Existence of Nth root, by Rudin soft question

Question

I have a quick question: Why does Rudin choose to let $t=\frac {x}{x+1}$ so that $t<1$ ?

Indeed, Rudin States:

Theorem: For every real $x>0$ and every integer $n>0$ there is one and only one real $y$ such that $y^n=x$

Proof: That there is at most one such $y$ is clear, since $0<y_1<y_2$ implies $0<{y_1}^n<{y_2}^n$.

Let $E$ be the set of all positive real numbers $t$ such that $t^n<x$. If $t=\frac {x}{x+1}$ then $0<t<1$. Thus $E$ is not empty.

He continues the proof which I understand, but how is it no loss of generality by making $t<1$ (I am guessing that there is a bijection between $[0;1]$ and $R$, if so how to make that rigorous, if not how does he allow himself to do such a thing).

Answer 1

At that point, all that Rudin wants to prove is that there is some $t>0$ such that $t^n<x$. Taking $t=\frac x{x+1}$ works, because $\frac x{x+1}<x$ and, since $\frac x{x+1}<1$,$$\left(\frac x{x+1}\right)^n\leqslant\frac x{x+1}<x.$$

Answer 2

That he chose $t = \frac x{x+1}$ is only to show that $E = \{t > 0|t^2 < x\}$ is non-empty. (because $\frac x{x+1} > 0$ and $(\frac {x}{x+1})^n < x$). Any other value would do.

He chooses $t$ to satisfy two conditions. 1) $0< t < 1$ and 2) $t < x$. Therefore $t^n < t < x$.

If $x > 1$ then any $t \le 1$ would do as $t^n < t \le 1 < x$.

If $x \le 1$ then any $t < x$ will do as $t^n < t < x$.

But if you want to avoid doing two cases and just find one case then any $t < 1$ and $t < x$ will do as $t^n < t < x$.

So if it were me I'd say. "Let $t < \min(1,x)$, say for example $t = .9*\min(1,x)$".

Rudin's a little more ... graceful.

If $t = \frac {x}{x+1}$ then $t < 1$ so $t^n < t$. And $t < x$ so $t^n < t < x$.

In any event. $E$ is not empty.