Why does 0.999… = 1? (Part 4)

In this series, I discuss some ways of convincing students that and that, more generally, a real number may have more than one decimal representation even though a decimal representation corresponds to only one real number. This can be a major conceptual barrier for even bright students to overcome. I have met a few math majors within a semester of graduating — that is, they weren’t dummies — who could recite all of these ways and were perhaps logically convinced but remained psychologically unconvinced.

Method #5. This is a proof by contradiction; however, I think it should be convincing to a middle-school student who’s comfortable with decimal representations. Also, perhaps unlike Methods #1-4, this argument really gets to the heart of the matter: there can’t be a number in between and , and so the two numbers have to be equal.

In the proof below, I’m deliberating avoiding the explicit use of algebra (say, letting be the midpoint) to make the proof accessible to pre-algebra students.

Suppose that . Then the midpoint of and has to be strictly greater than , since

Similarly, the midpoint is strictly less than :

(For the sake of convincing middle-school students, a number line with three tick marks — for , , and the midpoint — might be more believable than the above inequalities.)

So what is the decimal representation of the midpoint? Since the midpoint is less than , the decimal representation has to be Furthermore, the midpoint does not equal . That means, somewhere in the decimal representation of the midpoint, there’s a digit that’s not equal to . In other words, the midpoint has to have one of the following 9 forms:

midpoint =

midpoint =

midpoint =

midpoint =

midpoint =

midpoint =

midpoint =

midpoint =

midpoint =

In any event, is the largest digit. That means that, no matter what, the midpoint is less than , contradicting the fact that the midpoint is larger than (if ).