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The Theorem 1.21 at page 10 of Rudin's Book states that

For every real $x>0$ and every integer $n>0$ there is one and only one positive real $y$ such that $y^n=x.$ This number $y$ is written $\sqrt[n]{x}$ or $x^{1/n}$.

I do not understand the first sentence of the proof, which states that

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

Why is it clear?

I will appreciate any answers.

Sangcheol Choi
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1 Answers1

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Suppose that there were two distinct real numbers $y_1,y_2>0$ such that $y_1^n=x$ and $y_2^n=x$.

Because $y_1\neq y_2$, without loss of generality we have that $y_1<y_2$ (in other words, if it were instead the case that $y_2<y_1$, just relabel them). But as is pointed out in Rudin's argument, this implies that $y_1^n<y_2^n$, and therefore $x<x$, which is a contradiction; therefore our assumption that there were two distinct real numbers whose $n$th power was $x$ was false. Therefore there can be at most one positive real number $y$ such that $y^n=x$.

Zev Chonoles
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  • So, the meaning of "at most" one number $y$ means that there are no two or multiple distinct $y$ that meet the given condition. Thank you for the answer. – Sangcheol Choi Feb 09 '13 at 02:55
  • @Zev Chonoles It's true if $n$ is an odd number but if $n$ is an even number $y$ is not unique. – Yesid Fonseca V. Mar 09 '16 at 15:25
  • @Fonseca: $y$ is assumed to be positive, so regardless of whether $n$ is even or odd, the $y$ is unique. – Zev Chonoles Mar 09 '16 at 21:56
  • @ZevChonoles Yeah, after reading the proof I realized that such $y$ must be positive, however the Theorem doesn't specify, at least in my book (3rth Edition). – Yesid Fonseca V. Mar 10 '16 at 00:12
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    Just for clarification to future readers, the statement of this theorem in the the 3rd ed 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. – smokeypeat Aug 22 '17 at 20:31