1
$\begingroup$

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

Prove for positive x,y, and positive integer n

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

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!

$\endgroup$
1
$\begingroup$

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 $$\sqrt[n]{x}\sqrt[n]{y}=ab=\sqrt[n]{xy}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.