Check your answer: Building mathematical muscles

Posted by: Gary Ernest Davis on: October 18, 2010

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Teachers everywhere have to pay more attention to learning outcomes. Sometimes these outcomes are very specific to certain courses, even sequences of lessons, and other times they are more generic. One generic outcome that is valued by most teachers of mathematics is a demonstrated ability of, and propensity for, students toÂ check their answers.

What is behind “check your answer” is that students, of any age, should progressively become more independent learners, capable of knowing if their answers are correct.

Students need to build progressively stronger mathematical muscles.

This learning outcome was brought to mind strongly for me when I read the following review of the book The Magic of Numbers by Benedict Gross & Joe Harris:

This book is VERY readable — and easily approachable — BUT, suffers from a major defect for the SELF-LEARNER. There are *tons* of exercises that are recommended — which would be great except doing the exercises is a complete waste of time — you don’t know if you got the exercise right or wrong! Yup. Ok, Now you do it: 1. what’s 2^5 mod 11?, 2. 2^17 mod 23?….etc? Nice BUT if you do the exercises, you don’t know if you got the right answer or not, so its a waste of time.
Worse — the authors emphasize the need to do the exercises — but if you don’t know if you did them right or wrong, what’s the point? There are a fair number of examples that do give answers, but these are about 1/5th – 1/6th the problems and the answers are given in the text of the problem. So while you know the answer, you can’t really do the example, then see if you got the right answer — because it’s also there in front of you.
Having an answer key — even if it was for every even or odd exercise would be helpful. Non-answered questions could be for those classes where the teacher doesn’t trust the student to learn, but just copy answers, but if that’s the case, why bother taking a class? Just copying answers doesn’t help much in the real world nor on exams (which one would presumably have in a classroom setting).

But I bought this book for educational/recreational reading (just like some people like to participate in sports or go to a gym to exercise a body, many find mental exercise equally valuable for one’s “head”).

I would NOT recommend *against* the book, but it may go slowly, as the paucity answered problems make for slower going in terms of comprehension and retention (I find I more often have to reread prior chapter I understood, but didn’t get enough practice to retain in order to progress onto more complex subjects).

For self-learning, FEEDBACK is required — no feedback, then no learning if you are doing it right or wrong, but the format, typesetting, and conversational style on this book make it very readable. Just wish they had the answers (or at least odd numbered answers in the back).

The author of this review is arguing that the exercises in the book have no answers and therefore, adequate learning cannot take place, because that requires feedback, in the form of correct answers.

I feel for the author because it seems they were raised on a diet of textbooks that had answers in the back so that a reader could check if their own answers were correct.

What is wrong with this? Several things. First, the answers in the book might not be correct: sometimes that is the case. Second, taking the book answer as the standard for a correct answer takes the authority for obtaining a correct answer from the student and places it firmly with the author of the book. Thirdly, taking the book answer as correct takes from the learner the possibility of becoming independent by learning how to check if their answer is correct.

Let’s take a look at some examples the review author complains about:

1. what’s 2^5 mod 11?, 2. 2^17 mod 23?….etc?

The section of the book that contains exercises like these is in Chapter 18 on Powers, specifically, in section 18.5 Calculating Powers mod p.

The main tool for calculating powers modulo a prime number $p$ is Fermat’s little theorem (p. 199):

Pirre de Fermat

If $p$ is any prime number and $a$ is any number not divisible by $p$ then $a^{p-1} \equiv 1 \textrm{ (mod} p\textrm{)}$

To apply Fermat’s little theorem to the above exercises we observe, for the first exercise, that 11 is prime and 2 is not divisible by 11 so $2^{10}\equiv 1 \textrm{ (mod} 11\textrm{)}$ .

This does not seem to be a lot of help right off because we want $2^5 \textrm{ (mod} 11\textrm{)}$ and Fermat’s theorem tells us about $2^{10}= (2^5)^2 \textrm{ (mod} 11\textrm{)}$ .

Couldn’t we just calculate $2^5$ directly, and see what the remainder is after division by 11? $2^5=32$ and $2\times 11=22$ so $2^5\equiv 10 \textrm{ (mod} 11\textrm{)}$ .

This seems simple enough, but if we have been fed a steady diet of “check your answer with the back of the book” we might be feeling nervous because maybe we made a mistake. We aren’t getting feedback. How could we check by giving ourselves feedback?

One thing we could do is to use that fact that $2^{10}=(2^5)^2$ . If our calculation of $2^5 \textrm{ (mod} 11\textrm{)}$ is correct then $2^{10} = (2^5)^2\equiv 100\equiv 1 \textrm{ (mod} 11\textrm{)}$ which Fermat’s little theorem says is correct. That’s reassuring.

How might we have made a mistake? Could it be in the calculation of $2^5$ . Not really because we know that $2^1=2,2^2=4,2^3=8,2^4=16 \textrm{ and }2^5=32$ . So if we made a mistake in the calculation of $2^5 \textrm{ (mod} 11\textrm{)}$ it could only have been in working out the remainder after $2^5=32$ is divided by 11. Since $2\times 11=22 \textrm{ and } 3\times 11 = 33$ we see that $2^5=32=2\times 11+10$ so $2^5\equiv 10 \textrm{ (mod} 11\textrm{)}$ .

This seems length and drawn out, and to some extent it is for such a simple exercise. Yet it illustrates the benefit obtained from the cost of checking: we now feel much more confident that our answer is correct. If we looked it up in the back of the book we would not be building up our checking muscles, and we would stay that much mathematically weaker.

How about the second exercise the author of the book review mentions: what’sÂ 2^17 mod 23?

Curiously, when I looked through the book I could not find this exercise. What I did find was:

Exercise 18.6.3 Evaluate the power $2^{83} \textrm{ (mod} 23\textrm{)}$ .

Let’s do this exercise first and think about how we might check the correctness of our calculation without looking up an answer in the back of the book.

Using Fermat’s little theorem we see that $2^{22}\equiv 1 \textrm{ (mod} 23\textrm{)}$ . This because 23 is a prime number and 2 is not divisible by 23.

How does this help us? We need to relate the power 22 to the power 83. Well, $3\times 22= 66 \textrm{ and } 4\times 22=88$ so we use the fact that $83=3\times 22+17$ to get $2^{83}=2^{3\times 22+17}=(2^{22})^3\times 2^{17}\equiv 1\times2^{17} \equiv 2^{17}\textrm{ (mod} 23\textrm{)}$ .

Now the problem is to calculate $2^{17} \textrm{ (mod} 23\textrm{)}$ . The text just before this exercise gives us a hint to use the method of section 18.2

So now we see maybe where the problem “ what’sÂ 2^17 mod 23?” came from – the author of the review has maybe used Fermat’s little theorem and is stuck on calculating $2^{17}\textrm{ (mod} 23\textrm{)}$ .

One way to proceed is to use that fact, from $2^{22}\equiv 1\textrm{ (mod} 23\textrm{)}$ and 22=17+5, that $2^{17}\textrm{ (mod} 23\textrm{)}$ is the inverse of $2^5 \textrm{ (mod} 23\textrm{)}$ .

We know that from a direct calculation that $2^5\equiv 9 \textrm{ (mod} 23\textrm{)}$ so we just have to find the inverse of $9 \textrm{ (mod} 23\textrm{)}$ : in other words solve the equation $9x\equiv 1 \textrm{ (mod} 23\textrm{)}$ . We can try $2^4\equiv 16 \textrm{ (mod} 23\textrm{)}$ so $2^8\equiv (2^4)^2\equiv 1^2=256\equiv 3 \textrm{ (mod} 23\textrm{)}$ . Then squaring again gives $2^{16}\equiv (2^8)^2\equiv 3^2=9 \textrm{ (mod} 23\textrm{)}$ and therefore $2^{17}\equiv 2\times 9=18 \textrm{ (mod} 23\textrm{)}$ .

This gives us our final answer of $2^{83}\equiv 2^{17} \equiv 18 \textrm{ (mod} 23\textrm{)}$ .

If this seems long and complicated it is: it’s messy and we could easily have made a mistake. How can we check our answer?

One way is to absorb more fully the lesson of section 18.2 and construct a table of powers of 2 (mod 23) up to the $22^{nd}$ power. We do this starting with $2^0\equiv 1 \textrm{ (mod} 23\textrm{)}$ and then multiplying the previous result by 2 and keeping the remainder after division by 23. Here’s is a table built using Excel:

Sure enough, we see that $2^{83}\equiv 2^{17} \equiv 18 \textrm{ (mod} 23\textrm{)}$ .

This gives us a much more complete and satisfying feeling than looking up the answer in the back of the book. It helps us build our mathematical muscles, and learn to reason mathematically as an independent thinking person.