Why 0.99999… = 1, proof, and limits

If you took calculus, believe it or not, you have the tools to prove that 0.9999… = 1 .  Think back to the concept of a “limit.”  OK, I really don’t expect you to remember it, so please be patient as I remind you and hopefully give you some intuition *why* it works.

To prove it to yourself that 0.9999… = 1, consider that if they weren’t equal, there would be a number E that is greater than zero such that E = (1 — 0.9999…).  So now we have a game.  You give me a candidate value for E, say 0.0001, and then I can give you a number D of 9’s repeating which causes (1 — 0.9999…) to be smaller than E (in this case 0.99999 (D = 5), because 1 — 0.99999 < 0.0001 ).   Since we’re playing this game, you counter and make E smaller, say 10^(-10), and I turn around and say “make D = 11” (because  1 — 0.99999999999 < 10^(-10) ).  Every number E that you give me, I can find a D.  Specifically, if E > 10^(-X) for some positive integer X, then setting D = X will do it.  It’s a proof by contradiction.  There is no E that is greater than zero such that E = (1 — 0.9999…).  Therefore 0.999… = 1.

This game is the definition of a limit.  We define the limit as x approaches infinity of f(x) = y if for every epsion > 0 (E) there exists a delta (D) such that x > delta implies that |f(x) — y| < epsilon.  Here f(x) = SUM from n=1 to x of 9 * (10^(-n)) = 0.9… [with 9 repeated x times].  As it turns out, all limits are proof by contradiction.  If the two quantities we are comparing aren’t equal, then there would be a non-zero difference, and by playing this game “for every epsion, there exists a delta”, you prove that such a non-zero difference doesn’t exist, and therefore the two quantities are equal.

5 responses to “Why 0.99999… = 1, proof, and limits

  1. But we also remember from calculus class that the limit of f(x) as x->y need not be the same as f(y) if the function f has a discontinuity at y and as such I find this proof lacking since you haven’t demonstrated the lack of such a discontinuity.
    Consider a boolean function lt1(x) defined as:
    1 if x is less than 1
    0 if x is >= 1
    and see what happens when we pass your f(x) = SUM… as an argument. We see that as x goes to infinity lt1(f(x)) remains 1, while lt1(1) is 0, which suggests that, even in the limit of infinite nines, .99999… is somehow different than 1.
    Of course, that’s not the case, but limits are not the right mathematical tool to make the proof, since they only talk about what happens when you approach, not what happens when you get there.
    A simpler proof is achieved through algebra:
    Let x = 0.99999…
    10x = 9.99999…
    10x — x = 9.00000…
    10x — x = 9
    9x = 9
    x = 1

  2. [Disclosure, Eric was a college suite-mate of mine.… Thanks for commenting!]
    Let me give you the punchline: You have just argued that you can’t use the definition of a “limit” to what is shorthand for a limit!
    Taking a step back, your argument uses limits about a point, and indeed a function can have a discontinuity about that point. I am talking about infinite limits, which have a slightly different definition.
    But instead of getting into the details of the difference in the definition of a limit about a point and an infinite limit, let me point blank ask you: What does 0.9999… mean? What is it short hand for? It literally means the act of appending 9’s after the decimal point infinitely. Mathematically that means 0.9 + 0.09 + 0.009 + …, which we write in shorthand (imagine the correct mathematical notation):
    SUM_(from x=0 to infinity) of (0.9 * 10^x)
    But how do you calculate an infinite sum? That by definition is
    LIMIT_(from n=0 to infinity) of SUM_(from x=0 to n) of (0.9 * 10^x)
    0.99999… is SHORTHAND FOR AN INFINITE LIMIT! So I the infinite definition of a limit is the appropriate way to understand it. When we learn decimal notation, we haven’t learned the limit yet. The concept of a limit is hard, which is why 70% OF ELEMENTARY SCHOOL TEACHERS INCORRECTLY THINK THAT 0.9999… 1. I am passionate about this since this is one of the facts that could really help people understand infinite limits. It should be one of the first examples that allow students to go “a‑ha!” instead of just having them blindly memorizing the definitions.

  3. I suppose that you can think of 0.999… as representing an infinite limit, but that’s not how I tend to think of it. I think of it as really being an infinite number of nines after the decimal point. Infinite numbers really are different than finite numbers, even finite numbers in the limit as they approach infinity.
    Since we’re talking about pedagogy, a great book on this which is easily approachable to young people is George Gamow’s “One Two Three…Infinity”, which I read at the age of about 12 or 13 and I recommend as required reading to anyone interested in mathematics.

  4. I am not a mathematician, but I appreciate theoretical discussions about physical concepts. I admit I don’t remember from my calculus days the implications of this concept but I appreciate both of your comments.

Comments are closed.