x>0,\ y\in\mathbb R\ \Longrightarrow\ \exists\, n\in\mathbb N:\ nx>y\]

More practice problems: archimedean property

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