Saturday, December 15, 2007
0.99999.. equals 1: Oh no, not another 'Proof!'
For the remainder of this post, the statement 0.99999... = 1 will be denoted by S.
Over the course of my math education and my professional teaching career, S has occupied considerable time and provoked much thought on my part and reflection among my students, countless mathematicians and, now, the math blogosphere (see Polymathematic's famous series of posts!). Sane individuals (aka, non-mathematicians) remain skeptical about S, unwilling or unable to grasp the equality in the statement.
They argue: "0.99999... gets closer and closer to 1 but how can you say it EQUALS 1. There's always a gap!" Ah, the mystery of limits!
For many years now, I have been posting my 'proof' of S on various listservs, discussion groups (including MathShare, the one I moderate) and blogs. Here's the reaction I 've generally received: ____________________
That's right - silence. Because I like to put a positive spin on things, I take that to mean no has found a way to refute it! I've even occasionally heard a student say that this convinced her/him.
I don't want to bore the veterans out there who've heard and read all of the well-known arguments, most of which have 'holes'in them (or should I say, discontinuities!). Even using the basic formula for the sum of an infinite geometric series doesn't necessarily satisfy the Odd Thomases (sorry, I'm a Dean Koontz addict) who will continue to question the validity of the statement.
Any attempt to justify S necessarily requires (to paraphrase Liping Ma) a profound understanding of fundamental principles regarding the real number system and my argument is no different.
Enough already -- Here it is:
Non-Rigorous Explanation: If 0.9999... is less than one, then there must be a decimal between it and 1. But this is impossible!
Rigorous Explanation:
Step 1: Consider the sequence: 0.9, 0.99, 0.999,...
Since this is an increasing sequence of real numbers bounded above by 1, this sequence has a limit, L, namely its least upper bound. As many of you know, I am using the Completeness Axiom for the Reals (known by other names). An excellent reference for the axiomatic structure of the real number system can be found here.
This demonstrates that 0.99999... does exist (i.e., it is a real number). Thus,
0.99999... is the limit L of the above sequence. Verification of the existence of 0.99999... is what is often lacking in other demonstrations of S.
Step 2: L is either greater than 1, equal to 1 or less than 1. We need only consider the last 2 cases.
Step 3: Reasoning indirectly, assume that L<1. By the density property of the real numbers, there must exist at least one real, x, between L and 1. Since L is different from x, it must differ from it in some decimal place. The tenths place? No! Since x is less than 1 and greater than L, it must have 9 in the tenths place. The hundredths place? No, again for the same reason. Need I continue or do you see we've reached a contradiction? Therefore, our assumption that L is less than 1 is false. Thus, L = 1 or, equivalently, 0.99999... = 1. QED!
Ok, your turn! Feel free to critique the proof or present your own favorite argument for or against S. Also, would you consider using this type of argument when teaching this topic?
Posted by Dave Marain at 1:26 PM 14 comments
Labels: indirect proof, infinite repeating decimal, limits, real number system