Are my proofs elaborate, yet concise? Is there something missing, or am I adding too much? 

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> Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ an upper bound of $E$. Prove $\alpha \leq \beta$
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Since $\alpha$ is a lower bound of $E$, then $$\forall e\in E \exists \alpha \in L : \alpha \leq e$$ Then every $e$ is an upper bound for $L$, and so $L$ is bounded above. Call the least upper bound of $L$ $\sup L = e\_{0}$; we end up with the inequality $\alpha \leq e\_{0} \leq \beta$. QED.

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> Prove $\sqrt{15}$ is irrational.
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Suppose instead that $\sqrt{15}$ is rational. Then $$\exists p, q \in \Bbb Z : {\frac{p^2}{q^2}} = 15 \rightarrow p^2 = 15q^2.$$

This implies $15 | p^2$ and thus $15 | p$ (since there is no common factor between $q$ and $p$). But the only factors of $15$ are $3$ and $5$, and $p^2$ cannot be compromised of two factors. RAA.