Which is Always a Natural Number?

(A new question of the week)

I want to look at a question that came in recently that is, in one sense, very simple, but at the same time is quite challenging. It was given to a 12-year-old whose father asked us about it, and requires some skill in thinking about non-routine problems.

Here is the question:

Hello,

a/b=c/d is a proportion where a, b, c, d are natural numbers. Which one of the following is always a natural number? 

a) a/b    b) c/d   c) (a+b)/(c+d)   d) (ad)/(bc)

I have added parentheses in the choices, according to what I think was intended.

Notice a few things about this question:

  • First, that word “always” is what makes it challenging; although the student is not asked for a proof, it essentially requires being able to prove that an expression represents a natural number for all natural number values of the variables. It is unlikely that a student at this level would be familiar with proofs.
  • Second, it is a necessarily multiple-choice question, in that it could not be solved by doing some calculations and then choosing which of the answers matches the result. You are required to look at each choice and decide whether it can be proved.

How can you approach such a problem?

Doctor Rick took the challenge:

Hi, Jamal. As with the other problem you have submitted, we’d like to know what your son has thought about the problem so far. We want to help him do most of the thinking himself, because that’s how the mind is exercised. If he can’t think of anything to do yet, I can provide some initial thoughts, but then, ideally, we would like to interact with him in developing these ideas further (even if that must be through you).

This problem is not just an exercise checking the student’s ability to apply methods that have been learned; it will take some more creative thinking. A 12-year-old may not be ready to approach this abstractly from the start, so I suggest he start by trying out an example. Can he write a proportion of the type described, with particular natural numbers in place of a, b, c, and d? Can he check whether each expression among the options is a natural number?

If you show me your son’s example and the values he gets for each expression, it will give me a good idea of how well he understands the question. That’s a good place to start. If he can come up with a good example and it already shows him that only one answer can be correct, that’s great. He may then want to discuss why that expression will always be a natural number, and how we can be sure (that is, prove it), and how we could come up with the answer without using an example. There is a lot of potential learning in this problem, if he is interested!

This is definitely a case where the only way to help the student really learn what this question is intended to teach is to get him to do the work. Any answer we would give would take away the experience of working it out.

The recommendation given here is not only applicable to this particular unusual problem; it is the way to approach any unfamiliar problem. First, we think best concretely, so we often suggest experimenting with a problem using specific numbers rather than general variables, in order to get a feel for how it works. If nothing else, this is a way to get started rather than just staring at the problem and worrying about it.

Also, this idea is related to an important issue involved in proofs: To prove that something is always true can be a difficult task, but to prove that it is not always true requires only a counterexample — one case in which it is not true proves that it is not always true. So when you are asked whether a statement is always true, trying examples is an excellent way to start. As a result, this question may be answered merely by elimination, having shown that all but one fail. That in itself would be unsatisfying to me; I would really want to have at least some reason to believe that the remaining statement is always true. But trying to convince yourself that one thing is true is much easier than doing so with all four possibilities!

Jamal took some time to respond, first telling us that he had not asked his son about it, and then asking for help in understanding it himself. I then wrote to him, filling in some details about the process:

Let me demonstrate what Doctor Rick had in mind.

Here is the problem:

a/b=c/d is a proportion where a, b, c, d are natural numbers. Which one of the following is always a natural number?

  1. a) a/b    b) c/d   c) (a+b)/(c+d)   d) (ad)/(bc)

When you are faced with a very unfamiliar problem, experimenting (which I call playing with the problem) is a good way to start, just as children faced with an unfamiliar world start to understand it by playing — trying things and seeing what happens.

We have discussed this often in the past; for example, in 2017, Doctor Ian said this:

Insufficient Information In Question

... when you don't know what to do, the worst thing to do
is just throw up your hands and do nothing. Try things. Play around! See
what you can conclude from what you've been given. Try to establish what
you would need to know to be a step or two from a solution -- like working
a maze backward as well as forward.

And in 2003, I said this:

Playing with Equations to Solve Problems

So my approach is to experiment (the more adult word for "play"!) and 
make a conjecture (the more adult word for "guess"), and then prove 
that conjecture.

So, rather than starting by trying to prove something about all natural numbers, we can just try an example. A simple proportion of the type shown would be 1/2 = 2/4, so a=1, b=2, c=2, d=4. Putting those into the four given expressions, we get

A: a/b = 1/2: not a natural number

B: c/d = 2/4: not a natural number

C: (a+b)/(c+d) = (1+2)/(2+4) = 3/6 = 1/2: not a natural number

D: (ad)/(bc) = (1*4)/(2*2) = 4/4 = 1

So we have already found that the answer has to be (D), if it is anything. If all that matters is the answer, we can move on. But we shouldn’t stop there, if we are interested in math at all!

Notice, however, that if I had chosen a different example, I might not have had an answer yet. For example, suppose I used 2/1 = 4/2:

A: a/b = 2/1 = 2 is a natural number

B: c/d = 4/2 = 2 is a natural number

C: (a+b)/(c+d) = ((2+1)/(4+2) = 3/6 = 1/2: not a natural number

D:(ad)/(bc) = (2*2)/(1*4) = 4/4 = 1 is a natural number

So I would only have eliminated one possibility.

Now, you could try some more numbers and see if either continues to work; if so, you might think about why it works, and convince yourself that it will always be true. Ideally, after doing this you will realize that it could have been “obvious”, if you had thought about it the right way — but we rarely approach a problem the right way the first time!

The goal of play is to observe. As we try things, we discover facts. We might, for example, realize that since a/b = c/d, A and B are always the same — if either is a natural number, then both are. And we can easily pick numbers so that they aren’t. In other words, our initial random selection can change over into a deliberate selection of a counterexample.

Now, why is D correct (as opposed to merely not incorrect)? Can we see a way to prove it?

A student who has been learning about proportions will probably have learned that, if a/b = c/d, then ad = bc. (This was once described as “The product of the means equals the product of the extremes”, and is now typically called “cross multiplication”.) From that fact, we see immediately that (ad)/(bc) = 1, always. So we might have immediately seen that this is, as I said, “obvious”. But it is only obvious when you look from the right perspective. If you didn’t initially see this, then perhaps realizing that in every example the ratio is 1 might make you look more closely at why ad and bc might always be equal.

For more on this, see

Why Does Cross Multiplication Work?
Rule of Three
Means and Extremes

I concluded with general advice:

A major lesson of this sort of problem is: Just try something! Doing nothing can’t help you solve a problem; doing something may change it enough that you can see what to do. If you don’t see what you are looking for, move, and see if it is visible from a different perspective. You don’t have to know how to solve the problem; you just have to keep your eyes open for possibilities as you walk around it.

We have often said that we can steer you better if you are already moving — when we see your efforts, we can see what direction you are going, and correct you, but if you try nothing, or show us nothing, then we have no way to correct you (apart from urging you to start). But this advice (which I have given in a couple face-to-face tutoring sessions recently) goes a step beyond that: “Motion” not only reveals where you are going, but it also takes you to a new place where you may be able to see the problem better.

This happens in many situations. In simplifying a fraction, you may not know how to fully simplify all at once (using the GCF), but you can usually find some common factor and divide by that. The new numbers are smaller, making it easier to see the next common factor to use (or to know there are none). Moving closer to the goal makes it more visible. Other times, you may not be closer to the goal, but you will have changed your perspective, as in some of the combinatorics problems I have discussed recently. Looking at it in a different way, making a different but equivalent problem, can make new methods available.

Jamal replied:

Sir,

Issue was resolved

When you move, things are visible from different perspective. Thanks for excellent suggestion.

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