For a positive real x, there is always a positive integer whose reciprocal is smaller than x.

by RoRi

June 30, 2015

Prove that if $0 < x$ , then there is an $n \in \mathbb{Z}^+$ such that $1/n < x$ .

Proof. Since $x > 0$ , we know that for any $y \in \mathbb{R}$ there exists an $n \in \mathbb{Z}^+$ such that $nx > y$ (Theorem I.30, p. 26 of Apostol). Let $y = 1$ , then we have $nx > 1$ and so $\frac{1}{n} < x. \qquad \blacksquare$

Stumbling Robot

For a positive real x, there is always a positive integer whose reciprocal is smaller than x.

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