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Counterintuitive Maths: Part Two

By Richard Hughes

17
Mar. 10

Cross posted from Divisible By Pi.

Obviously this post has been in the works for a little bit now, with minor delays due to university and the Secular Society. While it’s obviously annoying that the article has been delayed (to me, anyway), there has been an unexpected silver lining because of it. You see, on Friday I went to a talk put on by the Melbourne University Maths and Stats Society (MUMS) that happened to be on exactly the topic I was planning to write on; through that talk, I learnt some interesting facts about the topic. For example:

  • The topic has its own FAQ.
  • This topic has generated enough talk to fill up 15 archives of talk/​discussion.
  • Plus an additional 9 archives of arguments.
  • The page has been edited over 3000 times.
  • And the debate is still raging on.

From the above, can you guess what bit of counterintuitive maths I’m going to discuss today? The answer is below the fold.

If you guessed the oft debated 0.\dot{9} = 1 debacle, you’ve guessed correctly. Why this topic seems to cause such a stir when it gets brought up is a bit of a mystery to me, and once you have a bit of maths under your belt the result seems obvious, but there’s no denying that from a layman’s point of view the result is counterintuitive and potentially even a bit weird.

I figure I should probably start up first with a few proofs of the result. I’m going to show two proofs here, one that is easy to follow but not necessarily as convincing as it should be, and one that uses a bit more maths (not too much, I promise) but that is much harder to argue against.

First up is a stock standard algebraic proof:

\displaystyle \begin{array}{r l} x &= 0.999\ldots \\ 10x &= 9.999\ldots \\ 10x - x &= 9.999\ldots - 0.999\ldots \\ 9x &= 9 \\ x &= 1 \\ \end{array}

Pretty easy to follow, right? But it raises some questions, such as:

  • How do we know that multiplication works like that on infinite decimals?
  • How do we know that subtraction is well-​​defined for infinite decimals?

It turns out, of course, that multiplication and subtraction can and do work like that; the problems aren’t so insignificant that we should just hand wave them away, however (indeed, it shows that the person is really thinking about the proof, which isn’t a bad thing).

So, instead of manipulating terms algebraically, this second proof is based on what we actually mean by a decimal representation, coupled with a little bit of analysis.

To begin with, take note that when we write a decimal such as 0.123, what we are really saying is:

\displaystyle 0.123 = \frac{1}{10} + \frac{2}{100} + \frac{3}{1000}

Or more generally:

\displaystyle 0.a_1 a_2 a_3 \ldots a_n = \frac{a_1}{10^1} + \frac{a_2}{10^2} + \frac{a_3}{10^3} + \cdots + \frac{a_n}{10^n}

If we’re dealing with an infinite decimal representation, then we get:

\displaystyle \begin{array}{r l} 0.a_1 a_2 a_3\ldots &= \displaystyle\frac{a_1}{10^1} + \frac{a_2}{10^2} + \frac{a_3}{10^3} + \cdots \\ \hfill &= \displaystyle\sum_{i=1}^{\infty}{\frac{a_i}{10^i}} \end{array}

Okay, so far so good. So, what is our representation for 0.999…? Easy!

\displaystyle 0.999\ldots = \displaystyle\sum_{i=1}^\infty{\frac{9}{10^i}}

Now, here is where the fun really comes in. Say you want to evaluate a sum of the form

\displaystyle ax + ax^2 + ax^3 + \cdots + ax^n = \displaystyle\sum_{i=1}^n{ax^i}

Well, it turns out that there’s a very simple formula that will let you do that:

\displaystyle ax + ax^2 + ax^3 + \cdots + ax^n = \displaystyle\sum_{i=1}^n{ax^i} = \frac{a(x^{n+1} - x)}{x-1}

It’s even easy to verify this formula — just note that

\displaystyle (x-1)(ax + ax^2 + ax^3 + \cdots + ax^n) = a(x^{n+1} - x)

Now, provided that \displaystyle |x|<1, we can easily extend this formula to give us the sum of an infinite series, giving us the result

\displaystyle ax + ax^2 + ax^3 + \cdots = \displaystyle\sum_{i=1}^{\infty}{ax^i} = \frac{ax}{1-x}

Now it’s just a matter of plugging in the numbers. Note that for 0.999…, a=9 and x=0.1, which gives us:

\displaystyle \begin{array}{r l} \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots &= \displaystyle\sum_{i=1}^{\infty}{\frac{9}{10^i}} \\ &= \displaystyle\frac{\left(\frac{9}{10}\right)}{1-\left(\frac{1}{10}\right)} \\ &= \displaystyle\frac{\left(\frac{9}{10}\right)}{\left(\frac{9}{10}\right)} \\ &= 1 \\ \end{array}

It’s hard to argue with real analysis — now you probably see why a bit of maths makes the result seem obvious. Nevertheless it’s easy to understand why some people find it a counterintuitive result. But what of people who refuse to accept the maths? Where do they actually go wrong?

It seems that dealing with the fact that the decimal continues on to infinity is one of the major issues people have — often there is an expectation that there is a final 9 ‘at infinity’, and that there is therefore some kind of infinitesimal difference between the two numbers. Alternately, you get people like the following:

I will try and explain your misconception if you are only willing to listen. You are entirely right that one third is the unique limit of the sequence 0.3, 0.33, 0.333, etc. But we are talking about a single number here, and not of a limit of a series. Yes, the limit is one third; but that is only an approximation. You may find it useful to read further explanations at Limit_(mathematics).

The idea that 0.\dot{3} \approx \frac{1}{3} and 0.\dot{9} \approx 1 is a popular one, but it’s also entirely incorrect — the limit of a sequence going to infinity is not an approximation to what will happen, but an exact value. Each finite length element in the sequence, i.e. 0.3, 0.33, 0.333, \ldots are approximations, but the infinite length 0.\dot{3} is exact. On one level, it’s just another situation where infinities behave differently to how we expect they shoud behave (more on that in the next installment, wink wink nudge nudge).

Another problem some people have involves confusing a particular representation of a number with the number itself. \frac{2}{2}, 1, 3-2 and 0.\dot{9} are all representations of the same number — the way it is written doesn’t change the number itself. This subtlety is lost on some people, however, and you wind up with arguments like the following:

Consider a function Flip(x) that has the result of pivoting all base ten digits around the ones place. This could be seen as a decimal equivalent of taking the reciprocal, or misreading base ten as base one tenth. (Doing this in another base would be a different function.)

[…]

So, what is Flip(0.999…)? Well, it is obviously …9990. But for various interesting reasons within standard mathematics, that number is actually negative ten! A totally different answer from Flip(1). So .999… can’t be equal to 1 because its flip value isn’t even close to Flip(1).

Ouch. In trying to get my head around what this guy was actually trying to say I think I pulled a brain muscle that might never fully recover. This guy’s reasoning is completely nucking futs.

The first issue that needs to be addressed with his ‘proof’ is the assertion that \ldots 9990 = -10. That assertion is extremely wrong, and a perfect example of why a little bit of knowledge can be a dangerous thing. See, in the real numbers there’s just no way known for that statement to be true. –10 is a well-​​defined number, we can use it, manipulate it, and just generally twist it around until it hollers Uncle. …9990, on the other hand, is not well-​​defined in the real numbers. In fact, it’s undefined, which is pretty much the opposite direction to where this guy was heading.

In other number systems, however, things are free to behave differently — so much so that in the 10-​​adic numbers, …9990 actually does equal –10. You can’t just transfer that over to say that the two numbers are therefore equal in the real numbers though. It’s like noticing that there are a lot of white people in Sweden, and concluding that the same must be true about Sierra Leone.

Sadly, that’s not even the biggest problem with his supposed proof. His biggest problem relates exactly to what I mentioned before: Confusing the representation of a number with the number itself. His statement that his proof shows that 0.\dot{9} \neq 1 is based on the implicit assumption that each number has a unique decimal representation — in reality, however, that’s simply not the case. The effect is that while he thinks he’s defined a function he actually hasn’t, because the same number (0.99… and 1) can be sent to two different values.

Of course, then sometimes people have reasons that don’t relate to the actual maths at all:

BUT, one reason many people don’t accept it is simply because in “real life” (as opposed to “mathematics”) there’s no such thing as 0.9999…

In “real life”, meaning reality, in order for a number like 0.999… to exist it would have to be constantly adding more 9s to itself.
In other words, in real life there are no infinitely large numbers (be it in length or size).
So in real life 0.999… doesn’t even exist to begin with, let alone to be “equaled” to 1.

Again, we see the misconception that 0.999… is some kind of process rather than a number in its own right. When you realise that, you realise that the statement that 0.999… doesn’t exist ‘in real life’ is nonsense — 0.999… = 1, and since we can identify ‘1’ in real life (philosophical arguments can go shove themselves), we can necessarily identify 0.999…

And then, just to finish on a crazy note, sometimes you get guys like this:

You have to be sure though when claiming that it is one, to point out that it is really (1 — .0000…1)
While the difference between .999… and 1 is infinitely nothing, it cannot be dismissed because it is everywhere.
.000…1 is the first thing there is greater than zero, and it’s between every change between every two numbers all the way up to one. It’s everywhere, but it’s nothing. I believe that it is the graviton number, but I cannot prove it.

I’m sure his Nobel Prize will be coming any day now… *backs slowly away*

Tags: counterintuitive, maths, Richard Hughes

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11 Responses to “Counterintuitive Maths: Part Two”

  1. 11
    Richard Hughes says:
    March 26th, 2010 at 3:52 pm

    Leon:

    I don’t think that for this particular proof an epsilon-​​delta proof is really necessary; that falls out of the fact that there’s actually a lot of maths required to properly define what series are and how they behave in the first place. Once you have that — which is implicitly assumed here, as I don’t want to give half a course on real analysis — you already have the background that tells you you can manipulate series in the necessary ways.

    Of course, the proof I’ve given here is not in any way as rigorous as it probably should be. But I don’t think you need to go all the way to an epsilon-​​delta proof in this case.

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