Baby Rudin, 2nd edition, chapter 1, exercise 4

Prove for positive x,y, and positive integer n


Doing this through induction on n seems reasonable enough (first prove $x^ny^n=xy^n$ with induction, and then use that) but rudin mentions theorem 1.37 which uses the LUB property to show uniqueness of $y$ for $y^n=x$. In 1.37 he uses binomial expansion and some fancy inequalities to contradict the < and > cases.

How can I use the LUB property to disprove the other two cases and show they must be equal? The closest I got was:

uniqueness follows from $0<y_1<y_2$ implying $y_1^n<y_2^n$ (I don't fully understand how order is implied but he uses this in 1.37 so I'm following suit if someone could link me something on this I would really appreciate it) which implies $\sqrt[n]{y_1}<\sqrt[n]{y_2}$

Let $z=xy$, and $E$ be the set of all reals $t$ such that $t<z$.

$t_0=1/(z+1)$ shows $E$ is not empty, $t=1+z$ shows that there exist bounds, so there is a lowest

Let $z$ be the lowest upper bound of $E$

Suppose $=\sqrt[n]{z}=\sqrt[n]{xy}<\sqrt[n]{x}\sqrt[n]{y}$

Choose k = (I have no idea)

$\sqrt[n]{z+k}=$ In 1.37 rudin uses a binomial expansion for this part $<\sqrt[n]{x}\sqrt[n]{y}$

At this point I'm currently looking into infinite expansions for $\sqrt[n]{z+k}$ to see if I can find some inequality to complete the proof, but I'm a bit in over my head and thought some expert help would be in order. Thank you!


You're working too hard. Let $a=\sqrt[n]{x}$ and $b=\sqrt[n]{y}$. Then $ab$ is positive and $$(ab)^n=a^nb^n=xy$$ By uniqueness of positive $n$-th roots, it follows that $$\sqrt[n]{x}\sqrt[n]{y}=ab=\sqrt[n]{xy}$$

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  • $\begingroup$ The distributive property of positive integer exponents would first have to be proved through induction and then this follows nicely, and now I understand how uniqueness (1.37) allows you to set your initial equalities (thank you!). Probably what he was going for now that I think of it. but still, is it possible with bounds? Just so curious. $\endgroup$ – Matthew Bourgeois Sep 23 at 3:45

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