Q:

Rudin's assertion that if $t = x/(1 + x)$ then $0 \leq t < 1$

I'm having trouble understanding one step in the proof of Theorem 1.21 in Rudin's Principles of Mathematical Analysis.
Theorem 1.21 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 he makes the following claim: Let $E$ be the set consisting of all positive real numbers $t$ such that $t^{n} < x$. If $t = \frac{x}{1 + x}$ then $0 \leq t < 1$. 
I don't understand how he got that inequality. If $t = 0$  that implies that $x = 0$ which is a contradiction since every $x > 0$. And if $x \rightarrow \infty$, then $t = 1$.

A:

Notice that in this proof $x$ is a fixed positive real number, and that we are assuming $t=\frac{x}{x+1}$.
Since $x>0$, we have $x+1>0$ so $t=\frac{x}{x+1}>0$ hence $t>0$ thus $t\geq 0$. Furthermore, since $x<x+1$ and neither of these are $0$ we have $t=\frac{x}{x+1}<1$. Putting these together gives $0\leq t< 1$.

