Have difficulty in understanding the proof below, in the book, by Traynor, titled: Introduction to analysis, at page #27; with title: The existence of roots — a consequence of completeness.
The proof is stated to show that irrational roots of $\sqrt{2}$ exist.
Theorem. Let $y$ be a positive real number. Then, for every $n\in \mathbb{N}$, there exists a unique positive real number such that $x^n = y$.
Proof.
(1) First, note that for positive numbers $a, b,\,\, a\lt b \implies a^n < b^n$.
This is proved by induction on $n$. (Exercise).(2) This implies uniqueness: $\,\,$Suppose $x_1^n = y\,\,$ and $\,\,x_2^n = y\,\,$, with $\,\,x_1, x_2\,\,$ positive but not equal. Then one must be smaller, by trichotomy. Say $\,\,x_1\lt x_2$.
Then $\,\,x_1^n \lt x_2^n$, so we can't have $\,\,x_1^n = x_2^n$. The contradiction shows $\,\,x_1 = x_2$.(3) Let $A = \{a \gt 0 : a^n \le y\}, B = \{b \gt 0 : b^n \ge y\}$. I claim that $A$ and $B$ are not empty and every element of $A$ is $\le$ every element of $B$.
Indeed, since $y \gt 0, \,\, 0 \lt \frac y{y+1}\lt 1$, so we have $$0 \lt (\frac {y}{y+1})^n \le \frac y{y+1}\lt y$$.
Thus, $\frac y{y+1}\in A$. In the same way, $y + 1 \gt 1$, so $(y + 1)^n \ge y + 1 \gt y$ and hence $y + 1 \in B$.
Now, if $a \in A$ and $b \in B$ we have $a^n \le y \le b^n $, so $a^n \le b^n$, and therefore $a \le b$.
This comes from step 1, because if $a\gt b$ we would have $a^n \gt b^n$.(4) Step 3 sets us up for our completeness axiom.
There must exist an $x$ with $a\le x \le b$, for all $a \in A$ and all $b\in B$. We will now show that for this $x, \,\, x^n = y$.
Let $\, 0 \lt a\lt x$. Then, $\, a \in A$.
For suppose not; then, $\, a^n \gt y$, which makes $\, a \in B$ and $\, x \le$ each element of $B$, a contradiction.
Similarly, if we let $\,b\gt x \,$, then $\, b \in B$.Thus, $$a^n \lt x^n \lt b^n, \,\,\,\,\,\,a^n \le y \le b^n.$$
If we multiply the second string of inequalities by $\,−1\,$, they turn around, so $$a^n \lt x^n \lt b^n\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,−b^n \le −y \le −a^n\,\,\,\, [Prob. 1$$ and adding gives $$\,a^n − b^n \lt x^n − y\lt \,b^n − a^n.\,\,\,[Prob. 2$$ In other words, $$|x^n − y| \le b^n − a^n \le (b − a)nb^{n−1}.\,\,\,[Prob. 3$$ Here we used the fact that $$b^n − a^n = (b − a)\sum_{i=1}^{n}b^{n-i} a^{i-1}\le (b − a)nb^{n−1}$$ Now take any $\epsilon$ such that $\,0 <\epsilon \lt x, \,a = x − \epsilon, \, b = x + \epsilon$.
Then $$\,b − a < 2\epsilon\,,\,b < 2x \,\,\,\, [Prob. 4 $$ and so $$|x^n − y| \le 2\,\epsilon \, n(2x)^{n−1}.$$ Hence, $$0 ≤ \frac{|x^n − y|}{2n(2x)^{n−1}} \le \epsilon.$$ Now $\,\epsilon \,\,$ here was arbitrary satisfying $\,0 \lt \epsilon\lt x$.
So, $$\frac{|x^n − y|}{2n(2x)^{n−1}} = 0.$$ Hence $|x^n −y| = 0$. But then $x^n −y = 0$, so $x^n = y$, which completes the proof.
The details of the problems problem numbers are stated below.
Prob. 1: Unable to show how multiplying $a^n \lt x^n \lt b^n$ by $-1$ leads to $−b^n \le −y \le −a^n$.
Prob. 2: How the equality sign gets removed upon adding the two inequalities?
Prob. 3: How the equality sign is re-introduced when modulus is taken of $|x^n − y|$?
Prob. 4: It seems like a typographical error is there in stating $\,\,b − a\lt 2\epsilon$.
$a = x − \epsilon, \, b = x + \epsilon$. Then $\,\,b − a = 2\epsilon\,\,$ & not $\,\lt 2\epsilon$.
Also, the correct ordering (equality, instead of less than) still brings the correct relation : $b < 2x$, as $\,\,b − a = 2\epsilon\implies b = x+\epsilon\implies b \lt x+x$, as $x\gt \epsilon$.
Also, have a feeling that the proof is not rigorous enough, as relies on so many assumptions and so request another method which is more rigorous for the same.
Problem 1: This is a basic theorem of ordered fields: If $a \le b$, then $-a \ge -b$. Proof: \begin{align*} a \le b &\implies -a + a \le -a + b \\ &\implies 0 \le -a + b \\ &\implies 0 + (-b) \le -a + b + (-b) \\ &\implies 0 + (-b) \le -a + 0 \\ &\implies -b \le -a. \end{align*} The given version is the double inequality version of this. Saying $a^n \le y \le b^n$ means $a^n \le y$ and $y \le b^n$. The above result implies $-y \le -a^n$ and $-b^n \le -y$, i.e. $$-b^n \le -y \le a^n.$$
Problem 2: Another basic theorem: if $a \le b$ and $c \le d$, then $a + c \le b + d$. Proof: \begin{align*} (a \le b) \land (c \le d) &\implies (a + c \le b + c) \land (b + c \le b + d) \\ &\implies a + c \le b + d. \end{align*} Now, let us suppose that $a + c = b + d$. This would mean (by trichotomy) that $$a + c \le b + c \le b + d = a + c \implies b + c = a + c = b + d.$$ Using cancellation law, this implies $a = b$ and $c = d$. So, if we have $a < b$ (i.e. $a \neq b$ as an additional assumption) and $c \le d$, then we naturally have $a + c < b + d$. This is why the equal sign has disappeared.
Problem 3: The equality sign shows only the possibility of equality. We can always add it back in, as it only weakens the statement. That is, we always have $a < b \implies a \le b$ (and not always the other way around).
As to why they've chosen to add the possibility of equality back in, I'm not sure. It doesn't add (nor does it detract) from the proof.
Problem 4: Agreed; it should be $b - a = 2\varepsilon$. However, this appears to be the only mistake in the proof. Aside from this, and Lord Shark's accurate comment that the Intermediate Value Theorem would make quicker, more elegant work of this, the proof seems fine.