At most one $n$th root of a positive real: Proof

Question

Seemingly very simple, but I am confused by a proof, which is fr0m Rudin's Mathematical Analysis book. The theorem is:

"For every real $x>0$ and every integer $n>0$ there is one and only one positive real $y$ such that $y^n=x$."

In the proof, it says quickly: "That there is at most one such $y$ is clear, since $0<y_1<y_2$ implies $y_1^n < y_2^n$."

It is not immediately clear to me why this is so. The ordering is preserved but I cannot see why this should imply that there is at most one $y$. What is the underlying idea here?

Answer 1

IF there were two $n$th roots, $y_1$ and $y_2$, then one would have to be less than the other, say $y_1 < y_2$ by trichotomy law. Raising them to the $n$th power preserves the order, as you say, so $x = y_1^n < y_2^n= x$. So $x<x$, contradiction.

Answer 2

If there are two such $y$, $y_1, y_2$, say, then $y_1< y_2$ or the other way round. Consequently $y_1^n < y_2^n$ (or the other way round), in particular only one of them may equal $x$

Answer 3

Usually the property of order you want to use is that multiplying by a positive number preserves order, which is not quite the same as taking powers, but it implies it. That is, order says that if $a < b$ and $c > 0$ then $ac < bc$. So if we have $0 < y_1 < y_2$ then you use it twice by noting that $y_1$ and $y_2$ are both positive $$ y_1^2 = y_1 y_1 < y_1 y_2 < y_2 y_2 = y_2^2 $$ The rest is induction on this idea. This often just gets used without mentioning, but it is always good to know how you'd go directly back to the axioms rather than just using intuition.