Derek: Clever, except that 10x is only 9 bar. Nice flubbery, though. :)

Gatac

Actually, Gatac, 0.9(bar) does equal 1. Derek's proof is valid.

There's an easier proof, if I remember correctly.

(as an English major, this will be quite sloppy. apologies.)

1/3 = 0.3(bar)
3 x 0.3(bar) = 0.9(bar)
3 x 1/3 = 1
therefore
0.9(bar) = 1

that kosher?

One third is not 0.3, it's 0.333333 (and an infinite amount of 3 after that).

Does exposing those make me a bad person?

Gatac

Yep -- in fact, it's my favourite way of making sense of the seeming paradox. Mathematically, though, the simplest proof is to observe that an infinite decimal expansion is really just shorthand for an infinite series -- in this case 0.9 + 0.09 + 0.009 + ... This has initial term 0.9 and constant ratio 0.1; when you plug them into the formula a/(1-r) for the sum of an infinite geometric series, you get 1.

A looser argument is that given any two numbers which don't have the same value, we can always find a third which is strictly between them. But there's no number bigger than 0.9999... and less than 1.0000... If there were, it would have to either start out like 0.9999... but then differ from it in some decimal place (which would make it less, not greater) or start out like 1.0000... and then differ somewhere (which would make it greater, not less). Thus they can't be different numbers, so they can only be equal.

Gatac: The notation (bar) in the posts above means infinite recurrence.

He didn't say it it is .3 . He said it was .3 (bar), or
_
.3

which means .3333333333333333...

And...I shoulda posted this when I originally wrote it.

[EDIT: You also shouldn't have deformed the thread with a huge number that didn't wrap. - Z]

*slams forehead*

Geesh, I totally ignored the bar.

Yes, that does make more sense now, even if I'm not particularly impressed with the findings. I'm *aware* that 9,9 (bar) is, for all intents and purposes, 10, but the proofs here feel strange. Particularly ijdgaf's, which basically gets two different numbers from the product of 3 times one third in what seems more like a rounding error to me than deep wisdom.

Gatac

we should have a thread with the best math proofs out there huh?

I like that :wink:

I seem to remember liking a proof I once saw for (sin x)', but I don't quite remember what it is. Calculus is getting further and further away.

I wish MathML were more ubiquitous so we could carry on these conversations with proper notations.

Proof: women are evil.

Assumption: Time is money.
Therefore, time = money.
Women = time x money
Women = (money)^2

But money is the root of all evil: money = (evil)^1/2

Women = ((evil^1/2)^2

:. women = evil.

My ex-girlfriend showed me that one. She found it immensely entertaining.

My geometry teacher showed us a list of twentyish methods of proving things. I'll see if I can find it.


Annnnnd....nope. Oh well.

Anyway, one of them, that I like, is Proof By Illegibility, where you scrawl it out so as to be unreadable. Then when the teacher asks you what it says, you claim it's whatever the right aswer is.

My Algebra II teacher used to regale us with stories of his time in engineering school. One of his favorites was the one about the guy who got so hopelessly lost in his formulae that he proved the given.

here you go proofs of DOOM

or this proofs of DOOM^2

Thanks, ddoof. Been looking for those.

Depends on the teacher. If it's me, you get no credit for illegible work, whether you meant the right answer or not. ;)

Proofs: I always liked proving that n^0=1, personally.

^ I liked the proof I heard for 0!.

If 3! = 3*2*1 = 6, and 2! = 2*1 = 2, then 2! can also be expressed as 3!/3. Thus any n! = (n+1)!/(n+1), which is a rather redundant thing to say since really n! is just the product of all the numbers from 1 to n, right?

But what about 0!?! (Will Zeke kill me for that?) Multiplying the numbers from 1 to 0 implies the answer would either be 0, or undefined if 1 must increment. But using the other formula, we see that 0! = (0+1)!/(0+1) = 1!/1 = 1. And sure enough, 0! = 1. But I've always found that proof kind of silly.

There's another one I've heard like this:

Since N! = (N)(N-1)(N-2)...(3)(2)(1)
And (N-1)! = (N-1)(N-2)...(3)(2)(1)
Then N! = (N)(N-1)!
If N = 1
Then 1! = (1)(0)! = 1
So, 0! = 1

:D! = (:D)(:))!