Exercise 1.4.7

Question

Finish the proof of Theorem 1.4.5 by showing that the assumption $\alpha^{2}>2$ leads to a contradiction of the fact that $\alpha=\sup T$

Ulisse Mini · Accepted Answer

Recall $T = \{t \in \mathbf{R} \mid t^2 < 2\}$ and $\alpha = \sup T$. suppose $\alpha^2>2$, we will show there exists an $n \in \mathbf{N}$ such that $(\alpha - 1/n)^2 > 2$ contradicting the assumption that $\alpha$ is the least upper bound. We expand $(\alpha - 1/n)^2$ to find $n$ such that $(\alpha^2 - 1/n) > 2$ $$2 < (\alpha - 1/n)^2 = \alpha^2 - \frac{2\alpha}{n} + \frac{1}{n^2} < \alpha^2 + \frac{1 - 2\alpha}{n}$$ Then $$2 < \alpha^2 + \frac{1 - 2\alpha}{n} \implies n(2 - \alpha^2) < 1 - 2\alpha$$ Since $2 - \alpha^2 < 0$ dividing reverses the inequality gives us $$n > \frac{1-2\alpha}{2 - \alpha^2}$$ This contradicts $\alpha^2 > 2$ since we have shown $n$ can be picked such that $(\alpha^2 - 1/n) > 2$ meaning $\alpha$ is not the least upper bound.

Exercise 1.4.7

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Exercise 1.4.7

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