# Using Dedekind Continuity to Show That $\sqrt[\leftroot{-2}\uproot{2}n]{a}$ Exists for Each $n\in\mathbb{N}$ and Any $a\in\mathbb{R}^{+}$.

I'm doing the exercises of Kuratowski's Intro to Calculus. Chapter 1 Exercise 5 says:

Applying the continuity principle prove that $$\sqrt[\leftroot{-2}\uproot{2}n]{a}$$ exists for each $$n\in\mathbb{N}$$ and any $$a\in\mathbb{R}^{+}$$.

I have no clue on how to solve this problem just using the principle of continuity. The following is the principle of Dedekind continuity as explained by Kuratowski:

1.5 The principle of continuity (Dedekind)

Among the properties of real numbers, we shall mention the so-called continuity principle. This principle states that if we divide the set of all real numbers into two non-empty sets $$A$$ and $$B$$ such that any number belonging to the set $$A$$ is less than any number belonging to the set $$B$$, then there are two possibilities: either there exists a greatest number of the set $$A$$, or there exists a least number of the set $$B$$. In other words, if we divide the straight line in two parts $$A$$ and $$B$$ in such a way that any point of the part $$A$$ lies on the left side of any point of the part $$B$$, then either there exists a last point in the part $$A$$ or a first point in part $$B$$. No "gap" may be found in this "cut" which we have performed. This implies the continuity of the set of real numbers distinguishing this set from the set of rational numbers, where such gaps exist; e.g. dividing the set of rational numbers in two parts such that all rationals $$\lt\sqrt{2}$$ belong to the first part and all other rational numbers, i.e. numbers $$\gt\sqrt{2}$$ belong to the second part, it is easily seen that any number of the first part is less than any number of the second part, but there does not exist any rational number which would be the greatest in the first part or the least in the second one (this follows from the fact that $$\sqrt{2}$$ may be approximated arbitrarily accurate by the rational numbers from below and from above, e.g. by the decimal expansions).

• The first part just says $\sup A=\inf B$ and $\Bbb R=(-\infty,\sup A)\cup [\inf B,+\infty)\ \lor \Bbb R=(-\infty,\sup A]\cup(\inf B,+\infty)$. In your case, $A, B\subset\Bbb Q, \sup A=\inf B=\sqrt{2}\notin\Bbb Q\implies \sup A\notin A\ \land\ \inf B\notin\Bbb Q$. – Invisible May 22 at 15:26

Take $$A$$ to be the set of real numbers $$r$$ such that $$r^n < a$$ and $$B$$ to be the set of real numbers $$r$$ such that $$r^n > a$$.

• I'm sorry, I don't follow.. How does this cut imply the existence of $\sqrt[n]{a}$? – Alex D May 22 at 15:07
• I think you meant $r^n<a$. – Invisible May 22 at 15:22
• OP wrote $\sqrt[\leftroot{-2}\uproot{2}a]{n}$ in his post. – green frog May 22 at 15:25
• I switched up the letters while typing I'm sorry.. – Alex D May 22 at 15:26
• @AlexD You can proceed as follows. First suppose it doesn't exist. Then you can partition the reals into these two sets $A, B$. Then consider the real number $x$ that is $sup A$ and $inf B$. – green frog May 22 at 15:31