For all $0<a\in \Bbb R$, there exists a unique $0<x\in \Bbb R$ such that $x^2=a$

Question

For all $0<a\in \Bbb R$, there exists a unique $0<x\in \Bbb R$ such that $x^2=a$.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

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My attempt:

1. Existence

Let $X=\{p\in \Bbb Q\mid 0 \le p \text{ and } p^2<a\}$.

First, $0\in X \implies X \neq \emptyset$. Second, $X$ is bounded from above by $a+1$. Thus supremum of $X$ exists. Let $x=\sup X$. We next prove that $x^2=a$.

By definition, $x^2=x \cdot x=\sup \{p \cdot q \mid p,q \in \Bbb Q \text{ and } p<x \text{ and } q<x\}$. Then it is easy to show that $x^2=\sup \{q^2 \mid q \in \Bbb Q \text{ and } 0<q<x\}$. On the other hand, $0<q<x$ $\implies$ $0<q<p$ for some $p \in X$ $\implies$ $q^2<p^2$ for some $p \in X$ $\implies$ $q^2<a$ $\implies x^2 \le a$.

Assume the contrary that $x^2 < a$. Then $\delta=a-x^2>0$.

Let $N = \min \left \{n \in \Bbb N \mid 2 \cdot \dfrac{x}{n} < \dfrac{\delta}{2} \text{ and } \dfrac{1}{n^2} < \dfrac{\delta}{2} \right \}$. It follows that $x^2 <\left (x+\dfrac{1}{N} \right)^2 < a$.

Since $\Bbb Q$ is dense in $\Bbb R$, there exists $p'\in\Bbb Q$ such that $x<p'<x+\dfrac{1}{N}$. It follows that ${p'}^2< \left(x+\dfrac{1}{N}\right)^2<a$. Hence $p' \in X$ and thus $p' \le \sup X =x$. This is a contradiction.

As a result, $x^2=a$.

2. Uniqueness

If $0<x<x'$, then $a=x^2<{x'}^2$. If $0<x'<x$, then ${x'}^2<x^2=a$. Hence such $x$ is unique.

Answer 1

Well, I think your 'Thus supremum of $X$ exists.' is a bit imprecise. You should argue that the supremum exists in $\mathbb{R}$, due to its completeness .. (e.g. in $\mathbb{Q}$ it does not exist)