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What is the square root of pi?
Me ^
Posted on 12/08/2001 2:26:08 PM PST by ambrose
What is the square root of pi?
TOPICS: Miscellaneous; Your Opinion/Questions
KEYWORDS: cheesewatch; moosewatch
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To: ambrose
"What is the square root of pi?
Rhubarb?
181
posted on
12/08/2001 5:19:32 PM PST
by
hgro
Comment #182 Removed by Moderator
To: Tennessee_Bob
Ooooohhhh, thanks!
To: ambrose; AmericaUnited
So, what did y'all do with Ambrose? Did he ever pick a winner? Come back and grade the papers? Stare into the void too long? Or what?
My pick for best answer = #170 Why not post something more weighty like "Does anyone have the time?"
To: Faraday
Cute page, thanks.
To: AmishDude
I've never read through the proof of transcendality for either e or pi. For pi even to prove irrationality is difficult; for e on the other irrationality is quite easy.
To: lewislynn
"One must wonder how long before the Libertarians here will insist that taking anyone's square root violates the Fourth, Fifth, and Tenth Amendments...." -Duke
"Suggesting Republicans, like Democrats have no problem violating rights of individuals for whatever reason as long as YOU or Republicans say it's OK?" -lewislynn
Thank you for so aptly demonstating my point, that a libertarian would a civil rights violation somewhere, somehow.
"The Republican fear of Libertarians here is astounding, especially when you consider there would likely be no elected Republicans without the libertarian vote..."
Thank you for confirming that almost all Democrats owe their victories to libertarians voting their "principles."
To: BADJOE
That rule's been superceded by a late-breaking development!
188
posted on
12/08/2001 5:28:01 PM PST
by
xzins
To: stands2reason
"Try squaring the circle....."
Silly - He doesn't to square the circle, he wants to square root the ratio of the circle! 8<)
To: Aurelius
Ditto for me. I would guess the proof of e's irrationality comes from the expansion of the sum of 1/(i!) and claiming that if e were rational then the denominator would be divisible by all possible positive numbers.
To: *moosewatch; *cheesewatch
What if it is a moose-meat pie au gratin?
191
posted on
12/08/2001 5:30:16 PM PST
by
Fixit
To: AmishDude
There are several ways. Easiest is to take the (alternating) series for 1/e (= 1/2! - 1/3! + 1/4! - ...) and multiply through by n!, for any arbitrary n. Then after multiplication the first (n-1) terms are integers and the remaining terms add to a number with magnitude less than 1. So n!/e is not an integer for any n.
To: Aurelius
Works for me.
To: go star go
what's wrong with you? all pies are round and most are not made from roots! BWAHAHAHAHAHAHA! You know, I used to know the answer to that question....but not anymore!
194
posted on
12/08/2001 5:42:28 PM PST
by
KLT
To: AmishDude
I saw a lecture recently where they talked about normed spaces. There, you could make pi many different things depending on the norm. Could be anything from 3 to 8. Lucky we don't live in those worlds, though.
You know, they never did mention pi on that series, mostly beer ...
(Actually, after taking a couple of years of advanced math in college, I pretty much lived in Norm's world.)
To: Aurelius; AmishDude
Got to admit that I've never studied set theory, but the discussion here got me looking into cardinality. In particular, I found this on a site:
"Are there as many even integers as integers? Since we can match every integer n to a single even integer 2n, we must concede that there are the same number of each. The matching is called a one-to-one correspondence. Infinite sets can have one-to-one correspondences with "smaller-looking" subsets of themselves."
While I understand (I think) this correspondence, what is wrong with the statement: "For every even integer, I can generate two unique integers, i.e. given n(even), return n and n+1."
196
posted on
12/08/2001 5:49:10 PM PST
by
Faraday
To: go star go
There's ALWAYS SWEET POTATOE PIE!!!! ROFLMAO:-))))
To: nomasmojarras
Wouldn't 5 have more to worry about?
198
posted on
12/08/2001 5:53:09 PM PST
by
mercy
To: Faraday
The answer is: What is 2*infinity?
Well, it's infinity. The problem comes when you add them up and see what you have.
Let N be the cardinality of the integers. Then N=2*N and N=N^2. But what is different is that N != 2^N
To: editor-surveyor
You get the prize!
The square root of two is also an irrational number!
That is a number that cannot be written as the ratio of two integers.
If a number cannot be written as the ratio of two integers
(like pi) then it follows that its square root also cannot!
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