Show $\sqrt{2} = \text{sup}A$ where $A = \{x \in \mathbb Q \;\colon x^2 < 2\}$

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I'm starting to learn real analysis and am having trouble understanding the supremum. I'm trying to work through this proof and think I need to show using the Archimedean property that there would be a contradiction if $\sqrt{2} < \text{sup}A$ or if $\sqrt{2} > \text{sup}A$. I'm not sure how to show this contradiction.