Roots of Unity

Theorem ($n$th Roots)

For every $x>0,x\in \mathbb{R}$, and any $n\in \mathbb{N},$ there exists a unique $y>0,y\in \mathbb{R}$ such that $y^n=x$.

Proof: (Uniqueness)

Recall if $0<a<b$ and $n\in \mathbb{N}$, then $a^n<b^n$. This can be proved with induction. Suppose $\exists y_1\ne y_2\in \mathbb{R}$ such that

$$y_1^n=x=y_2^n$$

Either $y_1<y_2$ or $y_2<y_1$. In either case, by our fact from above, $y_1^n\ne y_2^n$ .

Proof: (Existence)

We want to show that $\exists y\in \mathbb{R},y>0$ such that $y^n=x$. Let

$$A=\{z\in \mathbb{R}\mid z^n<x\}$$

Note,

• $A\ne \varnothing ,0\in A$
• $A$ is bounded above. Indeed, $z>x+1⟹z^n\ge z>x⟹z\notin A⟹x+1$ is an upper bound. Then, $y:=\sup (A)$ exists by the Completeness Axiom. Note $y\ge 0$. We claim $y^n=x$.

Suppose for the sake of contradiction, that $y^n\ne x⟹y^n>x\vee y^n<x$.

Case 1: $y^n<x$

1. $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +a{b}^{n-2}+b^{n-1})$

Our goal is to find some $r\in (0,1)$ such that $(y+r{)}^n<x$. If we can do this, then $y+r\in A$ and $y+r>y=\sup (A)$, a contradiction.

We want to show, $y^n<(y+r{)}^n<x$ as this is equivalent to $(y+r{)}^n-y^n<x-y^n$. Using fact $1$ we have:

$$r\left((y+r{)}^{n-1}+(y+r{)}^{n-1}y+\cdots +(y+r)y^{n-2}+y^{n-1}\right)<x-y^n$$

Since $r\in (0,1]$, we can get a uniform bound on these terms.

• Indeed, the first term is $(y+r{)}^{n-1}<(y+1{)}^{n-1}$.
• The second term $(y+r{)}^{n-2}y<(y+1{)}^{n-2}y<(y+1{)}^{n-2}(y+1)=(y+1{)}^{n-1}$ is bounded by the same term.
• The second last term, $(y+r)y^{n-2}<(y+1)(y+1{)}^{n-2}=(y+1{)}^{n-1}$
• The last term, $y^{n-2}<(y+1{)}^{n-1}$

So, if

$$r(n(y+1{)}^{n-1})<x-y^n⟺r<\frac{x-y^n}{n(y+1{)}^{n-1}}$$

Choose $r$ to be half of the RHS. Then we are fine. However, the RHS may be larger than $1$, so we can say

$$r=\min \left\{\frac{1}{2}\text{RHS},\frac{1}{2}\right\}$$

Case 2: $y^n>x$ Sketch: Find $r\in (0,1]$ such that $y-r$ is still an upper bound for $A$. Then $y-r<y$ which is contradiction with the Supremum. We want $r<\min \{1,y\}$ such that

$$x<(y-r{)}^n<y^n$$

This means $y-r$ is an upper bound for $A$. This is the same as

$$y^n-(y-r{)}^n<y^n-x$$

We can use the identity $1$ from earlier to show an upper bound.

$$y^n-(y+r{)}^n=r(y^{n-1}+y^{n-2}(y-r)+\cdots +(y-r{)}^{n-1})$$

So, we can minimize this by maximizing $r$. So, each term is smaller than $y^{n-1}$. We get

$$y^n-(y-r{)}^n\le r(n{y}^{n-1})$$

We can let

$$r=\min \left\{\frac{1}{2},\frac{1}{2}\cdot \frac{y^n-x}{n{y}^{n-1}}\right\}$$

Definition

Let $x>0$ and $n\in \mathbb{N}$. Then, we denote $x^{1/n}$ as the unique $y\in \mathbb{R},y>0$ such that $y^n=x$.

Corollary

If real $a,b>0$ then $(ab{)}^{1/n}=a^{1/n}\cdot b^{1/n}$.

Proof:

By definition, $(ab{)}^{1/n}$ is the unique $y>0$ such that $y^n=ab$. Similarly, $a^{1/n}$ and $b^{1/n}$ is the unique $y_1$ and $y_2$ such that $y_1^n=a$ and $y_2^n=b$.

So,

$$(a^{1/n}\cdot b^{1/n}{)}^n=(y_1\cdot y_2{)}^n=y_1^n\cdot y_2^n=ab$$

Then

$$(a^{1/n}\cdot b^{1/n}{)}^n=n⟺a^{1/n}\cdot b^{1/n}=n^{1/n}$$

and we are done.