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Produced by Charles Wells     Revised 2017-04-03

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DOING MATH: Useful attitudes and behaviors

CONTENTS Self-monitoring Zooming and chunking Rewrite according to the definitions Look ahead Think of examples Trial and error Guessing

Self-monitoring

When you are doing math, watch yourself doing it.

The idea is to install in your brain a Watcher  who watches what you do without being judgmental. Math Ed people call this self-monitoring.

Self-monitoring enables you to discover what works and what doesn't work. When you practice self-monitoring successfully while doing math, then in the long run you will learn more about doing math than you will with any other practice suggested on abstractmath.org.

Abstractmath.org describes mistakes other people make, and sometimes reading about them may help you. But when you watch yourself doing math, you are learning about what you do, not about what other people do.

Example

Suppose you are told that your college has $6$ times as many students as professors and that the total number of students and professors is $1400$.  How many students does your college have?

You start working.  Let $s$ be the number of students and $p$ be the number of professors.  So \[6s=p\text{   and   }s+p=1400\] Then $s+6s=1400$, so $7s=1400$, so $s=200$. So the college has $200$ students. Panic!  That can’t be right!  Let’s see, $6s=p$ and $s+p=1400$.  So plug $p=6s$ into $s+p=1400$ and you get $s+6s=1400$, so $7s=1400$ but that's the same thing I got before!

But now the Watcher  says:  You did the same thing twice and got the same answer.  Maybe it’s time to try something else.  Like for example checking what you did!

Well, $6s=p$ is right, that’s $6$ students for every professor.  No wait a minute, "$6s=p$" means that if there are $100$ students there are $600$ professors. Aargh. I got it backward. Rowrbazzle! I should have written $6p=s$…

Notes

• This example is well-known to math ed people. The conceptual blunder is thinking of $p$ and $s$ as labels instead of as variables.  The math ed people call this the “student-professor problem”.  (Seems to me every math misconception is a student-professor problem?)
• Self-monitoring is a great help in other aspects of life besides math!  Typically, people do it in certain types of activity and not in others (different for different people). Sometimes people avoid thinking about or grappling with problems that they have in certain areas of life, with the result that they don’t solve their problems in that area.

Zooming and chunking

Zooming and chunking is a technique that almost anyone with practice at dealing with abstract math will use to analyze a problem.

Example

This example requires only first year calculus. It is also discussed in Representations of Continuous Functions

Suppose your research has come up with the function \[f(x)=.0002{{\left( \frac{{{x}^{3}}-10}{3{{e}^{-x}}+1} \right)}^{6}}\]

defined for real numbers $x$.  You want to understand how this function behaves.  If I were looking at it, I would go through some process like this:

• Is $f(x)$ defined everywhere? (Meaning, is the denominator never  $0$?)  Yes, ${{e}^{x}}$ is positive for all real numbers $x$, so $3{{e}^{-x}}+1$ has to be positive.
• In fact, I just noticed that $3{{e}^{-x}}+1$ must always be greater than $1$.  This means that $f(x)$ is always less than $.0002{{({{x}^{3}}-10)}^{6}}$.
• $f(x)$ is a positive number times a $6$^th power, so it is always nonnegative. 
• Could $f(x)$ be zero?  Yes, the numerator is zero when $x=\sqrt[3]{10}$, and that is the only place.  (Every real number has exactly one real cube root.)
• When $x$ is positive, $3{{e}^{-x}}+1$ is extremely close to $1$, so when you get very far out on the $x$ axis, $f(x)$ looks just like ${{({{x}^{3}}-10)}^{6}}$.  This means it gets big very quickly as you go to the right.
• …Well, big except near $\sqrt[3]{10}$ (about $2.1544$), where it has to be zero.  So it must dip down to $0$ shortly before it gets to $2.1544$ and that start going up again very fast.
• That means it has to have a local minimum at $\sqrt[3]{10}$.  Ye gods, I don’t want to take that derivative… But I don't have to, I already know there is one local minimum and where it is.
• What about negative $x$?  Well, there the denominator $3{{e}^{-x}}+1$ is very large and grows a heck of a lot faster than any polynomial, so the graph to the left of the $y$-axis ought to hug the $x$-axis (while staying just above it).
• …and so on.

It is worth taking a very close look at what goes on in a mathematician’s head when they go through a process of analysis like this. What happens with me is that I zoom in and out and when I have zoomed out I see things in chunks:

Chunking is a kind of psychological inverse to substitution. See this example.

In more detail

In each step, I am selective in what details I pay attention to and lump the stuff I am not paying attention to into “something”, perhaps with a qualification such as “something $\geq0$”.   The lumping part is called chunking or encapsulation in the math ed literature. (Computer science has a related idea of encapsulation.)

1) Zoom out. $f(x)=.0002{{\left( \frac{\text{something}}{\text{something }>0} \right)}^{6}}$ so it is always defined.

2) Zoom in on numerator: $f(x)=.0002{{\left( \frac{{{x}^{3}}-10}{\text{something }>1} \right)}^{6}}$ so $f(x)<.0002{{\left( {{x}^{3}}-10 \right)}^{6}}$.

3) Zoom out further (ignoring the fact that what is between the parentheses is a fraction): $f(x)=.0002{{\left( \text{something} \right)}^{6}}$, which is $.0002{{\left( \text{something else} \right)}^{2}}$, so it is always nonnegative.

4) Remember a math fact.  Every real number has exactly one cube root, so ${{x}^{3}}-10=0$ when $x=\sqrt[3]{10}$ and nowhere else. 

5) Blur out the denominator, keep everything else exact. Since \[f(x)=.0002{{\left( \frac{{{x}^{3}}-10}{\text{something nonzero}} \right)}^{6}}\] and the only time a fraction is zero when its numerator is zero, $x=\sqrt[3]{10}$ is the only $x$ for which $f(x)=0$. 

And so on…

People are different from each other

Not every mathematician would do this in exactly the way I did.

• Many would have lumped steps (1) and (2) into one step, noticing right away that the denominator was not only positive but greater than 1.  I wrote the steps the way I did because that was in fact the order in which I noticed things. 
• I could have referred to .0002 repeatedly as “something positive”.  In fact when I was going through the process I always thought of it as .0002, so I wrote it that way.  You might do it differently.
• In step (3) I referred to the fact that ${{\left( \text{something} \right)}^{6}}={{\left( \text{something else} \right)}^{2}}$.  I could have noticed that the particular expression \[\text{something else }=\text{ }{{\left( \text{something} \right)}^{3}}\] but that is irrelevant. I did not try to remember more detail than I had to.

I have described a way of understanding something complicated by focusing on one detail and lumping all the other stuff into “something”, or “something with some property”.  This is the zoom and chunk method. 

The graph

The graph tells less than you think

Tthe graph does not show you some aspects of the graph that the analysis above does show. You can’t tell just from looking at the graph that the function is asymptotic to $0$ in the negative direction and increases rapidly to the right of $\sqrt[3]{10}$.  (It appears to do these things but you could be fooled.)  You can’t tell it is zero only at one place, either.

Acknowledgments

Thanks to Olaf Stackleberg for corrections.

Rewrite according to the definitions

Every definition provides a method of proof:

Example definition

Definition: For any integer n:

• $n$ is positive if $n\gt0$.
• $n$ is negative if $n\lt0$.
• $n$ is nonnegative if $n\ge0$.

This definition gives a precise meaning to the words positive, negative and nonnegative. Any question about whether a given integer is positive or negative must be answered by checking this definition.

A theorem based on the definition

Theorem: $0$ is nonnegative.

Proof

The definition of “nonnegative” says $n$ is nonnegative if $n\ge 0$.  So to prove $0$ is nonnegative, we must show that $0\ge 0$.  But that is true because $0=0$.

Similarly, the statement “0 is positive” is false because  “$0\gt0$” is false.

When you are starting to learn about a part of abstract math, the method of rewriting what you want to prove using the definitions is the first thing you should do when faced with a statement to be proved.  In most cases, that will not be enough to give the proof!  But it is a start, and it may suggest what to do next.

Baby examples

The proof above that $0$ is nonnegative is a baby example. Abstractmath.org uses baby examples a lot. Their value is that you can concentrate on what is being illustrated without getting stuck in unfamiliar math. Some students sneer at baby examples. “I already knew that!” But the example above does not exist to teach you something about $0$. It exists to teach you how, when you come across a definition new to you, you use examples to understand it.

There are many different kinds of examples in math, discussed in Varieties of Mathematical Prose, section 2.3.1.

Other examples of proof by rewriting

• The proof in the abmath section about the direct method is a less babyish example of rewriting by definition.
• Another one is the detailed proof in the chapter on presentation of proofs. 
• Notice how often the definition of “divides” is used in these examples. That is typical of proofs in the beginning of a subject.
• More advanced proofs will typically not refer to the definitions much but instead draw on many theorems that have already been proved.

Warning

Whether they use rewrite by definition (at least in their heads) or not is a sharp dividing line between newbies to abstract math and those with even some successful experiences with it. Those with more experience do it automatically and even unconsciously, and some of those who read the description I give here will react like:

But that’s trivial!!  Why did you devote so much space to that??  It’s OBVIOUS.

See ratchet effect.

As math educators who pay attention to their students know, it is not obvious to some beginners. Rewriting according to the definition may seem trivial, but: 

Look ahead

When performing a calculation to solve a problem, you may want to

Example

Suppose you have a right triangle with legs $a$ and $b$ and hypoteneuse $c$.  The Pythagorean Theorem says that

Problem

Derive the trig identity

from the Pythagorean Theorem.

Solution

• This is what you are given: \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]
 
• Look ahead to what you want to prove:  ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
• By using the definition of sine and cosine, this means you want to prove
  \[\frac{{{a}^{2}}}{{{c}^{2}}}+\frac{{{b}^{2}}}{{{c}^{2}}}=1\]
• You can do that this way:  Take the given equation \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\] and divide through by ${{c}^{2}}$, getting \[\frac{{{a}^{2}}}{{{c}^{2}}}+\frac{{{b}^{2}}}{{{c}^{2}}}=\frac{{{c}^{2}}}{{{c}^{2}}}\] But if $c\neq0$, then $\frac{{{c}^{2}}}{{{c}^{2}}}=1$, so the proof is finished. (What happens if $c=0$?)

How did I know to divide by ${{c}^{2}}$?
Answer:  I looked ahead to see what I needed to prove!

Pep Talk

Dividing both sides of a correct equation by a nonzero number gives you another correct equation. 

• You don’t have to have a specific reason for doing it. 
• The new equation doesn’t know how it got there and doesn’t care.
• It is correct and can be used as part of a proof. 

See method addiction and rabbits.

I learned this example from David A. Olson.

Think of examples

One characteristic of people that are good at math is that when they are faced with a new concept, they immediately start thinking of examples:

Example

Suppose you just learned the definition of prime number:

Definition

A positive integer $n\gt 1$ is prime if its only positive divisors are $1$ and $n$. (See definition of prime.)

You might have thoughts like this go through your head:

• $1$ is not a prime because it is not greater than $1$
• $2$ is prime
• $3$ is prime
• $4$ is not a prime because $2$ is a divisor
• $5$ is prime
• $6$ is not prime because $2$ and $3$ are divisors.
• $7$ is prime
• $8$ is not prime because $2$ is a divisor…

• Hold on! If $n$ is prime it has to be odd, because otherwise $2$ is a divisor! Wait: $2$ is a prime anyway even if it is even. So the only even prime is $2$. (You have now discovered your first theorem.)
• So if $n$ is odd it must be prime, because it is not divisible by $2$. (You just discovered what you think is a theorem.)... Hold on! It might be divisible by something else. That’s right, $9$ is odd but it is not a prime! (You just disproved what you thought was a theorem.)
• And so on…

It is vital to generate example of any new concept (if you can!) because that is the fastest road to understanding the concept. See [Selden & Selden]

Trial and error

Example

Suppose you need to know the largest integer n for which $n!\lt109$. One way to do it is to calculate: $4! = 24$, $5! = 120$, so the answer is $n=4$. When I gave a problem that came down to this calculation in my discrete math class, most students solved it correctly, but several wrote apologies on their paper for doing it by trial and error. Apologies are not necessary!

In the case of this problem it is probably the easiest way to do it. Even if you need to know the largest integer for which $n\lt 70,000,000$ it makes sense to do it using trial and error with a calculator or a program such as Mathematica or Maple.

You don't have to start at $5!$ and work your way up.

• Take a wild guess: $20!=2,432,902,008,176,640,000$. Way too big.
• Cut your guess into half: $10! =3,628,800$. Way too small.
• Split the difference: $15!=1,307,674,368,000$. Still awfully big.
• Try $12!=479,001,600$. Nope, go down again.
• $11!=39,916,800$. Got it!

This experiment may have made you realize how fast the factorial grows..

Guessing

Guessing at the answer to a problem and then using a theorem to prove it is correct is legitimate. Some students don't believe this!

Example

You need to find $\int{\sin x \cos x\,dx}$. You vaguely remember that it was $\sin^2 x$. But that is not a proof. So you differentiate $\sin^2 x$ using the chain rule and you get $2\sin x\cos x$. Oops. That's twice what it should be. So the answer must be $\frac{1}{2}\sin^2 x$. Check again (always check again!). The derivative of $\frac{1}{2}\sin^2 x$ is $\frac{1}{2}\cdot 2\cdot \sin x\cos x$ which is $\sin x\cos x$. The fundamental theorem of calculus tells you that $\frac{1}{2}\sin^2 x$ must be the correct answer.

You need not use a specific method (for example substitution or integration by parts) to get an integral. Any way of coming up with the answer is OK as long as you can check it out (in this case by using differentiation).

A teacher might insist that you solve an integral problem by using substitution (for example). Of course, if you practice doing five integrals by substitution you will get better at it. That didn’t stop me from being irritated by such practices when I was a student.

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