My Top Tip for Tackling Tough Math Problems

I recently came across an algebra problem which doesn’t require any advanced math skills to solve, but still takes quite a bit of thinking power and logic. It’s a great example of how math problems can be hard at any level, whether you are in high school or a math professor somewhere. Here is the problem.

x, y and z are distinct non-zero real numbers such that

Show that x + 1/y can only have one of two values.

The challenge with this problem is there is no ‘textbook’ approach to solving it. We can rely only on our powers of exploration and logical deduction. Have a go at this yourself and see where you get to. Once you’ve solved it or given up on it, read on to see how I approached it.

My approach

One of the first things I always do with a math problem is itemize and clarify what facts we are given to work with. In this case we are not given very much — basically we have a set of three equalities to work with (do you see that there are three equalities and not two?). In addition, sometimes critical information is given in words that we can often gloss over and not take into account. In this case, the words ‘distinct’ and ‘non-zero’ are critical. Mathematically, they tell us that x ≠ y, x ≠ z and y ≠ z. On top of this we are told that none of x, y and z are zero. These facts are likely to be critical to our solution, because otherwise there would have been no need to state them. So I expect to use these facts in one or more places in how I solve this problem.

But these facts don’t help me with an obvious method. One of the things I always tell my students is if you don’t know how to approach it, do something, anything and see if an approach reveals itself. In this case I am going to do something, and I am going to pick the easiest thing I can do, which is to do some rearranging of the equalities. Let’s pick this equality:

Now I am going to do this simple rearrangement:

I find this interesting because I have an x-y on one side and a y-z on the other. So if I did this for the other two equalities, I will end up with the following three rearranged equalities:

Now I notice that each of x-y, y-z and x-z appear once on both sides of these equations, as long as you negate a couple of terms. Here’s what I mean:

And if I multiply the left hand side and right hand side of these three equations together, noting that two negative terms when multiplied create a positive term, we have this:

Using deductive logic

There’s something nice about this derived equality, but what can we deduce from it? First, note that since x, y, z are all non-zero, the denominator on the right cannot be zero (I’ve used one of my facts — yay!). So this derived equality is telling us that there are two numbers a and b, with b ≠ 0, such that a = a/b. The only way this is possible is if:

1. a = 0. But in our case a is (x-y)(y-z)(x-z) which is not zero because we were told that x, y and z are distinct. So we can rule this out. (I’ve used my other fact — yay again!)
2. a ≠ 0 and b = 1. So in our case we can deduce that x²y²z² = 1

We can also note that x²y²z² = (xyz)² and so xyz can only have two values: 1 or -1.

We haven’t yet fully solved our problem, but it seems like we are getting somewhere because we have used all our facts, and we have identified a constraint where a term can only take one of two values.

Wrapping it up

So let’s go back to our original equalities, with our newly discovered fact that xyz = 1 or xyz = -1.

Now let’s take this term x + 1/y which the question wants us to look at. Let’s say that it is equal to some number c. Since y is non-zero (used it again 😀), we can multiply by y as follows:

Now, from our original set of equalities, we can also say that y + 1/z = c. But we also know that xyz = 1 or xyz = -1 so we can say that either 1/z = xy or 1/z = -xy.

In the first case, this gives the following logic:

So now we have the following two equations to work with:

If we subtract the second equation from the first, we get:

If y ≠ 1, we can safely divide through by -(1-y) which gives us c = -1. If y =1, then the only way we would have distinct x, y, z is if z =-1/2 and x = -2, also yielding a value of c = -1. In the second case where 1/z = -xy, we end up with these two equations:

Adding them up and following similar logic, we can conclude that c = 1. Hence x + 1/y can only take the values 1 or -1.

What did you think about this solution? Did you see how we arrived at a solution by playing around until something revealed itself? Feel free to comment.