I’m not sure where the “subtly dividing by zero” parts comes in. My understand of the fault with this lies in the fact that differentiating a sum is not (necessarily) the same as summing the differentiation.
In saying that x*x = x + x + … + x (x times), you’re assuming that x is a natural number, as sums are undefined otherwise. There’s no differentiation operator on N -> N, because N is not metrically dense (I think that’s the right term). That’s where the error is.
You could probably create a similar but much more slick fallacious proof by using a differentiation operator on Q -> Q (which is metrically dense, but incomplete), but I’ve never seen one.
By “a member of the subtly dividing by zero” school, I was analogizing. You introduce an equation along the way that causes problems. (Just as the divide by zero proofs introduce equations showing a - b = 0 or whatever.) Specifically, you say that 2x = c, where c is a constant. This is where the problem comes in, and you can express the problem in a variety of different ways.
Q.E.D.:
Yep. Another good example why that’s all too simple. He’s just changing the thing on the left side to something that exactly equals whats on the left side. It’s substitition, same as the last one you tripped over.
Why does x have to be a natural number for x*x = x + x + x + … x times to be true?
Why can’t I say 2.5*2.5 = 2.5 + 2.5 + 1.25 = 2.5 + 2.5 + … 2.5 times?
Metrically dense? No clue.
Why is N -> N not metrically dense but Q -> Q is? Either way, you obviously know a lot more math than I. No worries, though. And no need to try and explain as I’d probably not understand it. Maximum C
I never said 2x = c anywhere in my equations. Or was that another analogy? The equation 2x = 1 + 1 + 1 + 1 + … x times is not a constant on the right side.
What “proofs” for 2 = 1 don’t have the subtlety you describe? Don’t they all introduce an equation that causes problems? What equation did I introduce to cause the problems? To me, in my “proof”, I performed an illegal operation, which may or may not be what you mean.
To both, if my explanation of the failure of this “proof” is incorrect, I welcome the feedback. I just don’t understand the feedback I’ve received.
The summation operator is only defined for sums with a non-negative integral number of terms. That’s just the way it is–we haven’t needed a more general definition.
A metric space (M, d) is metrically dense if for any e > 0, there are points p and q such that d(p, q) < e. If you don’t know what a metric space is, look it up on Mathworld.
The derivative is based on limits, and one of the key features of a limit is that it’s unique. In other words, if as x -> c, f(x) -> a and f(x) -> b, then a = b. In a space that’s not metrically dense, the limit need not be unique. That’s why you can’t have a differentiation operator on a non metrically-dense space.
If A is not a complete space, then you can define a differentiation operator on A -> A, but A -> A may not be closed under differentiation.
I’ll try to think of a simple example; give me some time.
You did say 2x = c. Saying “1 + 1 + 1 … 1 x times” is not the same as saying “x”. Your own proof shows why, but let me show a simplier example.
x = x
x = sigma, as n = 1 to x ( n )
If we derive both sides, we get
1 = sigma, as n = 1 to x ( 0 )
Which is nonsense.
I don’t think the reasoning behind this need be as esoteric as ultrafilter is giving, though he’s absolutely correct. X is a variable, and you can’t treat it like a finite sum of constants, because that will always be constant… you see what I’m saying?
1 + 1 + 1 + 1 + x times isn’t constant in as much as x isn’t constant. Your simpler example seems to be the exact reason I gave for why this “proof” fails. Then again, I could just be metrically dense about all this.
If there is an error in my explanation of the failure I’d still like to know. It seemed rather simple when I first saw this “proof”, but now I’m wondering if I really did understand the problem.
Look at this! green_dragon started a perfectly good thread for bitching about how awful math is and the mathematicians have come in and hijacked it to do math in! That proves something. I’m not sure what, but it isn’t mathematical. Why don’t you guys go get your own stinkin’ thread?!
Are you shitting me? How the hell can you call 1 + 1 + 1 + … x times constant?! See that X there you nimrod? See, that is what we call a v a r i a b l e. Know what that means, or is anything above two syllables just a touch too much for you?
Just why the fuck isn’t the explanation I gave fucking good enouh?
Fuck you, SparrowHawk. We’ll hijack any damn thread we want. Well, the real mathematicians might… I’m just playing along here.
I also notice no-one has answered my maths question, being too busy with 1+1=3 type faux-proofs.
In case anyone cares anymore, I had my first exam today. Didn’t go too great, the evil evil lecturer decided to up the difficulty from previous years by like a factor of 10. GRRRR.
Got a coupla easy ones on Friday and then it’s the mysterious and not a little scary world of Quantum Mechanics on Monday with a side order of Chaos Theory on Wednesday. small and pathetic whimper help!
I was referring to this question posted near the bottom of page 1…
“if anyone wants to prove the Riemann-Lebsgue Lemma for me I’d appreciate it because there’s a good chance I’ll have to do it in an exam in about 36 hours…”
Which I don’t believe anyone has done yet, ummm Maxxy…
(and yes, it did come up, I wrote something down which hopefullly should get me a mark or two…)
Green Dragon: Pray that it being harder means lower boundaries.
I should know R-L, but can’t remember it. Google helped a bit, surprisingly.
Maximum C Sum[sub]1[/sub][sup]x[/sup]1 is the same as x, where it is defined. The equivalent for x not +ve integer is Integral[sub]1[/sub][sup]x[/sup]1ds (s arbitrary) which is equal to x and can be differentiated by x fine (but due to a fairly simple calculation with the defn of differentiation, needs a 1.dx/dx=1 term added to the obvious Int[sub]1[/sub][sup]x/supds =0 term)
