An argument of infinite proportions (pi)
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Sorry for the double post
@krisidious said:
yeah I agree Bruce...
infinity is a hard thing for people to grasp... but some things are unanswered, we never know for sure. or at least not while we ask the question. math may or may not be natural. I see it more as our description of the universe...
from when the single cell splits 1+1=2 to π the whole universe is math... but it may have an edge? an end? that 13 dimension theory calls for multiple universes would that not mean that infinite was not so infinite after all?
Even if the 13 dimension theory was true (I kind of like that theory) the concept of infinity would still hold true for things like numbers, who occupy no real space physical or otherwise, and so has no bounding limits regardless of any limits the universe might have.
Like I told that kid, you can always add one more, and in the same regard for the negative end of the number line you can always subtract one.
I think the only limit for numbers would be the physical limit of the size of the universe not having enough room if filled with sheets of paper, stacked up and spread out in multiple stacks across the entire universe, filling every square nanometer, to write down all the numbers! :egrin:
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Would you guys agree with this post? I just sort of figured it all out as I went along:
@unknownuser said:
@unknownuser said:
Etchii wrote:PI is non terminating, non repeating. (I find the non repeating part hard to believe... in my head if you were patient enough to keep dividing and ended up with 10^10000000000000000000000000000000000000000000000000000000000 digits...somewhere there has to be a pattern. Even if it is a 10 billion digit sequence.)
1/3 is non teminating, repeating.divide 1/3
.33333333333333333333333333333333333333333333333333333333333333333333333 (see it will never end, just keep dividing for your entire life...enjoy a legacy of 3s)
@unknownuser said:
Masta Squidge wrote:
Actually i understand your logic, there are 10,000,000,000 possible combinations of the numbers 0-9 when only using 10 places, ten digits to the power of ten. Which either you figured out or its a fluke that you said 10 billion.
However the reason why that may be untrue, is that in order for the pattern to repeat, you would need the first 10 billion digits and the second ten billion digits to be in the same order.
But the problem comes when dealing with 10 billion digits, as you then have 10 billion places to take into account, which pushes the number to 10,000,000,000^10, and im not willing to figure that one out, as the character limit of this post probably wouldnt touch .1% of the digits.What I mean is that the more places you have, the more possible combinations of 0-9 you have. When this is taken into account it is easy to see why its posisble that it will never repeat.
Lets not forget that a pattern of ten billion digits repeated twice then has 20 billion^10 combinations.... pusching that even farther out of reach. -
I just re-read that and my typing is horrible when I'm on a roll. -
@masta squidge said:
He stated to me that since someone made it, that means it HAS to have an end.
Nobody made Pi or 1/3, they are ratios. Perhaps if he believes that God made the universe he will believe that God made those ratios. If God did it, perhaps he will believe that the ratios have no ending when expressed as numbers.
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@schreiberbike said:
@masta squidge said:
He stated to me that since someone made it, that means it HAS to have an end.
Nobody made Pi or 1/3, they are ratios. Perhaps if he believes that God made the universe he will believe that God made those ratios. If God did it, perhaps he will believe that the ratios have no ending when expressed as numbers.
Good point, in fact, I dont think I can think of any response for that.
But I told him half a dozen times that nobody "made" the numbers, hes just one of those thick headed kids who by a strange coincidence is also a conspiracy theorist.
Somehow he managed to bring a discussion about mathematics around to a discussion of what I believe happened on 9/11, the went totally off explaining how the government did it.
Sorry, but I personally don't care how it happened, what I care about was the fact that it happened at all, and shouldn't have. But that's got nothing to do with this now does it!I love the interwebz, its where everyone has a voice, even the people who act so dumb that logically (based on their portrayed intelligence) they shouldn't be capable of understanding how to type.
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There's a really good joke about the wisdom of arguing on the Internet, but it is not PC at all, so I won't post it.
I did once have a long series of respectful discussions about global warming and actually changed some people's minds. I find that pretty amazing.
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Please by all means message this joke to me! Unless its the one about the Special Olympics of course lol, because that is all too familiar.
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My thinking is that pi, must in fact, end .
(for it must provide a shape).
Whilst 1/3 does not have to provide a shape.
it is not his work.
can dream.
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@juanv.soler said:
My thinking is that pi, must in fact, end .
(for it must provide a shape).
Whilst 1/3 does not have to provide a shape.
it is not his work.
can dream.ooh circular reasoning (pun intended)
That makes sense, however there is nothing to say that a ratio cannot work properly without an end, only infinitely increasing levels of accuracy (which on its own is a paradox, because you cant do better than perfect right?)
I would be interested in running that by someone in the profession of calculating stupid numbers to places beyond any usable level of accuracy. Something about that statement, while it makes sense right now, doesn't exactly sit right in the back of my mind and I cant place it.
Anyways, I'm off to bed so I can survive work tonight, please by all means continue this because I love reading these sorts of things.
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@unknownuser said:
That makes sense, however there is nothing to say that a ratio cannot work properly without an end, only infinitely increasing levels of accuracy (which on its own is a paradox, because you cant do better than perfect right?)
Sorry, I dont get you.
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So who was the genius that coined the phrase "easy as pi"?
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A number isn't just something made up it's the answer to an equation. Under that definition if you take that answer to equations can sometimes be quantified as between a real & imaginary number and infinity then numbers can go on forever. Numbers can be limitless.
You should tell him 2 isn't really 2 and see what he says.
Pi is limitless, Pie is not... especially around guys from the south. -
Okay, maths is very much my weakness, but I'm confused by your stance. In the other forum you wrote:
"I say they never end as pi is an irrational (which by definition never ends) and that 1/3 also never ends as it is a non terminating repeating decimal."
But in your post here you state:
"Now, anyone who paid attention in math class in middle school knows that he is wrong, that pi is called an irrational number for a reason, and that 1/3 does in fact end, thats why its called a repeating, terminating decimal."
For the record I agree that neither number ends- especially 1/3 as a decimal. Your "opponent" ain't worth the argument if he seriously reasons that 1/3 expressed as a decimal logically ends in a 3!
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What are you guys smoking? :egrin: Pi r not Sq...Pi art round!
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@masta squidge said:
someone in the profession of calculating stupid numbers...
Recently it was reported that some university professor here in Canada completed a massive seven or eight year long calculation using a supercomputer. He was able to absolutely prove that a game of checkers will always end in a draw provided neither player ever makes a mistake or cheats. It is only the mistakes and cheats that swing the outcome away from a draw.
Its great to see my tax dollars at work.
I can't understand why anyone would think 1/3 would ever end. The repeating character of the ratio seems easy enough to comprehend as unending. I hope that university professor doesn't decide to tackle the 1/3 number next.
Regards Ross
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@unknownuser said:
...and that 1/3 does in fact end, thats why its called a repeating, terminating decimal."
I believe that is what we call a typo
I believe it should read "does in fact NOT end, thats why its called a repeating, NON-terminating decimal.Yeah, I slipped, it was past my bedtime.
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@masta squidge said:
Please by all means message this joke to me! Unless its the one about the Special Olympics of course lol, because that is all too familiar.
Yep, that's the one.
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Not that it will help, but here is a mathematical proof that pi is irrational:
@unknownuser said:
Niven's proof
The proof uses the characterization of π as the the smallest positive zero of the sine function. Like all proofs of irrationality, the argument proceeds by reductio ad absurdum.
Preparation: Suppose that π is rational, i.e. π = a / b for some integers a and b ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function
f(x) = {x^n(a - bx)^n \over n!},\quad x\in\mathbb{R},and denote by
F(x) = f(x) + \cdots + (-1)^j f^{(2j)}(x) + \cdots + (-1)^n f^{(2n)}(x). \,the alternating sum of f and its first n even derivatives.
Claim 1: F(0) = F(π)
Proof: Since
f(x)=b^n{x^n(\pi - x)^n \over n!}=f(\pi-x),\quad x\in\mathbb{R},the chain rule and mathematical induction imply
f^{(j)}(x) = (-1)^j f^{(j)}(\pi - x),\quad x\in\mathbb{R},for all the derivatives, in particular
f^{(2j)}(0)=f^{(2j)}(\pi)\,for j = 1, 2, ...,n and Claim 1 follows from the definition of F.
Claim 2: F(0) is an integer.
Proof: Using the binomial formula to expand (a – bx)n and the index transformation j = k + n, we get the representation
f(x)={1\over n!}\sum_{j=n}^{2n}{n \choose j-n}a^{2n-j}(-b)^{j-n}x^{j}.\,Since the coefficients of x0, x1, ..., xn − 1 are zero and the degree of the polynomial f is at most 2n, we have f (j)(0) = 0 for j < n and j > 2n. Furthermore,
f^{(j)}(0)={j!\over n!}{n \choose j-n}a^{2n-j}(-b)^{j-n}\quad\mbox{for } n\le j\le 2n.Since j ≥ n, the fraction of the two factorials is an integer. The same holds for the binomial coefficient, as can be seen from its combinatorical interpretation or Pascal's triangle. Hence f and every derivative of f at 0 is an integer and so is F(0).
Claim 3:
\frac12 \int_0^\pi f(x)\sin(x)\,dx=F(0)Proof: Since f (2n + 2) is the zero polynomial, we have
F'' + F = f.\,The derivatives of the sine and cosine function are given by (sin x)' = cos x and (cos x)' = −sin x, hence the product rule implies
(F'\cdot\sin - F\cdot\cos)' = f\cdot\sin.By the fundamental theorem of calculus
\frac12 \int_0^\pi f(x)\sin(x)\,dx= \frac12 \bigl(F'(x)\sin x - F(x)\cos x\bigr)\Big|_{x=0}^{x=\pi}.Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization of π as a zero of the sine function), Claim 3 follows from Claim 1.
Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < π (because π is the smallest positive zero of the sine function), Claims 1 and 3 show that F(0) is a positive integer. Since
x(\pi -x) = \Bigl(\frac\pi2\Bigr)^2-\Bigl(x-\frac\pi2\Bigr)^2\le\Bigl(\frac\pi2\Bigr)^2,\quad x\in\mathbb{R},and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have
\frac12 \int_0^\pi f(x)\sin(x)\,dx\le \frac{b^n}{n!}\Bigl(\frac\pi2\Bigr)^{2n+1},which is smaller than 1 for large n, hence F(0) < 1 by Claim 3 for these n. This is impossible for the positive integer F(0).
Not that I understand it, but there it is.
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Aha...
Well, thanks - at least for that very last line (now I can hope I'm not alone in the universe who does not understand it).
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