# prove existence of nth roots for non-negative real numbers

I would like to prove the following result: "Let $$x,y \geq 0$$ be non-negative reals, and let $$n,m \geq 1$$ be positive integers. If $$y = x^{\frac{1}{n}}$$, then $$y^{n} = x$$." This is lemma 5.6.6 (a) from the book Analysis 1 by Terence Tao.

The nth-root is defined as follows. $$x^{\frac{1}{n}}:=$$sup$$\{y\in \mathbb{R}: y\geq 0$$ and $$y^{n}\leq x\}$$.

Previously, the following lemma has been proven. "$$\textbf{Lemma 5.6.5:}$$ "Let $$x\geq 0$$ be a non-negative real, and let $$n\geq 1$$ be a positive integer. Then the set $$E:= \{y\in \mathbb{R}: y\geq 0$$ and $$y^{n}\leq x\}$$ is non-empty and is also bounded above. In particular, $$x^{\frac{1}{n}}$$ is a real number."

Given lemma 5.6.5, all we need to show is that $$y^{n} and $$y^{n}>x$$ lead to contradictions. For example, in the case where $$n=2$$ and $$y^{2} we can find an $$\varepsilon>0$$ such that $$(y+\varepsilon)\in E$$ just by expanding $$(y+\varepsilon)^{2}$$ and choosing $$\varepsilon$$ appropriately, contradicting the assumption that $$y = sup E$$.

I am familiar with how this result is proven using either the identity $$b^{n} - a^{n} = (b-a)(b^{n-1} + b^{n-2}a + ... +a^{n-1})$$, which is used for example in Rudin's real analysis book, or the binomial theorem. However, I am trying to prove the result using only some hints given in the textbook. The hints are as follows:

1) Review the proof that $$\sqrt2$$ is a real number (the proof follows the exact outline above). 2) Proof by contradiction. 3) The trichotomy of order. 4) Proposition 5.4.12

$$\textbf{Proposition 5.4.12:}$$ "Let $$x$$ be a positive real number. Then there exists a positive rational number $$q$$ such that $$q\leq x$$, and there exists a positive integer $$N$$ such that $$x\leq N$$."

I have tried to prove the result using only the four hints given above, but I haven't been able to get anywhere. The four hints are given for the whole lemma, which consists of more than the above statement, so it is not clear that all hints are meant to be used for this particular statement. Previously, properties of exponentiation have been proven for real numbers and integer exponents, so these can be used in the proof.

There is a similar question here Help with a lemma of the nth root (without the binomial formula), but my question is not answered there (neither has it been answered in any other similar posts I have read).

My attempts have been centered around the following idea: Assume $$y^{n} < x$$. Then $$x-y^{n}>0$$, which implies the existence of $$q\in \mathbb{Q}^{+}$$ such that $$q\leq x -y^{n}$$. We could also assume that $$0 to get $$q^{n}\leq x-y^{n}$$, though it is not clear to me that this helps. If we assume that $$(y+\varepsilon)^{n} \geq q^{n} + y^{n}$$ for all $$\varepsilon>0$$, then we could get a contradiction by taking the limit as $$\varepsilon$$ tends to zero. However, limits are not developed until the next chapter. Instead, I have been trying to find $$\varepsilon$$ directly, especially by trying to use hint number four, without any luck (I think including all the messy trials here would make an already lenghty post unreadable).

Any help would be greatly appreciated. Please excuse the lengthy post. Thank you very much to those that take the time to read this post.

$$\textbf{Edit:}$$ I posted my attempt at a solution below. I also realize that I do not really need to use proposition 5.4.12 to find a rational $$q$$. I could for example work with the real number $$x-y^{n}$$ ($$y^{n]-x$$) directly.

Here is my attempt without using combinatorials. The trick is to replace $$(y + \varepsilon)^n$$ and $$(y - \varepsilon)^n$$ with $$y^n + \delta$$ and $$y^n - \delta$$ respectively.

Let $$E = \{z \in \mathbb{R} : (z \geq 0) \land (z^n \leq x)\}$$. So $$y = x^{1 / n} = \sup(E)$$. Suppose for sake of contadiction that $$y^n \neq x$$. Then by Proposition 5.4.7, exactly one of the following statements is true:

(I) $$y^n < x$$. Now we want to show that $$\exists\ \varepsilon \in \mathbb{R}$$ and $$\varepsilon > 0$$ such that $$(y + \varepsilon)^n < x$$. Because $$y < y + \varepsilon$$, so we have $$y^n < (y + \varepsilon)^n$$. Let $$\delta = (y + \varepsilon)^n - y^n$$, then $$\delta > 0$$. By Corollary 5.4.13, we can find an $$N \in \mathbb{N}$$ and $$N > 0$$ such that $$\delta < 1 \times N$$. By Proposition 5.4.14, $$\exists\ q \in \mathbb{Q}$$ such that $$\delta < q < N$$, which means $$\delta / q < 1$$, and we have \begin{align*} (y + \varepsilon)^n &= y^n + \delta \\ &= y^n + q \delta / q & (q \neq 0) \\ &< y^n + q. & (\delta / q < 1) \end{align*} This means if we can show that $$\exists\ q \in \mathbb{Q}$$ and $$q > 0$$ such that $$y^n + q < x$$, then we can show that $$\exists\ \varepsilon \in \mathbb{R}$$ and $$\varepsilon > 0$$ such that $$(y + \varepsilon)^n < x$$. We can show such $$q$$ exists because by Proposition 5.4.14 $$\exists\ q \in \mathbb{Q}$$ and $$0 < q < x - y^n$$. So we must have $$\varepsilon \in \mathbb{R}$$ and $$\varepsilon > 0$$ such that $$(y + \varepsilon)^n < x$$. But this means $$y + \varepsilon \in E$$ and $$y + \varepsilon \leq y$$, a contradiction.

(II) $$y^n > x$$. Now we want to show that $$\exists\ \varepsilon \in \mathbb{R}$$ and $$\varepsilon > 0$$ such that $$(y - \varepsilon)^n > x$$. Because $$y > y - \varepsilon$$, so we have $$y^n > (y - \varepsilon)^n$$. Let $$\delta = y^n - (y - \varepsilon)^n$$, then $$\delta > 0$$. By Proposition 5.4.13, we can find an $$q \in \mathbb{Q}$$ and $$q > 0$$ such that $$q < 2q \leq \delta$$. Then we have $$\delta / q > 1$$ and \begin{align*} (y - \varepsilon)^n &= y^n - \delta \\ &= y^n - q \delta / q & (q \neq 0) \\ &> y^n - q. & (\delta / q > 1) \end{align*} This means if we can show that $$\exists\ q \in \mathbb{Q}$$ and $$q > 0$$ such that $$y^n - q > x$$, then we can show that $$\exists\ \varepsilon \in \mathbb{R}$$ and $$\varepsilon > 0$$ such that $$(y - \varepsilon)^n > x$$. We can show such (q) exists because by Proposition 5.4.14 $$\exists\ q \in \mathbb{Q}$$ and $$0 < q < y^n - x$$. So we must have $$\varepsilon \in \mathbb{R}$$ and $$\varepsilon > 0$$ such that $$(y - \varepsilon)^n > x$$. But this means $$y - \varepsilon$$ is an upper bound of $$E$$ and $$y - \varepsilon < y = \sup(E)$$, a contradiction.

From all cases above we get contradictions, so $$y = x^{1 / n} \implies y^n = x$$.

Here is my attempt at a solution. Note that for the case $$y^{n} > x$$ I hoped that we could use the result proved in the first induction by setting $$y=k+\varepsilon$$, but so far I haven't been able to prove that there exists a pair $$(k,\varepsilon)$$ such that $$y=k+\varepsilon$$ and $$(k+\varepsilon)^{n} - k^{n} are satisfied simultaneously.

We shall prove the following by induction: For any non-negative real number $$y$$ and for any positive rational number $$q$$ there exists $$\varepsilon>0$$ such that $$(y+\varepsilon)^{n} - y^{n} < q$$. The case $$n=1$$ is obvious. Now suppose the statement has been proven for $$n=k$$. We must show that it holds for $$n=k+1$$. Note that $$(y+\varepsilon)^{k+1} - y^{k+1} = (y+\varepsilon)((y+\varepsilon)^{k} - y^{k}) + y^{k}\varepsilon$$. Let $$q_{0}$$ be a positive rational number smaller than $$q/2(y+1)$$. Such a number exists by proposition 5.4.14. By our induction hypothesis, there exists $$\varepsilon_{0}$$ such that $$(y+\varepsilon)^{k} - y^{k} < q_{0}$$. There also exists $$\varepsilon_{1}$$ such that $$\varepsilon_{1} < 2y^{k}$$. Hence, letting $$\varepsilon =$$min$$(1, \varepsilon_{0}, \varepsilon_{1})$$, we get that $$(y+\varepsilon)^{k+1} - y^{k+1} < (y+1)((y+\varepsilon)^{k} - y^{k}) + y^{k}\varepsilon < q/2 + q/2 < q$$. This completes the induction.

But this shows that there exists $$\varepsilon>0$$ such that $$(y+\varepsilon)^{n} < q + y^{n}\leq x$$, which implies that $$(y+\varepsilon)\in E$$. Thus, $$y$$ is not the supremum of $$E$$, a contradiction.

Next, suppose that $$y^{n} > x$$. Note that this implies that $$y>0$$, since $$y^{n} = 0$$ if and only if $$y=0$$. Then, there exists a positive rational number $$q$$ such that $$y^{n}-x\geq q$$. Thus, if we can show that there exists $$0 < \varepsilon < y$$ such that $$y^{n} - (y-\varepsilon)^{n} < q$$, we are done. Lacking a more elegant solution at the moment, let's do the same induction procedure as above. We want to prove that for any positive real number $$y$$ and any positive rational number $$q$$ there exists $$\varepsilon$$, with $$0<\varepsilon < y$$, such that $$y^{n} - (y-\varepsilon)^{n} < q$$. The base case $$n=1$$ is obvious. Next, suppose we have proven the statement for $$n=k$$. Note that $$y^{k} - y^{k+1} = (y-\varepsilon)(y^{k} - (y-\varepsilon)^{k}) + \varepsilon y^{k} < y(y^{k} - (y-\varepsilon)^{k}) + \varepsilon y^{k}$$. By proposition 5.4.14 (there exists a rational between any two reals) there exists a positive rational number $$q_{0}$$ such that $$q_{0} < q/(2y)$$. By our induction hypothesis, we know there exists $$\varepsilon_{0}$$ such that $$y^{k} - (y-\varepsilon)^{k} < q_{0}$$. Also, let $$\varepsilon_{1} < q/(2y^{k})$$. Then, letting $$\varepsilon =$$min$$(y, \varepsilon_{0}, \varepsilon_{1})$$, we get $$y^{k} - y^{k+1} = (y-\varepsilon)(y^{k} - (y-\varepsilon)^{k}) + \varepsilon y^{k} < y(y^{k} - (y-\varepsilon)^{k}) + \varepsilon y^{k} < q/2 + q/2 = q$$. This closes the induction. Hence, using this $$\varepsilon$$, we get that $$-(y-\varepsilon)^{n} < q - y^{n} \leq -x$$, which implies that $$(y-\varepsilon)^{n} > x$$. Hence $$y-\varepsilon$$ is an upper bound for $$E$$, which contradicts the fact that $$y$$ is the least upper bound for $$E$$.

Since both $$y^{n} and $$y^{n}>x$$ lead to contradictions, we conclude that $$y^{n}=x$$.