Question: For every two real numbers $a$, $b$ such that $0 \leq a < b$, show there exists a positive real number $x$ such that $a < x^2 < b$.
I am aware that there are a few analogous versions of this question posted. They are not exactly the same and I also wanted to know if I'm expressing the idea correctly because I think I'm confusing myself with one step. Probably my notation.
Attempt
1) I Know: $0 \leq a < b$.
2) I Want: to show there exists $x > 0$ such that $a < x^2 < b$
My Idea:
I want to introduce two sets: $A = \{x\in \mathbb{R}: a \leq x^{2} \}$ and $B = \{x \in \mathbb{R}: x^{2} \leq b \}$. From this I would apply the LUB priniciple two each of these sets. I was also thinking perhaps I may have to show that $sup(A) = inf(B)$ to establish $x^{2}$ between the two values.
Let's start with $A = \{x\in \mathbb{R}: a \leq x^{2} \}$. The same idea will follow for $B$ with minor adjustments. So For $A$ I need to show that
i) $A \neq \emptyset$
ii) $A$ is bounded below.
Establishing i):
From a previous exercise I established that if $x < y$ there exists a rational number $r$ such that $x < r < y$. In this case let $a = x$ and $x^{2} = y$ and thus $a \leq r \leq x^{2}$. Thus $A \neq \emptyset$.
ii) Towards contradiction suppose $a$ is not a lower bound of $A$. Then there exists $x \in A$ such that $x^{2} < a$. This contradicts the definition of my set and as such $A$ is bounded below. (This conclusion feels really flimsy to me.)
Now to culminate, I need to establish that $inf(A) = a$.
From establishing the two properties above I can conclude that my set $A$ had an infimum by the least upper bound principle. Let us call this infimum, $G$. I want to establish that $inf(A) = G = a$. There are two cases: (I feel I took the right "idea", but went about it wrong)
i) if $a = G^{2}$. If this is the case we are done.
ii) if $a < G^{2}$. To establish this, suppose towards contradiction that $ a > G^{2}$.
Then $$a > G^{2} \Rightarrow G < \sqrt{a} < a < x^{2}$$ It was previously established that $a$ is a lower bound, BUT I have also assumed that $G$ is the greatest lower bound. This is a contradiction, therefore $a < G^{2}$.(I am also concerned about how I went about concluding this, it doesn't feel "correct")
So we have established the existence of a value $G$ that bounds the set $A$ from below.
I would now do the same for the set $B$ and establish a bound from above and since both of these sets are non empty I would have the existence of my $x^{2}$. I will of course flush this argument out on my own in the same fashion that I did for $A$, but just wanted to state it here for completeness.
Concerns The three parts I commented on in my argument are in bold and parentheses. Feedback on my approach would be appreciated along with any helpful ideas.