One of those was mine. I ended up conceding the point and changing my mind.
Was it you, Lib? I don’t remember your participation. Maybe you were just completely overshadowed by those who needed more convincing.
If you try to subtract .999999… from 1.000000… using the standard rules for subtraction of infinite decimals (which are somewhat less than obvious if you think about it), you get 0.00000000… and if you don’t accept that that is 0, then there is nothing I can do for you. But it is inherent in what you mean by an infinite decimal. And yes, if you accept that 1/3 = .333333…, then 3 * 1/3 = .99999999…, again according to the rules for multiplying infinite decimals, which again are far from evident. I once heard an hour lecture whose entire content was that you could carry out arithmetic operations with infinite decimals, at least in principle.
How about a proof that the square root of 2 is a nonterminating nonrepeating decimal (cannot be represented by fractions) or in other words is not rational?
Assume sqrt(2) is rational.
Its rational so it can be represented by some two positive integers a and b (b!=0) as:
sqrt(2) = a/b
such that a/b is in lowest terms (that is, there are no common integer factors other than 1 between them). If it is not in lowest terms, divide out the common factors until it is in lowest terms, and use that a and b.
Since there are no common integer factors other than 1, then specifically 2 is not an integer factor common between them.
What this means is that they cannot both be even numbers, or you’d be able to divide both by two and still have integers on the top and bottom.
Then,
2 = a^2 / b^2
Keep in mind that since a and b were integers, a^2 and b^2 are also integers.
2b^2 = a^2
This says that a^2 is twice b^2. But if you double any number (b^2 in this case) then it becomes even (because it has a factor of 2).
So a^2 is even.
But if a^2 is even, a must be even, because any odd integer multiplied by itself is odd (since it has no even factors to square).
So a is even.
Since both a and b cannot be even, b must be odd.
b is odd, so b^2 is also odd.
However, b^2 = a^2 / 2.
Since a is even, it means it can be represented by 2k for some integer k.
b^2 = (2k)^2 / 2 = 4k^2 / 2 = 2k^2
It doesn’t matter if k^2 is odd or even at this point – we multiply it by 2, so 2k^2 is obviously even.
But this implies that b^2 is even. We’ve already proven it must be odd.
We’ve arrived at a contradiction. This means one of our assumptions must be wrong.
The only assumption we’ve made is that sqrt(2) is rational. Therefore it is not rational, and cannot be represented as a fraction a/b for integers a and b.
So it is an irrational number between 1 and 2 that cannot be represented as a simple fraction and therefore has no simple repeating decimal form.
I just love this proof for some reason. 
I’m one of them. But I am, of course, fully aware that 0.99… = 1. In fact, it’s what I would have guessed, and seeing a proof of it was not surprising at all to me. The reason I dislike the proof is that it blindly applies algebra to something that is, in the end, based on calculus. I certainly wouldn’t put it among the classic proofs of mathematics, like the one that William_Ashbless has graced us with. Speaking of which, if I had done it, I would have explicitly invoked the Fundamental Theorem of Arithmetic. I like that theorem, because non-mathematicians are often so shocked that Arithmetic has theorems.
Please so grace us. I’d love to hear an alternate proof. 
Thank you! I thought I was the only one driven crazy by the hordes of penis flexers who jump in for their moment of glory every time a super-triv math question arises. :smack:
We’re here to educate, discuss, and hopefully manage to correct ignorance in ourselves and others. Sometimes it’s not easy to come to the ‘truth’ since zealous belief, stubborn philosophies, wildly differing opinions, and occasionally a lack of available data cloud the issue. At least with math we have some not inconsiderable tools to work with.
I’ll be a ‘penis-flexer’ if it helps educate you (or me), sir. Pardon the disturbing unintended image that creates. Mayhaps you shouldn’t open any more math discussions?
There’s no alternate proof; the Fundamental Theorem is implicit in your designation of integers as exclusively “even” or “odd”. Pretty fundamental, I know. Hence the name.
KarmaComa: Maybe we’re not trying to impress each other; maybe we just think it’s cool.
You’re upset about people answering questions on a messageboard whose major purpose is to disseminate information?
That’s…fascinating.
Actually, I’ve found two alternate proofs at least by doing a search on ‘root of 2 is irrational’ on search engines…
http://www.maths.soton.ac.uk/~gan/MA111/Slides/sld049.htm
http://www.maths.soton.ac.uk/~gan/MA111/Slides/sld051.htm
Those two from the same site.
There’s another more direct proof based on some sort of series, but I can’t find it.
Since we’re giving irrationality proofs, how about the more general
THEOREM: Suppose that a[sub]0[/sub], …, a[sub]n-1[/sub] are integers and that x is a root of
x[sup]n[/sup] + a[sub]n-1[/sub]x[sup]n-1[/sup] + … + a[sub]0[/sub] = 0.
Then x is either an integer or irrational.
Note that since sqrt(2) is a root of x[sup]2[/sup] - 2 = 0 and is certainly not an integer, this implies that sqrt(2) is irrational.
Proof of theorem: Suppose x = b/a, where a and b are integers with no common factor. Then
b[sup]n[/sup]/a[sup]n[/sup] + a[sub]n-1[/sub]b[sup]n-1[/sup]/a[sup]n-1[/sup] + … + a[sub]0[/sub] = 0.
Multiply throughout by a[sup]n[/sup]:
-b[sup]n[/sup] = a[sub]n-1[/sub]b[sup]n-1[/sup]a + … + ba[sup]n-1[/sup] + a[sub]0[/sub]a[sup]n[/sup]
Now a divides each term on the right-hand side, so a divides the left-hand side, and a divides b[sup]n[/sup]. Since a and b are coprime, this implies that a = 1 ( again this uses the fundamental theorem of arithmetic). We have shown that if x is rational then it is an integer, and this proves the theorem.
I’m afraid you misspelled “our best customers.”
Although I have complete confidence that some of them have also flexed a penis, their own or that of a loved one, from time to time.
Use the [ code ] [ /code ] tag (minus the internal spaces of course):
1 + 1
---------
2 + 1
-----
2 + 1
- ...
2
Hey, I enjoy a good display of intellectual penis-flexing
[sub]But what is the distaff equivalent? Intellectual boob-flashing?[/sub]
Can anyone demonstrate a reasonable proof that that continued fraction equals root 2? I’m fascinated. (Much as I took some algebra and calculus courses, I don’t remember any proof techniques involving continued fractions…)
Sure. Add 1 to it and call your result x. Note, by direct substitution, that:
x = 2 + 1 / x
Or: x[sup]2[/sup] - 2x - 1 = 0
So x = 1 - sqrt(2), or x = 1 + sqrt(2). But x is clearly positive. So there you go.
Well, we can certainly make the result intuitively plausible.
Write u = sqrt(2), for typographical convenience.
Note that 1 < u < 2, and that (u-1)(u+1) = 1. Thus
u = 1 + (u-1)
= 1 + 1/(u+1)
= 1 + 1/(2 + u-1)
= …
William_Ashbless, thanks for the “sqrt 2 is irrational” proof above; that’s fascinating. I’m more of a mathematics enthusiast than an actual mathematician, but I was able to follow that clearly.
And (my) ignorance retreats another step…
I should say that both my proof and Achernar’s have the same failing, namely that they assume that the infinite continued fraction converges. Strictly, all we have shown is that if the fraction converges, then it converges to sqrt(2). In fact, the proof that all simple continued fractions converge is quite simple, but it involves a lot of fractions and subscripts, which are a real pain to type out in vB. If you want to see it, I’ll consider it tomorrow ( if no-one does it in the meantime).