I hope you could help me with the following task:
Show the existence of a root function: $$\sqrt{*} : \mathbb{R}_{>0} \to \mathbb{R}_{>0}, a\to \sqrt{a}$$ with the property $$\sqrt{a}^2 = a $$ for all $a \in \mathbb{R}$. First have a look at the two subsets $X = \{ x \in \mathbb{R}_{>0} | x^2 < a\}$ and $Y = \{ y \in \mathbb{R}_{>0} | y^2 > a\}$.
Now apply the completeness axiom to show that there is a $c \in \mathbb{R}$ with $x \leq c \leq y$ for all $x \in X$ and $y \in Y$. Use the following information: $$\epsilon \in \mathbb{R}$$ with $$0 < \epsilon < 1$$ the following statements apply: $$c + ε ∉ X, c - ε ∉ Y $$ Conclude that $$c^2 ≥ a$$ respectively $$c^2 \leq a$$
Our completeness axiom is: $$\forall X,Y \subseteq \mathbb{R}: (( X \neq \emptyset \wedge Y \neq \emptyset \wedge \forall x \in X \forall y \in Y : x \leq y ) \implies ( \exists c \in \mathbb{R} \forall x \in X \forall y : x \leq c \leq y )) $$
I've seen some proof but they all use supremum and infimum which we haven't learned yet. Neither maximum and minimum.
So here is my idea: It's given by the task that "$x^2 < a$" and "$y^2 > a$". So $$x^2 < a < y^2$$ With transitivity we can say: $$x^2 < y^2$$ This also means "$x < y$" since we are in $\mathbb{R}_{>0}$. We have an element "$a$". With "$\frac{a}{2} < a < 2a$" we can say that $X$ and $Y$ are not empy. So we can apply the completeness axiom. There must be a "$c$" between "$x$" and "$y$". So $$x \leq c \leq y \Rightarrow x^2 \leq c^2 \leq y^2$$ With "$c + \epsilon \notin X$" and "$c - \epsilon \notin Y$" we have that $$c^2 + \epsilon^2 + 2\epsilon c \geq a$$ and $$c^2 + \epsilon^2 - 2\epsilon c \leq a$$ I hope the proof is correct so far. Now I can't find the right steps to finish the proof. I hope you can help me out. Please keep in mind, I'm at the beginning of all this proving hullabaloo, so keep your answers non trivial
Thank you in advance
Let $a > 0$. We define $X = \{x \ge 0, x^2 < a\}$ and $Y = \{y \ge 0, a < y^2\}$. Then
From this, the completeness axiom says that there exists $c\in \mathbb{R}$ such that $\forall x\in X, \forall y \in Y, x \le c \le y$.
Let now $\epsilon \in (0, 1)$ be a real number. As $c+\epsilon > c$, one has $c+\epsilon \notin X$, hence $(c + \epsilon)^2 \ge a$. This implies $a \le c^2 + 2 \epsilon c + \epsilon^2 \le c^2 + \epsilon(2 c + 1)$.
Suppose that $c^2 < a$, then $a-c^2>0$ and we can choose $\epsilon = \min(1,\frac{a-c^2}{2 (2c+1)})$, so that $\epsilon (2c+1)\le \frac{a-c^2}{2}$ and finally $a \le c^2 + \epsilon(2c+1)\le \frac{a+c^2}{2}$. This implies $a \le c^2$, a contradiction.
It follows that necessarily, $a \le c^2$. Suppose now that $a < c^2$. Then one has $c > 0$ and for all $\epsilon \in (0, c)$, $0 < c-\epsilon <c$. In particular, $c-\epsilon \notin Y$, which means that $(c - \epsilon)^2\le a$. It follows that $c^2 - 2 \epsilon c+ \epsilon^2\le a$, hence $c^2 - a \le 2 \epsilon c-\epsilon^2\le 2\epsilon c$. Choosing $\epsilon = \min(c, \frac{c^2-a}{4 c})$, we get $c^2 - a\le 0$, a contradiction.
It follows from all this that $a = c^2$, QED.