Phil Bowermaster is wondering if there's something dodgy about the math here.

No, this is in fact a standard technique for determining the sum of an infinite series, which is in fact what 0.999... is (it could be expressed as the sum, from n=0 to infinity, of the expression 9 times 10 to the minus n). Perhaps, as one commenter notes, it's the word "precisely" that's hanging people up, but certainly that number is equal to one, whatever modifier you want to put on it or leave off.

[Update in the afternoon]

I'm not sure I follow the commenter's objection. He claims that no matter what you start out with as "a" you get a=1. I don't see that.

Try it with two, as suggested.

a = 2
10a = 20
10a - a = 20 - 2
9a = 18

Ergo, a = 2.

In fact, do it with 1.999...

a=1.999...
10a = 19.999...
9a = 18
a = 2

As I said, it's a standard technique for expressing repitends as whole numbers or fractions.

I think the author has walked us into a mathematical trick. no matter what you pick for a the problem looks like (10a - a)/9a = (10-1)/9 = 1, so a could be 2 and it would still equal 1.

Huh?

Where is the division coming from? You can prove that 1 equals 2, but it involves a division by zero. That's not happening here.

Yes, 0.999... = 1. This is confusing to people the first time they encounter it because most people only learn how to manipulate the symbols in their math classes, and therefore don't quite grasp the distinction between a numeral and the number it represents. The system of writing numbers in decimal form (or in a "base-N" system for any N) has the curious feature that certain numbers can be represented in two different ways. For example, one can be written either a "1" or "0.999...". But since the numerals look different, people find it hard to cope with the idea that they represent the same number. Hence the confusion of the commenter at the Speculist who ponders how he would use the symbol-manipulating techniques he learned in 3rd grade to multiply 0.999... by 9, whereby he imagines himself coming up with a 1 digit in the "last" decimal place.

My favorite one of these is:

x^2 = x + x + x + ... {x times} (by definition)

2x = 1 + 1 + 1 + ... {x times} (differentiation with respect to x)

2x = x (combining terms on right side)

2 = 1 (QED)

It's wrong, of course. But it's fun to baffle high school students who've had a month of calculus.

Here's another way to see that .9999... = 1:

1/3 = .333...
3 * 1/3 = .999... = 1

(works the same for 1/9 = .111...)

The most rigorous proof involves showing that there exists no number (real, or rational) that can exist between .999... and 1, which must mean that the two representations are equal to the same number. But that involves more elements of abstract algebra than most people are familiar with.

P.S. anon's problem above is that he has factored out the variable from the equation and proven 1=1.

His equation is:
(10a-a)/9a = 1, which is simply 1 = 1 and tells us no information about the value of a, other than it should not be 0.

P.P.S. For anyone trying to figure out what's wrong with Jane's calculus, here's a hint: the product rule.

Here's more of a hint: does the derivative of x*x equal 1*x or (1*x + x*1)?

The problem is that the equation

9a=9 has a definite, precise answer, and the answer is a=1

Now if you try to conjecture with the polynomial 9a on the left side of the equal, and try to make something with it on the left...

These people obviously are mistaking the equation 9a=9 with a polynomial 9a=?

0.999 does not equal 1.0 or 1.000 or 0.99.

The mistake they make is where they heard that for 2 decimal rounding at the hundredths position

a 0.99 will be rounded to 1.0, notice i left off the hundredths 0.

The otrher problem (possibly) they have is that math is definite, and logical and precise.

Which is difficult hen one brings emotion into play.

emotion is not precise.

Teaching software to behave according to the rules of repeating decimals involves some surprises. You can teach a computer that .999...=1, but it takes some effort. "=" behaves more like a quantum mechanical operation than a classical operation with floating point math.

This is essentially the same as the paradox of the arrow which supposedly can never reach its target, because first it must travel half the distance, then half the remainder, than half of that etc. The implication is that since an infinite number of such steps are required, the arrow can "never reach" its target. In binary (as opposed to decimal) notation this is the same as questioning whether 0.1111... (base 2) is equal to 1. One could substitute "nine tenths of" for "half" and reach the same alleged quandry. But there is no paradox here. In decimal, 0.999... indicates a limit sum, which is the sum of an infinite number of values which themselves approach a size of zero. Not all such limits converge to a specific finite value, but this one does. Maybe the difficulty lies in comprehending that 0.999... indicates only the limit value of this sum, not any partial sum along the way.

Hey Robin

Isn't x*x just x^2? Therefore the derivative is x and not x + x.

Dennis

(gotta read my DE books more often)

No, the derivative of x^2 is 2x, by the standard formula that the derivative of u^n is n*u^(n-1)du/dx, where u is some function of x. In the case of x^2, it's 2 times x to the first power (times dx/dx which is one).

Dennis, yes x*x is just x^2 (which is the intent, in Jane's example), and the derivative of x^2 is 2x. Similarly, the derivative of x*x, using the product rule, is 1*x + x*1, which is 2x as well. The trick Jane pulls is putting a variable off to the side and skipping it when taking the derivative, which produces erroneous results (essentially, when taking the derivative of x*x she only does half and ends up with 1*x, because she differentiates the first x (the x+x+x...) but doesn't finish by differentiating the "x-times" part).

As long as we're on funny proofs and repeating decimals...

I had a coworker once show me a counter argument to cantor's diagonal argument. I actually don't know why or if it's wrong, and it presents an interesting question about the whole concept of countable vs. uncountable sets. Note that he did NOT have an algorithm for mapping the integers to the real numbers, so he didn't convince me all the way that the reals are countable. It is an interesting potential whole in cantor's proof however:

The basic idea is this. We work in binary representation, because clearly it should not matter what base we choose for cantor's diagonal argument. Now we are going to work with the numbers in the set 0 to 1, exclusive of 0 and 1. We can do this because the reals are infinitely dense (I can map all the reals to the reals from 0 to 1, exclusive of 0 and 1). So far we haven't departed from cantor, except possibly to exclude 0 and 1 from our infinite set of reals, and this shouldn't matter.

Now, according to Cantor's argument, we can take any possible list of the reals and construct another real which is not in the list. The method is to make the first digit of the constructed number different than the first digit of the first real number in the list, the second digit of the constructued number different than the second digit of the second real number in the list, and so on down the line forever.

However, consider the following list of numbers (binary, remember):

0.0ssssssssss...(s means 0 or 1, I don't care)
0.s0sssssssss...
0.ss0ssssssss...
0.sss0sssssss...

Thus in my list, 0 is always on the diagonal. Since we are in binary, the number constructed by cantor's diagonal argument is therefore:

0.111111111111...

but this is 1.0. 1.0 was excluded from the set of reals we are working with, so we haven't succesfully constructed a number which shows the reals to be uncountable.

Anyone got a quick explaination of why this counterargument is wrong?

To Jeff:

You should first note that what you *have not* done is construct a counterargument to Cantor. What you have is a flawed proof of Cantor's theorem. That is, the fact that you have *not* successfully constructed such a number is very different from proving that one does not exist, which is what you would need to do for disproving Cantor's argument.

Now the reason the simple diagonal argument fails in the binary case is basically the argument you've shown. If you look at careful base-10 presentations of the Cantor proof, they deal with the 0.999...=1 problem by explicitly specifying that the numbers will be listed preferentially ending with 0.000... and that the constructed diagonal number will contain no 9s (or "n-1"s, in base n). For example, the number may be constructed as having digit k equal to 5, unless the corresponding kth digit of the kth number in the list is a 5, in which case it is a 4.

Of course this restriction is not possible in base 2. Now as you point out, the choice of base is not important; what this really means is that the base-10 arguments are perfectly valid, and doing the proof in any other base is just a means of confusing you.

But if you really want a base-2 proof, then you can do it by considering pairs of digits: that is, choose the first two bits of your number so that they are not the same as the first two bits of the first number, and are also not "11"; and so on. This is of course just a translation of the base-4 proof.

<runs screaming to the next thread>

Jeff;

The flaw is that you don't have to pick the primary diagonal, it's just convenient to do so. For instance, I could pick the diagonal that goes from the bottom left to the upper right. Or I could permute the rows and then look at the diagonal. I would still get a new number not in the table.

Rand

That's what I get for doing derivatives before coffee.

Dennis

(it could be expressed as the sum, from n=0 to infinity, of the expression 9 times 10 to the minus n)

As written, this is not true. It is equal to 9.9999.... It has to be either:
a) Sum(0..inf) 9x10^(-(n+1)) or
b) Sum(1..inf) 9x10^(-n)

As written, this is not true. It is equal to 9.9999...

I was wondering if anyone would catch that. It should have been from one to infinity.

Well! I'm glad my feeble mind can keep up! ;)

It actually took me a couple of readings to parse through what you were trying to say. I've forgotten so much math from college it's disgusting. I don't get to exercise that part of my brain nearly enough.