$\forall a\in \mathbb R^+ \space\forall n \in \mathbb N \space\space\exists x\in \mathbb R^+ \space\space x^n=a$

Question

I'm trying to prove a theorem. This theorem somehow states that Real Numbers have roots of every degree like $d \in \mathbb N$. By $\mathbb R^+$, I mean the Real numbers which are greater than zero.

Theorem :
$\forall a\in \mathbb R^+ \space\forall n \in \mathbb N \space\space\exists x\in \mathbb R^+ \space\space x^n=a$

Question :
Prove the theorem written above.

Note : I've seen some related examples like proving that there exists a real number like $x$ that $x^3=5$. In those examples, i knew what $a$ is. So i could conclude that a set like $A=\{x\in\mathbb R:x^3\le5\}$ has a smallest upper bound like $S$. Then i'd try to prove that $S=x$. But in this general form, I can't use the same method.

Answer 1

Hint

If $a>0$ and $n\in\mathbb N^*$, consider $x=a^{1/n}$. If $a=0$, just consider $x=0$.

Answer 2

Consider (for $a>0$) $$x:=e^{\frac 1n \ln(a)}.$$

Answer 3

The exponential map induces a group isomorphism $ \exp : \mathbb R \to \mathbb {R^{+}}^\times $, and $ \mathbb R $ is a divisible group.

Answer 4

Is sufficient to prove that the sequence $x_1=1+\frac{1}{n}(a-1)$ $$ x_{k+1}=x_{k}-\frac{x_{k}^{n}-a}{nx_{k}^{n-1}}=x_{k}-x_k\cdot\frac{1}{n}+\frac{a}{n\cdot x_{k}^{n-1}},\quad k\geq 1 $$ converge to a positive real number $r=\lim_{k\to \infty}x_k$. In fact, if $\lim_{k\to \infty}x_{k}=r$ we have \begin{align} \lim_{k\to \infty}x_{k}=\lim_{k\to \infty}x_{k+1} \implies & r=r-\frac{r^{n}-a}{n{r}^{n-1}} \\ \implies & \frac{r^{n}-a}{n{r}^{n-1}}=0 \\ \implies & r^{n}-a=0 \\ \implies & r^{n}=a \end{align}