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$B=a+(2a)^{-1}: a\in \mathbb Q, 0.1≤a≤5$.

What I tried to do is work on the limits, $0.1≤a≤5 \implies 1/10≤1/2a≤5$ than to find the supremum and infimum. I tried to find the limit for $a+(2a)^{-1}$ which was $1/10≤a+1/2a≤5+1/10$. Making the supremum and maximum to be $5+1/10$ and infimum and min to be $1/10$.

Is this correct?

amWhy
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Ria Park
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  • Welcome to Math SE! Please use MathJax when formatting, it makes it much easier to read and makes your question clearer. – Amaan M Feb 11 '21 at 04:52
  • You found out the range of $\frac{1}{2a}$ all fine. But while adding the two inequalities, add all the three parts simultaneously. Your answer is wrong. – DeBARtha Feb 11 '21 at 05:37

1 Answers1

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Plotting $y = f(x) := x + \frac{1}{2x}$ for $0.1\leq x\leq5$ reveals that your answer is incorrect. Although the supremum does appear to occur at the boundary of the interval $[0.1, 5]$, the infimum can be clearly seen to occur in the interior of the interval. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ you can find the infimum by applying the first derivative test for closed intervals to the function $f$.

Yakes
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