This is a revision and expansion of the entry on rabbits in the abstractmath article Dysfunctional attitudes and behaviors.

Rabbits

Sometimes when you are reading or listening to a proof you will find yourself following each step but with no idea why these steps are going to give a proof. This can happen with the whole structure of the proof or with the sudden appearance of a step that seems like the prover pulled a rabbit out of a hat . You feel as if you are walking blindfolded.

Example (mysterious proof structure)

The lecturer says he will prove that for an integer $n$, if $n^2$ is even then $n$ is even. He begins the proof: Let $n^2$ be odd” and then continues to the conclusion, “Therefore $n$ is odd.”

Why did he begin a proof about being even with the assumption that $n$ is odd?

The answer is that in this case he is doing a proof by contrapositive . If you don’t recognize the pattern of the proof you may be totally lost. This can happen if you don’t recognize other forms, for example contradiction and induction.

Example (rabbit)

You are reading a proof that $\underset{x\to2}{\mathop{\lim }}{{x}^{2}}=4$. It is an $\varepsilon \text{-}\delta$ proof, so what must be proved is:

• For any positive real number $\varepsilon $,
• there is a positive real number $\delta $ for which:
• if $\left| x-2 \right|\lt\delta$ then
• $\left| x^2-4 \right|\lt\varepsilon$.

Proof

Here is the proof, with what I imagine might be your agitated reaction to certain steps. Below is a proof with detailed explanations .

1) Suppose $\varepsilon \gt0$ is given.

2) Let $\delta =\text{min}\,(1,\,\frac{\varepsilon }{5})$ (the minimum of the two numbers 1 and $\frac{\varepsilon}{5}$ ).

Where the *!#@! did that come from? They pulled it out of thin air! I can’t see where we are going with this proof!

3) Suppose that $\left| x-2 \right|\lt\delta$.

4) Then $\left| x-2 \right|\lt1$ by (2) and (3).

5) By (4) and algebra, $\left|x+2 \right|\lt5$.

Well, so what? We know that $\left| x+39\right|\lt42$ and lots of other things, too. Why did they do this?

6) Also $\left| x-2 \right|\lt\frac{\varepsilon }{5}$ by (2) and(3).

7) Then $\left| {{x}^{2}}-4\right|=\left| (x-2)(x+2) \right|\lt\frac{\varepsilon }{5}\cdot 5=\varepsilon$ by (5) and (6). End of Proof.

Remarks

This proof is typical of proofs in texts.

• Steps 2) and 5) look like they were rabbits pulled out of a hat.
• The author gives no explanation of where they came from.
• Even so, each step of the proof follows from previous steps, so the proof is correct.
• Whether you are surprised or not has nothing to do with whether it is correct.
• In order to understand a proof, you do not have to know where the rabbits came from.
• In general, the author did not think up the proof steps in the order they occur in the proof. (See this remark in the section on Forms of Proofs.) See also look ahead.

Proof with detailed explanations

1. Suppose $\varepsilon >0$ is given. (We are starting a proof by universal generalization.)
2. Let $\delta$ be the minimum of the two numbers $1$ and $\frac{\varepsilon}{5}$). (Rabbit out of the hat. You can “let” any symbol mean anything you want, so this is a legitimate thing to do even if you don’t see where this is all going.{
3. Suppose $\left|x-2\right|\lt\delta$. (We are about to prove the conditional statement “If $\left| x-2 \right|\lt\delta$ then $\left| {{x}^{2}}-4 \right|\lt\varepsilon$” and we are proceeding by the direct method.)
4. Then $\left| x-2 \right|\lt 1$ by (2) and (3). (The fact that $\delta =\text{min}\,(1,\,\frac{\varepsilon }{5})$ means that $\delta \le 1$ and that $\delta \le \frac{\varepsilon }{5}$. Since $\left| x-2 \right|\lt \delta $, the statement $\left| x-2 \right|\lt 1$ follows by transitivity of “$\lt $”. This is another rabbit. WHY do we want $\left| x-2 \right|\lt 1$? Be Patient.)
5. By (4) and algebra, $\left| x+2 \right|\lt 5$. ($\left| x-2 \right|\lt 1$ means that $-1\lt x-2\lt 1$. Add $4$ to each term in this equation to get $3\lt x+2\lt 5$. This is another rabbit, but it is a correct statement!)
6. Also $\left| x-2 \right|\lt \frac{\varepsilon }{5}$ by (2) and (3). ((2) says that $\delta\le\frac{\varepsilon }{5}$ and (3) says that $\left| x-2 \right|\lt\delta$, so $\left| x-2 \right|\lt \frac{\varepsilon }{5}$ follows by transitivity.)
7. Then $\left| {{x}^{2}}-4\right|=\left| (x-2)(x+2) \right|\lt\frac{\varepsilon }{5}\cdot 5=\varepsilon$ by (5) and (6). End of Proof. (This last statement actually shows the algebra.)

Coming up with that proof

The author did not think up the proof steps in the order they occur in the proof. She looked ahead at the goal of proving that \[\left| {{x}^{2}}-4\right|\lt\varepsilon\] and thought of factoring the left side. Now she must prove that \[\left| (x-2)(x+2) \right|\lt\varepsilon\]

But if $\left|x-2\right|$ is small then $x$ has to be close to $2$, so that $x + 2$ can’t be too big. Since the only restriction on $\delta$ is that it has to be positive, let’s restrict it to being smaller than $1$. (The choice of $1$ is purely arbitrary. Any positive real number would do.)

In that case step (5) shows that $\left|x+2\right|\lt5$.. So how small do you have to make to make $\varepsilon$? In other words, how small do you have to make $\delta $ to make $\left| 5(x-2) \right|\lt\varepsilon$ (remembering that $\left| x-2 \right|\lt\delta $). Well, clearly $\frac{\varepsilon }{5}$ will do!

That explains her choice of $\delta$ be the minimum of the two numbers $1$ and $\frac{\varepsilon}{5}$. Notice that that choice is made very early in the proof but it was made only after experimenting with the sizes of $\left|x-2\right|$ and $\left|x+2\right|$.

You can check that if she had chosen to restrict $\delta $ to being less than 42, then she would need $\delta =\text{min}\,(42,\,\frac{\varepsilon }{47})$.

Acknowledgments

Thanks to Robert Burns for corrections and suggestions