$$x^2≤2$$ $$-\sqrt2\le x\le\sqrt2$$
How to proceed for this domain?
2026-03-29 14:32:25.1774794745
How to find infimum and supremum of $\{x∈\mathbb{Q}^+∣x^2\le2\}$
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$$A=\{x\in\mathbb{Q}^+:x^2\le2\}=\{x\in\mathbb{Q}:0\le x\le2^{\frac{1}{2}}\}\implies \inf A=0, \sup A=2^{\frac{1}{2}}$$