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.
Show $\sqrt{2} = \text{sup}A$ where $A = \{x \in \mathbb Q \;\colon x^2 < 2\}$
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real-analysis
supremum-and-infimum
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1You are going to want to show that for all upper bounds $u$ of $A$, $\sqrt{2} \leq u$ – 2017-01-23
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0Try the first case $\sqrt2<\sup A$. Once it's done, the second one follows from a similar argument. Apply the definition of supremum to obtain an element $x'\in A$ so that $(\sup A)^2 \ge x'^2 > 2$, so we get a contradiction. – 2017-01-23
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0The body of the Question should contain your definition of $A$. In particular although I've taken the initiative to edit the math notation in the title, the point that $A\subset \mathbb Q$ should be explicit. Advise me if I misinterpreted the problem. – 2017-01-23
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0@hardmath: if the question is about supremum, putting $\mathbb Q$ instead of $\mathbb R$ doesn't make much of a difference. – 2017-01-23
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0@MartinArgerami: The change I made was not that, so we can skip that point. My request is that $A$ should be defined in the body of the Question, not in the title only. – 2017-01-23
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0IMHO, defining $A$ in the body is better than in the title. – 2017-01-23
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0Yeah, I see how it would be better in the body. In the future, I'll be sure to do that. – 2017-01-23