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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 $t = 1$.

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He could probably write $0 < t < 1$; what he's written isn't incorrect, but as you note it is less precise than it could be. Do you want to know how to prove that? I don't see any other issues. You can't "evaluate $t$ at $x = \infty$". – Dylan Moreland Feb 1 '12 at 4:34
@DylanMoreland: So shouldn't he have written $0 < t \leq 1$? – Student Feb 1 '12 at 4:35
Since $x>0$ we have $x+1>0$ so $\frac{x}{x+1}>0$, and thus if $t=\frac{x}{x+1}$ then $t>0\geq 0$. Furthermore, since $x<x+1$ we have $\frac{x}{x+1}<1$ so if $t=\frac{x}{x+1}$ then $t<1$. – Alex Becker Feb 1 '12 at 4:35
@Jon If we had $t=1$, then $x=x+1$ so $0=1$, a contradiction. – Alex Becker Feb 1 '12 at 4:36
@AlexBecker: Thanks! Everything is so clear now. – Student Feb 1 '12 at 4:38
up vote 3 down vote accepted

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 $0\leq t< 1$.

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