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Show that for any $α ∈ R$, there exist infinitely many rational numbers $\frac{m}{n}$ with $|α − \frac{m}{n^2}| < \frac{1}{n}$.

So we know that $-1≤\frac{1}{n}≤1$ which implies $\frac{1}{n^2}≤1$.

Case $1$: if $m=n$ then $\frac{m}{n^2} = \frac{1}{n}$ so obviously we get $|α − \frac{m}{n^2}| < \frac{1}{n}$.

Case 2: if $m< n$ that implies $m< n^2$ which implies $\frac{m}{n^2}<1$ and if $n< n^2$ then $\frac{m}{n^2}<\frac{1}{n}$ so again $|α − \frac{m}{n^2} | < \frac{1}{n}$ makes sense.

I'm having trouble seeing how $m > n$ would come up with the same conclusion. (Am I going about this proof the right way?)

Zhoe
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learning
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1 Answers1

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Let $n$ be fixed, and imagine it to be large. There exist integers $p$ and $q$ such that $$\frac{p}{n^2}\lt\alpha\lt \frac{q}{n^2}.$$ Now consider the set $S$ of all integers $k$ such that $\frac{p+k}{n^2}\lt \alpha$. The set $S$ is non-empty, and bounded above. Thus $S$ has a largest element. So there is a largest $k$ such that $\frac{p+k}{n^2}\lt \alpha$.

It follows that $\frac{a+k+1}{n^2}\ge \alpha$, and therefore $$\left|\alpha-\frac{p+k}{n^2}\right|\le \frac{1}{n^2}.$$

Remark: The argument above does not have a close connection to the attempt in the post. Please note that it is $\alpha$ that is given. The point of the question is to show that there exist suitable $m$, $n$. In essence we showed how to construct $m$, given $\alpha$ and (it turns out) any $n\gt 1$.

André Nicolas
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