## The Big Zero

I want to try re-proving the little point from last time, that the chance of picking one specific number from the range of zero to one is actually zero. This might not seem like a big point but it can be done using a mechanism that turns out to be about three-quarters of all the proofs in real analysis, which is probably the most spirit-crushing of courses you take as a mathematics undergraduate, and I like that it can be shown in a way that you can understand without knowing anything more sophisticated than the idea of “less than or equal to”.

So here’s my proposition: that the probability of selecting the number 1/2 from the range of numbers running from zero to one, is zero. This is assuming that you’re equally likely to pick any number. The technique I mean to use, and it’s an almost ubiquitous one, is to show that the probability has to be no smaller than zero, and no greater than zero, and therefore it has to be exactly zero. Very many proofs are done like this, showing that the thing you want can’t be smaller than some number, and can’t be greater than that same number, and we thus prove that it has to be that number.

Showing that the probability of picking exactly 1/2 can’t be smaller than zero is easy: the probability of anything is a number greater than or equal to zero, and less than or equal to one. (A few bright people have tried working out ways to treat probabilities that can be negative numbers, or that can be greater than one, but nobody’s come up with a problem that these approaches solve in a compelling way, and it’s really hard to figure out what a negative probability would mean in the observable world, so we leave the whole idea for someone after us to work out.) That was easy enough.

Now to show the probability of picking exactly 1/2 can’t be greater than zero. It’s a bit harder than the first half, which is a curious phenomenon of proofs that divide so neatly into two conceptual halves — often proving one half is dramatically easier than the other. If you encounter this, do the easy half first, as you’ll want the partial credit, and occasionally it offers some insight into the harder half. (In this case it doesn’t, but it can happen.)

Proving this half is, I think, better done with a first draft that gets the idea across, and then a second draft that makes the case sound.

The first draft: well, whatever the probability of picking exactly 1/2 is, it’s got to be smaller than the probability of picking “some number between 1/4 and 3/4”, since, after all, there’s so many more numbers in that wider range. The range between 1/4 and 3/4 is a length of half the conceivable numbers (we might pick anything from 0 through 1), so, the probability of picking exactly 1/2 has to be less the probability of picking some number between 1/4 and 3/4, which is 1/2.

Similarly, the probability of picking exactly 1/2 has to be less than the probability of picking “some number between 2/5 and 3/5”. Between 2/5 and 3/5 are one-fifth of all the possible numbers we might pick, so the probability of picking a number in that range is less than or equal to 1/5. So the probability of picking exactly 1/2 has to be less than the probability of picking some number between 2/5 and 3/5, so it’s less than 1/5.

And we can work this out for any range of numbers that include 1/2: the probability of picking exactly 1/2 has to be less than the probability of picking “some number between 9/20 and 11/20”, which is going to be 1/10. The probability of picking exactly 1/2 has to be less than the probability of picking “some number between 49/100 and 51/100”, which is 1/50. Pick a range as small as you like that contains 1/2; the probability of picking exactly 1/2 has to be less than the probability of picking a number in that range. We could make a range smaller than any positive number we like, so, the probability of picking exactly 1/2 can’t be any bigger than zero.

That’s probably got you convinced. It’s probably convinced anyone that this is really true, although it’s not quite in presentable shape for a proof. Why not? Mostly because what’s given up there are a lot of examples, each of which is true, but they’re all examples of something. What’s that something? That something is, usually, what the proper proof is. It’s usually harder to think of that than to think of specific examples, like the first draft offered, and that’s why very often a proof is easier to work out that way and then clean up to the presentable second draft.

So, the second draft: pick any positive number; since it needs a name, call it ε. The probability that a randomly picked number between 0 and 1 is exactly 1/2 has to be less than the probability that this randomly picked number is between the number 1/2 – ε and the number 1/2 + ε. And the probability that the picked number is between that lower and that upper bound is at most 2ε.

So the probability we have picked exactly 1/2 is less than 2 ε. But since ε can be any positive number, this means that the probability has to be less than any positive number. The probability has to be no greater than zero.

Since the probability has to be no smaller than zero (the first part of the proof), and has to be no greater than zero (as we just did), then it’s got to be equal to zero.

This is an awful lot of mechanism for the point proved, but the tools used are potent. And the proof can be easily adapted to show the chance of picking any particular number is zero, not to mention that the probability of picking any finite collection of numbers (for example, of picking any of 1/2, or 1/3, or 2/3, or 4/5) is still zero. The advanced student is probably already thinking of even more remarkable things this routine can be used to prove. And if you’ve got this down, you’ve got at least introduced to the worst part of any mathematics major’s third undergraduate year.

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