Given a set $A = \left\{\dfrac{m} {n} + \dfrac{8n} {m} : m, n \in \mathbb{N}\right\}$ Determine its supremum and infimum.

My attempt: With $t:={m\over n}$ we have to consider $t+8/t$ on ${\bf Q}^+$. We have $t+{8\over t}\ge 2\sqrt{8}=4\sqrt{2}$. Of course equality is not possible but letting $t$ converge to $2\sqrt{2}$ we find that $4\sqrt{2}$ is approachable, hence the infimum. The supremum equals $\infty$ by sending $t\rightarrow \infty$

Does this approachment be valid/true?

  • $\begingroup$ As a side note, you can clearly see the set is unbounded on the high side by letting $n=1$ $\endgroup$ – Alan Jun 2 '18 at 13:16

You're absolutely right.

If $\frac{m}{n}=t$ then we have $t+\frac{8}{t}=k\in \mathbf{Q^+}$ which is quadratic in $t \in \mathbf{Q^+}$ solving for t we must have $\frac{k\pm\sqrt{k^2-32}}{2}\in\mathbf{Q^+}$ which allows us to discard one of the roots as $\sqrt{k^2-32}< k\ $ so we are only concerned with

$\frac{k+\sqrt{k^2-32}}{2}\in\mathbf{Q^+}$ This gives us the following:

$k^2-32>0\Rightarrow k>4\sqrt{2}$

where $\lim_{t\to 2\sqrt{2}}k=4\sqrt{2}$

Further if we consider $\lim_{t\to 0}k$ we get $\infty$ which means for all $t \in \mathbf{Q^+}$ there is a $t^{'} \in\mathbf{Q^+}$ such that $k^{'}= t^{'}+\frac{8}{t^{'}}>k$


Your approach is fine, except that I would say that it has no supremum, since it has no upper bound.


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