[proofplan] Fix a [real number](/pages/1303) $t > 0$. The reciprocal $1/t$ is positive, so the [Archimedean Property](/page/Archimedean%20Property) supplies a natural number $n$ larger than $1/t$. Since both $n$ and $t$ are positive, multiplying and dividing by them preserves the inequality direction and gives $1/n < t$. [/proofplan] [step:Apply the Archimedean Property to the reciprocal of $t$] Let $t \in \mathbb{R}$ satisfy $t > 0$. Define the real number $a := 1/t$. Since $t > 0$, we have $a > 0$. The [Archimedean Property](/theorems/737), applied to the positive real number $a$, gives some $n \in \mathbb{N}$ such that \begin{align*} n > a = \frac{1}{t}. \end{align*} [/step] [step:Convert the inequality $n > 1/t$ into $1/n < t$] Because $n \in \mathbb{N}$, we have $n > 0$. Since also $t > 0$, the product $nt$ is positive. Starting from \begin{align*} n > \frac{1}{t}, \end{align*} multiplying both sides by the positive real number $t$ preserves the inequality direction: \begin{align*} nt > 1. \end{align*} Dividing both sides by the positive real number $n$ again preserves the inequality direction: \begin{align*} t > \frac{1}{n}. \end{align*} Equivalently, \begin{align*} \frac{1}{n} < t. \end{align*} Thus there exists $n \in \mathbb{N}$ such that $1/n < t$, as required. [guided] We begin with the inequality obtained from the Archimedean Property: \begin{align*} n > \frac{1}{t}. \end{align*} The goal is to isolate $1/n$ on one side and $t$ on the other. Since $n \in \mathbb{N}$ and natural numbers are positive, $n > 0$. The hypothesis gives $t > 0$. Therefore multiplying or dividing by either $t$ or $n$ preserves the direction of a strict inequality. Multiplying the inequality by the positive real number $t$ gives \begin{align*} nt > 1. \end{align*} Now divide by the positive real number $n$: \begin{align*} t > \frac{1}{n}. \end{align*} Rewriting the same strict inequality with the smaller quantity first yields \begin{align*} \frac{1}{n} < t. \end{align*} Thus the natural number $n$ found by the Archimedean Property satisfies the required inequality. [/guided] [/step]