Given $x>0,n\in \mathbb{N}$ prove there is a unique $y>0$ such that $y^n=x$.

Question

Given $x>0,n\in \mathbb{N}$ prove there is a unique $y>0$ such that $y^n=x$.

Let $S=\{s\in\mathbb{R} : s^n\leq x\}$

Since $0=0^n<x$ then $0\in S$ thus $S$ is a nonempty subset of $\mathbb{R}$ and so there is a least upper bound of $S$

Let $y=lub(S)$

I want to prove that $y^n=x$

Suppose $y^n<x$

then for some $\epsilon>0$, $y^n<y^n+\epsilon<x$, Since there is no real number closest to $x$

then there exists a $\delta>0$ such that $\vert y+\frac{\delta}{2}-y\vert<\delta$

implies $\vert y^n - (y+\frac{\delta}{2})^n\vert<\epsilon$

then $(y+\frac{\delta}{2})^n<\epsilon+y^n<x$

so $y+\frac{\delta}{2}\in S$ but clearly $y<y+\frac{\delta}{2}$ thus $y\neq lub(S)$ a contradiction.

Answer 1

I can fix your approach. You want to find some $\epsilon>0$ such that $y^n<(y+\epsilon)^n<x$ in order to get a contradiction. Well, this is the same as finding $\epsilon$ such that $$(y+\epsilon)^n-y^n <x-y^n=\delta$$.

This is of course obvious, if we know continuity of functions, but what I understand is that you just want to prove this with just inequalities (trichotomy of real numbers) and the fact that the real numbers are complete (that your bounded sets have lower upper bounds).

Claim: $(y+\epsilon)^n-y^n<\epsilon n x^{n-1}$. Indeed, $$(y+\epsilon)^n-y^n=((y+\epsilon)-y)((y+\epsilon)^{n-1}+(y+\epsilon)^{n-2}y+\dots +y^{n-1})<\epsilon(nx^{n-1})$$

Using the claim, then we just need to take $\epsilon=\delta/(nx^{n-1})$ and we are done.

Answer 2

Firstly we have to prove $f(x)=$ $x^n$ is a 1-1 on interval (0,+$\infty$) For $n=1$ it is trivial. For $n>1$, we have to prove. Suppose f(a)=f(b). Then $a^n=b^n$. By factoring, $(a-b)(a^{(n-1)}+a^{(n-2)}b...........b^{(n-1)})=0$. As second factor can't be zero as $a,b>0$. So $a-b=0$, i.e., $a=b$.

Now we have to prove that every positive real no. has unique positive $n^{(th)}$ root., $y=(x_0)^{(1/n)}$. If n=1, $y=x_0$, simply Suppose $n>1$, and $x_0>0$. Consider a polynomial function $p(x)=x^n$. As $p(0)=0$ and $p(x)$ diverges to +$\infty$. So, there exist, $b$ s.t. $ p(b)>x_0$. By intermediate value theorem, there exist $y$ belong to$(0,b)$ s.t. $ p(y)=x_0$. Uniqueness follows from first proof.

Answer 3

The easiest way to solve this problem is to perform an analysis of the function $f : x \mapsto x^r$ on $\mathbb{R}_+$.

$f$ is differentiable, and $f'(x) = rx^{r-1}$. For any positive $x$, $f'(x)>0$ hence $f$ is sctrictly increasing, whence injective. So if we find $y$ such that $y^r = x$, it will be unique.

Let $S=\{s\in\mathbb{R} : s^n\leq x\}$, as in OP.

To prove the existence of such a $y$, use intermediate value theorem : it is easy to find $b$ such that $b^r > x$ (any $b > \sup(S)$ will do). Since $f$ is continuous and $f(0) = 0 < x$ and $f(b)>x$, there must be $y \in [0, b]$ such that $f(y) = x$.

Answer 4

Set

$f(y) = y^n; \tag 1$

note that $f(y)$ is continuous. This follows from the fact that the function $y$ is continuous, and products of continuous functions are continuous; more formally, we may use induction, observing that for $k \in \Bbb N$,

$y^{k + 1} = yy^k; \tag 2$

then if $y^k$ is continuous, $y^{k + 1}$ is; the base case is simply the case $k = 1$, i.e.

$y^2 = yy. \tag 3$

It is easy to see that $f(y)$ is strictly monotonically increasing on $[0, \infty)$, that is

$y_2 > y_1 > 0 \Longrightarrow y_2^n > y_1^n; \tag 4$

this of course follows from the elementary property of multiplication

$0 < a < c, 0 < b < d \Longrightarrow ab < cd. \tag 5$

Next, we have

$f(0) = 0, \tag 6$

as is trivially obvious, and if we choose

$z > 0 \tag 7$

such that

$1 < z, x < z, \tag 8$

then repeated application of (5) yields

$x < z^n = f(z); \tag 9$

in light of (6) and (9) it follows from the intermediate value theorem that

$\exists y \in [0, z], y^n = f(y) = x; \tag{10}$

the uniqueness of $y$ follows easily from (4), which shows that $y_1 \ne y_2$ implies that $y_1^n \ne y_2^n$; one may also argue algebraically that

$y_1^n = y_2^n \Longrightarrow y_1^n - y_2^n = 0$ $\Longrightarrow (y_1 - y_2)(y_1^{n - 1} + y_1^{n - 1}y_2 + \ldots y_1y_2^{n - 2} + y_2^{n - 1}) = 0, \tag{11}$

and since, unless $y_1 = y_2 = 0$,

$y_1^{n - 1} + y_1^{n - 1}y_2 + \ldots y_1y_2^{n - 2} + y_2^{n - 1} > 0, \tag{12}$

we conclude that

$y_1 - y_2 = 0, \tag{13}$

or

$y_1 = y_2, \tag{14}$

that is, the solution to $y^n = x$ is unique.