# Someone Else’s Homework: Some More Thoughts

I wanted to get back to my friend’s homework problem. And a question my friend had about the question. It’s a question I figure is good for another essay.

But I also had second thoughts about the answer I gave. Not that it’s wrong, but that it could be better. Also that I’m not doing as well in spelling “range” as I had always assumed I would. This is what happens when I don’t run an essay through Hemmingway App to check whether my sentences are too convoluted. I also catch smaller word glitches.

Let me re-state the problem: Suppose you have a function f, with domain of the integers Z and rage of the integers Z. And also you know that f has the property that for any two integers ‘a’ and ‘b’, f(a + b) equals f(a) + f(b). And finally, suppose that for some odd number ‘c’, you know that f(c) is even. The challenge: prove that f is even for all the integers.

Like I say, the answer I gave on Tuesday is right. That’s fine. I just thought of a better answer. This often happens. There are very few interesting mathematical truths that only have a single proof. The ones that have only a single proof are on the cutting edge, new mathematics in a context we don’t understand well enough yet. (Yes, I am overlooking the obvious exception of ______ .) But a question so well-chewed-over that it’s fit for undergraduate homework? There’s probably dozens of ways to attack that problem.

And yes, you might only see one proof of something. Sometimes there’s an approach that works so well it’s silly to consider alternatives. Or the problem isn’t big enough to need several different proofs. There’s something to regret in that. Re-thinking an argument can make it better. As instructors we might recommend rewriting an assignment before turning it in. But I’m not sure that encourages re-thinking the assignment. It’s too easy to just copy-edit and catch obvious mistakes. Which is valuable, yes. But it’s good for communication, not for the mathematics itself.

So here’s my revised argument. It’s much cleaner, as I realized it while showering Wednesday morning.

Give me an integer. Let’s call it m. Well, m has to be either an even or an odd number. I’m supposing nothing about whether it’s positive or negative, by the way. This means what I show will work whether m is greater than, less than, or equal to zero.

Suppose that m is an even number. Then m has to equal 2*k for some integer k. (And yeah, k might be positive, might be negative, might be zero. Don’t know. Don’t care.) That is, m has to equal k + k. So f(m) = f(k) + f(k). That’s one of the two things we know about the function f. And f(k) + f(k) is is 2 * f(k). And f(k) is an integer: the integers are the function’s rage range). So 2 * f(k) is an even integer. So if m is an even number then f(m) has to be even.

All right. Suppose that m isn’t an even integer. Then it’s got to be an odd integer. So this means m has to be equal to c plus some even number, which I’m going ahead and calling 2*k. Remember c? We were given information about f for that element c in the domain. And again, k might be positive. Might be negative. Might be zero. Don’t know, and don’t need to know. So since m = c + 2*k, we know that f(m) = f(c) + f(2*k). And the other thing we know about f is that f(c) is even. f(2*k) is also even. f(c), which is even, plus f(2*k), which is even, has to be even. So if m is an odd number, then f(m) has to be even.

And so, as long as m is an integer, f(m) is even.

You see why I like that argument better. It’s shorter. It breaks things up into fewer cases. None of those cases have to worry about whether m is positive or negative or zero. Each of the cases is short, and moves straight to its goal. This is the proof I’d be happy submitting. Today, anyway. No telling what tomorrow will make me think. 