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Section 4.3 Roots

In this class we will be working a lot with roots of numbers like \(\sqrt{2}\text{,}\) but like \(\sqrt{2}\text{,}\) that they exist at all takes justification.

We proceed by induction on \(n\text{.}\) When \(n=1\text{,}\) this is immediate. Now assume that our claim holds for \(n=k\text{;}\) we aim to prove the claim for \(n=k+1\text{.}\) If \(x\lt y\text{,}\) then \(x^k \lt y^k\text{,}\) and so \(x^{k+1} \lt y^{k+1}\text{.}\) Conversely, if \(x \geq y\text{,}\) then either \(x=y\text{,}\) in which case \(x^{k+1}=y^{k+1}\text{,}\) or else \(y \lt x\text{,}\) in which case \(y^{k+1} \lt x^{k+1}\text{.}\)

We proceed by induction on \(n\text{.}\) When \(n=1\text{,}\) this follows from the Week 3 Exercise. Now assume that the claim holds for \(n=k\text{;}\) we aim to prove the claim for \(n=k+1\text{.}\) If \(y^{k+1}\lt x\text{,}\) then \(y^k \lt x/y\text{,}\) and so by hypothesis, there is a real number \(z_0\) such that \(y\lt z_0\) and \(z_0^k \lt x/y\text{.}\) Thus \(y\lt x/z_0^k\text{,}\) so there exists a real number \(z_1\) such that \(y\lt z_1\lt x/z_0^k\text{.}\) Now if \(z\) is the lesser of \(z_0\) or \(z_1\text{,}\) then \(y\lt z\text{,}\) and \(z^k \lt x\text{.}\)

Let us prove the uniqueness first. Let \(y_1,y_2\gt 0\) be real numbers such that \(y_1^n=y_2^n=x\text{.}\) Using Lemma 4.3.1, we find that if \(y_1>y_2\text{,}\) then \(y_1^n>y_2^n\text{,}\) and if \(y_1 \lt y_2\text{,}\) then \(y_1^n \lt y_2^n\text{.}\)

We now aim to show that such a real number \(y\) exists. Consider the set

\begin{equation*} S = \{z\in \RR : (z>0)\wedge(z^n\geq x)\}\text{.} \end{equation*}

The set \(S\) is nonempty 1 , and by definition, \(0\) is a lower bound for \(S\text{.}\) Hence we may define \(y=\inf S\text{,}\) which is nonnegative. We now aim to prove that \(y^n=x\text{.}\)

Why?

If \(y^n \lt x\text{,}\) then by Lemma 4.3.2, there exists a real number \(z\) such that \(y\lt z\) and \(z^n\lt x\text{.}\) It follows from Lemma 4.3.1 that \(z\) is a lower bound for \(S\text{,}\) but this now contradicts the fact that \(y\) is the greatest lower bound of \(S\text{.}\)

On the other hand, if \(y^n \gt x\text{,}\) then \((1/y)^n \lt 1/x\text{,}\) so there exists a real number \(z\) such that \(1/y \lt z\) and \(z^n \lt 1/x\text{.}\) Thus \((1/z)^n \gt x\text{,}\) whence \(1/z \in S\text{.}\) But \(y \gt 1/z\text{,}\) which contradicts the fact that \(y\) is a lower bound for \(S\text{.}\)

We can now define any rational power of a positive number: if \(\frac{m}{n}>0\) is rational and \(x>0\text{,}\) we define

\begin{equation} x^{\frac{m}{n}} = \left(x^{\frac{1}{n}}\right)^m\text{.}\label{e_x_m_n}\tag{4.3.1} \end{equation}