# Proving the Distributive property of exponents and radicals using bounds $X^(1/n)$

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 $$0 implying $$y_1^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.

$$t_0=1/(z+1)$$ 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 $$\sqrt[n]{x}\sqrt[n]{y}=ab=\sqrt[n]{xy}$$