Comments on AngryMath: Arguing Infinite Decimals

To me, that seems like a fairly roundabout way of ...

2011-11-11T03:22:45.487-05:00

To me, that seems like a fairly roundabout way of saying that for consistency's sake, we accept the distributive law for real numbers (and thus combining like terms, such as that for all x, 10x-x = 9x). :-)

I'm roleplaying as a high school math prodigy....

2011-11-10T08:50:34.162-05:00

I'm roleplaying as a high school math prodigy. I agree that all my life, teachers have told me 10x-x=9x, without proof (none have ever mentioned the axioms of a complete ordered field!) Suddenly a teacher wants to prove something to me-- that 0.999...=1-- and that's great, but along the way, he uses a different fact which was never proved, the distributive law. And indeed, according to my calculations-- multiplying 9 by 0.999...-- it seems as if 10x-x is *NOT* 9x. Unless 8.999... happens to equal 9, but how can I establish that?

There are 3 ways out of this dilemma:

1. explicitly assume the axioms of a complete ordered field and generalize our discussion to any isomorphic copy thereof. The problem: the definition in this blog post of "0.999..." doesn't make sense. If reals are taken to be Dedekind cuts or Cauchy sequence equivalence classes, what's a "digit"? It requires a lot more work.

2. Work in an explicit model of the reals (e.g., Dedekind cuts) and explicitly prove the distributive law as well as the other axioms. Again, "0.999..." becomes very difficult to define, UNLESS the model we are using is Stevin's Construction.

3. More realistically for high school: Don't try to prove 0.999...=1. Instead, say something like: "Numbers are a tool, and as a tool, we would like them to have the property that distinct numbers have nonzero difference. The difference between 1 and 0.999... seems to be 0, so in order to force numbers to have the distinct-numbers-have-nonzero-distance property, we simply DEFINE 0.999... to be 1. Because otherwise, as you see by this example about 9 times 0.999..., very basic facts like the distributive law would break." Finally the math prodigy is able to understand, and it makes beautiful perfect sense at last, without having to go into graduate level math.

But Step #3 is a subtraction (not a multiplication...

2011-11-10T00:35:53.960-05:00

But Step #3 is a subtraction (not a multiplication). You agree that for all real x, 10x-x = 9x?

>So I don't follow this: "if you work ...

2011-11-09T20:29:06.644-05:00

>So I don't follow this: "if you work it out 'naively', is not 9 at all but 8.999...". Can you explicate what you think the "naive" work is, because I don't see it?

Certainly. We're calculating
9*(.999...)
We multiply 9 by .9 to get 8.1. To this we add 9 times .09 which is .81, giving 8.91. And so on and so on-- we get 8.999...

A paper which I keep meaning to read, though sadly I haven't gotten around to it yet, is: "The real numbers as a wreath product" by F. Faltin, N. Metropolis, B. Ross and G. C. Rota. If I understand right, this paper addresses issues like the "infinite carrying problem" in your comment.

Glowing Face Man -- Thanks for commenting. It'...

2011-11-09T01:34:03.634-05:00

Glowing Face Man -- Thanks for commenting. It's interesting, because as I was musing about it last night, I was in fact thinking that the decimal subtraction in Step #3 might be weaker part of the demonstration.

My take is this -- the standard decimal subtraction algorithm starts on the far-right, which does not exist here. So I think we're making use of this idea: If you know in advance that there are no carries (lower digit always less than or equal to any upper digit), then we can alternatively do all place-value subtractions in any arbitrary order; and of course, in all the digits after the point, 9-9=0. So I'd get the result of 9.000..., and if anything, the unstated assumption I could identify is that that's the same as 9.

So I don't follow this: "if you work it out 'naively', is not 9 at all but 8.999...". Can you explicate what you think the "naive" work is, because I don't see it?

This algebra proof of 0.999...=1 is a pet peeve of...

2011-11-08T22:22:34.722-05:00

This algebra proof of 0.999...=1 is a pet peeve of mine. I hold that it's circular. In going from (2) to (3) you skipped some steps:

(2.5) 10x - x = 9.999... - 0.999...
(2.6) (10-1)x = 9

We've used the distributive law. But the distributive law does not work when numbers are naively treated as strings of digits. Indeed, (10-1)x, or 9x, if you work it out "naively", is not 9 at all but 8.999..., and we to assume that 8.999...=9 *in the middle of a proof of 0.999...=1* is a blatant example of circular reasoning.

The correct way to do things is to define numbers as *equivalence classes* of strings of digits, under the equivalence relation where things are identified if one has an infinite tail of 9's and the other is the simplified form thereof. Indeed, this is Stevin's Construction, one of the alternate constructions of R, named after the inventor of decimals (see: http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Stevin.27s_construction )

Under Stevin's construction, 0.999...=1 *by definition*. And this is the only sensible way to prove 0.999...=1. All the other ways involve circular reasoning. (The infinite series proof? It assumes limits are unique-- which is NOT a true property of numbers if by "number" we naively mean "string of digits". It can be made rigorous, but only through a detour which is well outside a high school course.)