Are there positive solutions to $x^q- a = 0$ other than $\sqrt[q]{a}?$ I'm reading a book that says the following (translated):
"Before we proceed, let us remember that, given a real number $a > 0$ and a integer $q > 0$, the symbol $\sqrt[q]{a}$ represents a positive real number such that its $q$-power equals to $a$, that is to say it's the only positive solution of $x^q-a=0$".
My whole problem is to show that there is one, and only one, positive real root to $x^q-a=0$. In the case $x^2 - a = 0$ I already couldn't go further. Any ideas? Thanks in advance.

Suppose that we know there is a positive number $\sqrt[q]{a}$ such that $(\sqrt[q]{a})^q-a=0$, but we don't know yet if there is a different positive number $x$ such that $x^q-a=0$. Using the identity
$$
w^n-z^n=(w-z)(w^{n-1}+w^{n-2}z+\dots+z^{n-1})
$$
we see that the equation $x^q-a=0$ is equivalent to
$$
\left(x-\sqrt[q]{a}\right)\left(x^{q-1}+x^{q-2}(\sqrt[q]{a})+\dots+(\sqrt{a})^{q-1}\right)=0 \, .
$$
One of the bracketed terms on the LHS must be equal to zero for the above equality to hold. Since the second bracketed expression is positive, we must have $x-\sqrt[q]{a}=0$, and so $x=\sqrt[q]{a}$.
