# Part of a proof in Spivak's Calculus, 4th ed.

• a7d07c8114
In summary, the conversation discusses a proof by Spivak that there is no square root function on the complex numbers. The proof involves a least-upper-bound argument, which the speaker has a question about. The speaker also provides an algebraic solution to the proof but notes that it does not take into account the possibility of a negative square root. The least-upper-bound argument is necessary to eliminate this possibility.

## Homework Statement

I'm following along in Spivak's Calculus, 4th ed. In the chapter on complex functions (page 547 in my copy), Spivak outlines a proof that there is no "square root function" on the complex numbers; he leaves one step to the reader, claiming that it is a standard least-upper-bound argument. I've got a question about that step, but I'll outline the rest of the proof as well for context:

Claim: There is no continuous function f defined on the complex numbers such that for all complex numbers z, $(f(z))^2=z$. In fact, there is no such function even for just the complex numbers z with $\vert z \vert = 1$.

Proof:
Suppose, aiming for contradiction, that such f exists. Without loss of generality, take f(1)=1, since f can always be replaced with -f. This takes care of both cases of $(f(1))^2=1$.

***This is the step left to the reader. Spivak says the argument is "a standard type of least upper bound argument.") ***
We claim that for all $\theta$ with $0\leq \theta < 2\pi$, $f(\cos\theta + i\sin\theta) = \cos(\frac{\theta}{2})+i\sin\theta(\frac{\theta}{2})$.

With the equality established, taking the limit as $\theta$ goes to $2\pi$, f goes to -1. Taking this limit is equivalent to taking the limit of f(z) as z goes to 1, so $\lim_{z\rightarrow 1}f(z)=-1$. But since f(1)=1, f is not continuous at 1, a contradiction.

My small issue with the proof: I did the step left to the reader completely algebraically; I'll show what I did below. What I don't get is the need for any sort of least-upper-bound argument.

## The Attempt at a Solution

My solution to the claim left to the reader:

Since by assumption $(f(z))^2=z$, we have $(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$. Applying the trigonometric identities $\cos 2\theta = \cos^2\theta-\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$, we get
$$(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$$
$$=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos \frac{\theta}{2}$$
$$=(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2$$.

Taking square roots then proves the claim.

Seeing as I was able to prove the claim just via some trigonometric identities, I'm not sure why Spivak says the claim follows from a least-upper-bound argument. I'm not disputing that there is one, but is there a reason why Spivak mentions it specifically when such an easy algebraic argument exists? It seems to go against his style throughout the rest of the book. Maybe the l.u.b. argument is even simpler? Or am I just overlooking something?

a7d07c8114 said:

## Homework Statement

I'm following along in Spivak's Calculus, 4th ed. In the chapter on complex functions (page 547 in my copy), Spivak outlines a proof that there is no "square root function" on the complex numbers; he leaves one step to the reader, claiming that it is a standard least-upper-bound argument. I've got a question about that step, but I'll outline the rest of the proof as well for context:

Claim: There is no continuous function f defined on the complex numbers such that for all complex numbers z, $(f(z))^2=z$. In fact, there is no such function even for just the complex numbers z with $\vert z \vert = 1$.

Proof:
Suppose, aiming for contradiction, that such f exists. Without loss of generality, take f(1)=1, since f can always be replaced with -f. This takes care of both cases of $(f(1))^2=1$.

***This is the step left to the reader. Spivak says the argument is "a standard type of least upper bound argument.") ***
We claim that for all $\theta$ with $0\leq \theta < 2\pi$, $f(\cos\theta + i\sin\theta) = \cos(\frac{\theta}{2})+i\sin\theta(\frac{\theta}{2})$.

With the equality established, taking the limit as $\theta$ goes to $2\pi$, f goes to -1. Taking this limit is equivalent to taking the limit of f(z) as z goes to 1, so $\lim_{z\rightarrow 1}f(z)=-1$. But since f(1)=1, f is not continuous at 1, a contradiction.

My small issue with the proof: I did the step left to the reader completely algebraically; I'll show what I did below. What I don't get is the need for any sort of least-upper-bound argument.

## The Attempt at a Solution

My solution to the claim left to the reader:

Since by assumption $(f(z))^2=z$, we have $(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$. Applying the trigonometric identities $\cos 2\theta = \cos^2\theta-\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$, we get
$$(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$$
$$=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos \frac{\theta}{2}$$
$$=(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2$$.

Taking square roots then proves the claim.

OK, but you can't take square roots just like that. For example, you know that if $x^2=y^2$, then $x=y$ or $x=-y$.

In this case, you know that

$$(f(\cos\theta+i\sin\theta))^2= (\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2$$

this is fine. But this only implies that

$$f(\cos\theta+i\sin\theta)= \cos\frac{\theta}{2}+i\sin\frac{\theta}{2}$$

OR$$f(\cos\theta+i\sin\theta)= -\cos\frac{\theta}{2}-i\sin\frac{\theta}{2}$$

You must still prove that the last possibility cannot occur. This is where the lower bound argument comes in (i.e. let $\theta$ be the smallest number such that the last possibility occurs, ...)

## 1. What is the purpose of a proof in Spivak's Calculus?

The purpose of a proof in Spivak's Calculus is to provide a logical and rigorous explanation of a mathematical statement or theorem. It serves as evidence to support the validity of a mathematical concept or idea.

## 2. How do you write a proof in Spivak's Calculus?

To write a proof in Spivak's Calculus, you must first clearly state the theorem or statement you are trying to prove. Then, you must logically and systematically provide a series of steps or arguments that lead to the desired conclusion. It is important to use precise and concise language, as well as mathematical symbols and notation, to convey your reasoning.

## 3. What is the difference between a direct proof and an indirect proof in Spivak's Calculus?

A direct proof in Spivak's Calculus uses logical reasoning and known facts or definitions to directly arrive at the desired conclusion. An indirect proof, on the other hand, uses contradiction or contrapositive reasoning to prove a statement by showing that its negation or opposite is false.

## 4. How do you know when a proof is considered complete in Spivak's Calculus?

A proof is considered complete in Spivak's Calculus when it has satisfied all the necessary conditions and has provided a convincing argument for the desired conclusion. This includes clearly stating the theorem, providing a logical and valid argument, and using proper mathematical language and notation.

## 5. Are there any common mistakes to avoid when writing a proof in Spivak's Calculus?

Some common mistakes to avoid when writing a proof in Spivak's Calculus include using vague or imprecise language, skipping steps or assuming prior knowledge, and making unsupported claims or assumptions. It is important to be thorough and precise in your reasoning and to provide clear explanations for each step in the proof.