Prove that $ y = x^{\frac{1}{n}} \Rightarrow y^n = x $

Question

Tao, Analysis I, exercise 5.6.1

I have to prove the following where $x,y \in \Bbb R^+, \ n \in \Bbb Z^+$ $$ y = x^{\frac{1}{n}} \Rightarrow y^n = x $$

Hints: review the proof of Proposition 5.5.12. Also, you will find proof by contradiction a useful tool, especially when combined with the trichotomy of order in Proposition 5.4.7 and Proposition 5.4.12.

My attempt:

Assume that $ y = x^{\frac{1}{n}} \Rightarrow y^n > x $. By definition: $$x^{\frac 1n}= \sup \{ y \in \mathbb R, s.t. y \geq 0, y^n \leq x \}$$ Thus $ y^n > x $ contradicts the above definition, since $x$ is upper bound of $y^n$.

Does this part seem ok? I next need to get to a contradiction starting from here:

Now assume that $ y = x^{\frac{1}{n}} \Rightarrow y^n < x $

Answer 1

Let $a := x^{1/n}$.

Then $a^n > x$ does not conflict with the definition of the set, since it merely states that $a^n$ is an upper bound.

Here is my version of the proof following the same mechanism than for

Proposition 5.5.12. There exists a positive real number $x$ such that $x^2 = 2$.

and using the result of

Exercise 5.4.4. Show that for any positive real number $x > 0$ there exists a positive integer $N$ such that $x > 1/N > 0.$

Assume that $a^n > x$. Then there exists $\varepsilon > 0$ such that $(a - \varepsilon)^n > x$ and $a - \varepsilon$ is not in the set $\{y \in \mathbb{R}: y \ge 0 \; \wedge \; y^n \le x\}$. This contradicts the fact that $a - \varepsilon$ is not an upper bound, since $a$ is least upper bound, by definition.

Assume that $a^n < x$. Then there exists $\varepsilon > 0$ such that $(a + \varepsilon)^n < x$, and therefore there is an element in $\{y \in \mathbb{R}: y \ge 0 \; \wedge \; y^n \le x\}$ larger than $a$. But this contradicts that $a + \varepsilon > a$ is an upper bound.

In effect, $a^n \le x$ and $a^n \ge x$, equivalent to $a^n = x$.

---

To see that there exists an $\varepsilon > 0$ such that $(a + \varepsilon)^n < x$, note that $a$ is bounded by $x$, if $x > 1$. And it is bounded by $1$ otherwise. This yields

$$ (a + \varepsilon)^n = \sum_{k=0}^n \left( \begin{array}{c} n \\ k \end{array} \right) a^{n-k} \varepsilon^k = a^n + \varepsilon \left( \sum_{k=1}^n \left( \begin{array}{c} n \\ k \end{array} \right) a^{n-k} \varepsilon^{k-1} \right) < \begin{cases} a^n + \varepsilon \left( \sum_{k=1}^n \left( \begin{array}{c} n \\ k \end{array} \right) x^{n-k} \varepsilon^{k-1} \right) \; \text{if} \; x > 1, \\ a^n + \varepsilon \left( \sum_{k=1}^n \left( \begin{array}{c} n \\ k \end{array} \right) \varepsilon^{k-1} \right) \; \text{otherwise}. \end{cases} $$

In both cases there exists a positive real number $y$ such that $(a + \varepsilon)^n < a^n + \varepsilon y$.

Now if $a^n < x$, we have that $x - a^n =: \delta > 0$. But for every real number $\delta$ there exists a natural number $N$ such that $\delta > N^{-1}$. Choosing $\varepsilon = y^{-1}N^{-1}$ yields $x - a^n > \varepsilon y \Rightarrow x > (a + \varepsilon)^n$.

Equivalently, to see that there exists an $\varepsilon > 0$ such that $(a - \varepsilon)^n > x$, expand $(a - \varepsilon)^n$:

$$ (a - \varepsilon)^n = a^n + (-\varepsilon)\sum_{k = 1}^n \left( \begin{array}{c} n \\ k \end{array} \right) a^{n-k}(-\varepsilon)^{k-1} = a^n + \varepsilon \sum_{k = 1}^n (-1)^k \left( \begin{array}{c} n \\ k \end{array} \right) a^{n-k}\varepsilon^{k-1}. $$

Splitting the sum in a positive and a negative part:

$$ (a - \varepsilon)^n = a^n + \varepsilon \left( \sum_{i=2k}^n \left( \begin{array}{c} n \\ i \end{array} \right) a^{n-i}\varepsilon^{i-1} - \sum_{j=2k - 1}^n \left( \begin{array}{c} n \\ j \end{array} \right) a^{n-j}\varepsilon^{j-1} \right), $$

$$ (a - \varepsilon)^n > a^n - \varepsilon \left( \sum_{j=2k - 1}^n \left( \begin{array}{c} n \\ j \end{array} \right) a^{n-j}\varepsilon^{j-1} \right) $$

But, again, since $a$ is bounded by $\max(1, x)$, there is a positive real number $y$ such that $(a - \varepsilon)^n > a^n - \varepsilon y$.

Now if $a^n > x$, we have that $a^n - x =: \delta > 0$. Following the same argument than above, $a^n - x > \varepsilon y \Rightarrow -x > \varepsilon y - a^n \Leftrightarrow x < a^n - \varepsilon y < (a - \varepsilon)^n$.

Answer 2

All numbers in this answer are assumed to be positive.

We present the following general result from real analysis:

Proposition 1: Let $f: (+0, +\infty) \to (+0, +\infty)$ satisfy the following conditions:

$\tag 1 f \text{ is strictly increasing}$ $\tag 2 (\forall \, \gamma, \text{ integer } k) (\exists \, a, b) \bigr [ a \lt b \land f(b) - f(a) \lt k^{-1} \land f(a) \lt \gamma \lt f(b) \bigr ]$

Then $f$ is a bijection (i.e. an automorphism of the total ordering structure). Moreover,

$\tag 3 f^{-1} (\gamma) = \text{sup}(\{a \mid f(a) \lt \gamma\}) = \text{inf}(\{b \mid f(b) \gt \gamma\})$

The proof is left as an exercise.

All that remains is to show that $f(x) = x^n$ satisfies $\text{(1)}$ and $\text{(2)}$.

Hint: Use the identity

$\quad b^n - a^n = (b - a) \, \displaystyle{\sum_{i=1}^n a^{n - i} b^{i - 1}}$