Why Memorize? Part 2

Note: This essay and those that following are based on posts I made on Internet usegroups in 2000-2001. At the time, I got precisely the pushback that I expected, which is why I posted in the first place. Let's see if I can present my thesis any clearer this time.

Note: The link to the next essay in this series is at the bottom of this page.

Why Memorize? Part 2

Abstract: Throughout this series of essays on memorization, I will pursue two themes at the same time. The first theme is to define memorization and to advocate for why we need to do it. The second is to define understanding and to advocate for why we need to do that, too. On the controversy of the memorization vs. understanding debate in education, I will not take sides, except to debunk them both. At every level of one's formal educational completion, one should both know more (have memorized more) and understand more than one had at the beginning of it. We need to stop framing this debate in the form of an either-or fallacy and recognize that we need both.

Section 1: You can only claimed to have learned it, if it's in your memory.

Okay, one more time:

Memorization is any process by which information has been stored
in the brain so that the conscious mind can retrieve it both instantly
and accurately. Any piece of information so stored is said to be memorized.
Now, I don't really care how you get information lodged into your memory -- just do it. If you want to use rote (or repetition), do so. There seems to be a conflation of two very different meanings of the word 'rote'. In one sense, it means the use of repetition to memorize, in another sense, it refers to an entire learning strategy, which lends itself keenly to the use of strawmen arguments against any use of rote learning.

I'll tell you a heuristic you should memorize, if you ever want to get good at solving algebra word problems:

Every total is equal to the sum of its parts.

So, the following is a list of general heuristics I cobbled together to help me solve algebra word problems. Take notice first that they're all posed to direct one to find an equation to write down, and second, what could be easier than to identify a part of something or a whole of something?

People often overlook the obvious. -- Dr. Who
  1. If there are any totals or parts of a total lying around, put them into an equation (or into an inequality, if appropriate). Then
    Total = Part1 + Part2 + ... + Partn.
  2. Is there some invariant quantity Q evidently holding from the initial state to the final state of a before-and-after process? If so, write
    Qinitial = Qfinal.
  3. Is there a common or problem-specific formula to use? Such as from physics, chemistry, etc., or from mathematics, like from geometry, statistics, or from number theory, such as for the summation of a series or for a weighted average of a set of numbers, or the greatest common factor or least common multiple of two or more numbers.
  4. Is there a proportion given? A proportion is the stated equality of two ratios:
    a/b = c/d.
  5. Are there one or more linear or quadratic relations given or detected? If so, write them down.
  6. Is there a function given, dependent on one or more discrete variables, each over finite ranges? If so, the problem might be solvable by exhaustive search. For example, the point of the problem may be to find all allowable values of the function satisfying certain given constraints, or it may be a simple matter of finding a minimum or maximum value of the function on its domain. This may not seem like a legitimate algebra problem, but I have found such problem types in algebra problem sets. So, as a practical matter, it's good to keep it in the list of heuristics.
  7. And, add as many additional heuristics as you like.

So, my list of heuristics is all about understanding how to solve word problems, and not just to throw a few memorized formulas at them. Sometimes a familiar problem can be solved by using a common formula, but what do you do when the problem doesn't conform to the requirements of the formula? Then you have to know how to go to first principles and solve the problem like an experienced pro.

Now I want to show the reader the two nonsense 'heuristics' that I used in high school:

  1. Read the problem, and then decide what are the 'critical' variables.
  2. Write down the equations that contain the 'critical' variables.
On comparing this latter list of 'heuristics' to my prior one, it's easy to see how very different they are. According to the prior list, one looks for the word equations presented in the word problem, and then begins the process of refining them until they are converted to pure algebraic form. Along the way, one determines what the 'critical' variables should be.

So, this list of heuristics should address the common problem of students who just don't get it about working word problems: They say that they don't even know how to get started. Well, with this set of heuristics, now they do! Any ten-year old should be able to recognize a part or a total in a word problem. Start there. And if you find either of those, that leads immediately to an equation to write down. But what equation? I'll restate it:

Every total = the sum of its parts.
And we're now one step closer to having a system of equations to solve. Once the student knows what a proportion is, if he or she finds one, he or she can immediately write down its equation. So, one scours the word problem for these key words in order to extract from the problem a set of word equations that can then be translated into a set of algebraic equations to be solved simultaneously, which is a separate skill from finding the equations in the first place.

Section 2: Problem-solving heuristics for the mixed-rate problem.

Let's revisit the mixed-rate problem I presented in the last essay:

I define a mixed-rate problem in the abstract as the situation where two or more 'machines' work together, usually at different rates, to accomplish one task or job. If the rate at which the nth machine works to complete the job on its own is Rn in units of 1 job / hour (to give it a convenient time unit), then we can solve for the time it will take the nth machine working by itself to complete the job as given by

1 job = Rn T.
On solving for T, we get
T = 1 / Rn,
where I have suppressed the units. Now, here's where it gets interesting. Say we now have two machines working together at rates R1 and R2. If the two machines start and stop at the same time, how long will it take them to complete the job? Well, we'll begin by defining this common work time as T. Then, the equation to write down is this
1 [job] = R1 T + R2 T = (R1 + R2) T.
On solving for T, we get
T = 1 / (R1 + R2).
But what if the rates for each machine is instead given to us in inverted units, so that R1 and R2 are replaced by r1-1 and r2-1, respectively, then this last equation would instead be expressed as
T = 1 / (1/r1 + 1/r2),        or        1/T = 1/r1 + 1/r2,
and this result is also commonly expressed as
T = r1r2 / (r1+r2).

Now that we've gone over this example again, what are these abstract 'machines' I keep talking about in real life? How about two printers, working at different rates, to accomplish a certain print job? How about two lawn mowers working together to mow a lawn? How about two painters working together to completely paint a house? How about two pipes working together at different rates to fill a tank? How about two proof readers working together to completely proof a manuscript? How about two fonts at different point sizes working together to fill a page?

By the way, if r1=r2=r, then the total time would be (1/2)r, which makes total sense.

Section 3: A real problem to solve.

Let's solve an actual mixed-rate problem, right here, right now.

Consider the following problem: Printer #1 can print 100 copies of a document in 3.4 hours and Printer #2 can print out the same print job in 2.5 hours. How long will it take for the print job to complete if both printers work on the job together, starting and stopping at the same time?

We introduce the shorthand 'part of job done by' = PJDB. Then our highest-level equation is formed on the basis that the total of one job is the result of the combined effort of two cooperating 'machines'. Obviously, there is a total of ONE job being done by both machines. Hence,

1 job = (PJDB Printer 1) + (PJDB Printer 2).          (1)
Let R1 be the average rate at which Printer 1 can work, which is 1 job/3.4 hours. Likewise, R2 is the average rate at which Printer 2 can work, which is 1 job/2.5 hours. Now, the most general expression we can write for the refinement of the last equation is
1 job = R1T1 + R2T2,          (2)
where T1 and T2 are the respective times that Printer 1 and Printer 2 are operating. For the current problem, these two times are equal and they are equal to the total time the print job takes. Let's call this time T and suppress units, to get
1 = (R1 + R2)T,              (3)
Solving for T, we get
T= 1 / (R1 + R2) = 1 / (1/3.4 + 1/2.5) [hr].      (4)
Thus T = 1.44 hours.

Section 4: Avoiding fragile knowledge.

Let's go back to that quote from Feynman:

I don't know what's the matter with people: they don't learn by understanding; they learn by some other way--by rote, or something. Their knowledge is so fragile!
So, I have spent a great deal of time attempting to demonstrate what I think Feynman meant when he claimed that people's knowledge is often fragile. In the case of solving algebra word problems, if you can't produce on your own the formulas you use to solve them, then your problem-solving abilities in this are fragile, because they can be easily broken. For example, what if Printer 2 starts ten minutes after Printer 1? Yes, that would definitely break the 'formula' we already arrived at. But now we have a new total to account for: The total time that Printer 1 works is equal to the interval it works by itself plus the interval of time it works with Printer 2. So, instead of T = T1 = T2, we write down, T = T1 and, employing another example of the total = the sum of its parts:
T1 = 1 / 6 + T2,               (5)
where I converted 10 minutes to 1/6 hour to keep the units the same.

One rule I follow to avoid getting confused in my efforts to solve word problems is to not go in just one BIG step from a word equation to its algebraic form. That's why I started my solution with the pedantic Eq. (1) and then added Eqs. (2) and (3) before writing down Eq. (4). And if you skip all these 'extra' steps when you are teaching the subject, you can confuse your students.

Algebra Word Problems (Scheme). A general heuristic for solving algebra word problems. Many solved problems presented and the techniques extended to Stoichiometry. (This is word problems for chemistry).

Section 5: The virtues of knowing a lot.

Of course, by knowing a lot, I mean remembering a lot, and that is the same as having memorized a lot. So how much needs to be memorized? My reply is, As much as you can.

In 1984, a year after I graduated from college, I studied the theory of algebraic equations. One of the things I learned was the proof of the Cardano formula for the roots for the general cubic equation. In one or two years after that, I was studying a subject that I thought was completely disconnected to the subject of algebraic equations over the complex numbers. I was experimenting just for fun with extracting roots of hyperintegers (10-adic numbers). In the effort, I came across a cubic equation which I could solve that had no quadratic term. I immediately understood that if I re-interpreted the number set to be the complex numbers over a Clifford algebra of one dimension, that the equation I could solve would be a solution to the so-called "reduced" cubic used in the complex numbers for the solution to the cubic I had studied the year before. Thus I had discovered a completely novel solution for the roots to the general cubic equation, and this was published in a journal with other results from my co-authors. (By the way, the algebra I worked in is called by many names: one is the bicomplex numbers and another is the unipodal numbers.)

The point is this: Had I not placed in my memory that the reduced cubic equation is just as general as the "general" cubic in terms of the roots it could solve for, then I might have discarded the entire discovery as a mere uninteresting "special" case. To me it was a lucid example of Pasteur's "chance favoring the prepared mind." I determined to learn from this fortuitous example in my own life that one trait of success is to have an immense amount of things accurately in memory to call upon instantly in both foreseen and unforeseen situations.

"Oh, just one more thing, ma'am!" Why Memorize, Part 3 .