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Find the infimum and supremum of the set: $$A=\{\frac{m-n-1}{mn+4m+3n+12}:m,n\in \mathbb N\}$$ My mind was drifting and I entered a circle by factorizing the algebraic fraction: $\frac{m-n-1}{mn+4m+3n+12}=\frac{m-n-1}{m(n+4)+3(n+4)}=\frac{m+3-n-4}{(m+3)(n+4)}=\frac{1}{n+4}-\frac{1}{m+3}$

I thought I could fix either $m$ or $n$ and examine different cases, but I haven't come up to anything formal according to the axioms of the field $\mathbb Q$, so I started believing this is alternating without a satisfying proof. The task is in the unit before the limits. What would be a wise first step?

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$$\lim_{n\to\infty}\frac{1}{n+4}-\frac{1}{1+3}\leq\frac{m-n-1}{mn+4m+3n+12}=\frac{1}{n+4}-\frac{1}{m+3}\leq\frac{1}{1+4}-\lim_{m\to\infty}\frac{1}{m+3}$$$$\implies \inf(A)=0-\frac{1}{4}=-\frac{1}{4},\ \sup(A)=\frac{1}{5}-0=\frac{1}{5}.$$

Remember, that the infimum of a decreasing sequence is its limit and the supremum of a increasing sequence is its limit.

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So, now we can write $$A = \{\frac{1}{n+4} - \frac{1}{m+3}: m, n \in \mathbb{N} \}$$

Observe that the condition $m, n \in \mathbb{N} $ can be translated as $m, n\geq 1$ belonging to the set of integers.


We know that the infimum of $A$ (a subset of $\mathbb{Q}$) is the greatest element of $\mathbb{Q} $ less than or equal to all elements of $A$. This means we have to find the minimum value of an element in $A$ (why?). This can be found by putting $m = 1$ and $n \rightarrow \infty$, to get infimum = $-\frac14$.


Similarly, the supremum of $A$ of $\mathbb{Q} $ is the samleest element of $\mathbb{Q} $ larger than all elements of $A$. This means we have to find the maximum value of an element in $A$ (why?). This can be found by putting $n =1, m\rightarrow \infty$ to get supremum = $\frac15$.

Hope this helps you.

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  • $\begingroup$ Thank you. My answer why we should find the minimum & maximum of A is to check whether infimum & supremum exist in $\mathbb Q$ and if so, then the greatest minorant and the least majorant are in $\mathbb Q$ $\endgroup$ – Praskovya2.718281828 Nov 3 at 12:38

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