Easy Proof of Irrationality of Pi

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climbhi

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[SOLVED] Easy Proof of Irrationality of Pi

Ooops, I found a critical error in my proof. Please ignore my stupidity if you already read it. So seeing as my easy "proof" failed perhaps someone here can help me out. Do any of you know of any easy proofs for the irrationality of pi? Thanks.
 
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  • #2
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I don't, sorry. But one thing I find interesting about pi is that many cultures over the millennia have had it, although they all used different exact values for it.
 
  • #3
HallsofIvy
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I hate to bother you but, in fact, many cultures have used many different APPROXIMATE values for pi. There is only one exact value.
 
  • #5
climbhi
Originally posted by bogdan
http://pi314.at/math/irrational.html

Not the simplest...but not too complicated...
Thanks, I'm stuck at the first part where they say it is not hard to show that... but I think I followed most of it after that. Do you know of any proofs which do not use calculus? I'm looking for one to show to someone who has had math up to but not including calculus. Thanks!
 
  • #6
Locutus
I have not taken calculus either, so although this proof may not be the mathematical standard, it seems to make sense.

[pi]= circumference/diameter. Imagine inscribing polygons inside a circle of radius [squ]2. If the first polygon you inscribe is a square, the perimeter of such square would be 8, and diameter 2[squ]2. Therefore, [pi](square) = 8/2[squ]2 ~= 2.8284.

If the polygon you inscribed was a hexagon, diameter would still be 2[squ]2 but circumference of the hexagon would be 6[squ]2. Now, [pi](hexagon) = 3. As you continue to increase the number of sides for the inscribed polygon, your values for pi continue to approach the irrational 3.1415926536......

Since a circle has infinite number of sides, however, [pi] can never be fully equated by this method, making it possible for one to create polygons with sides that approach infinity in number, while getting closer to [pi] using this method. Since a polygon with infinite sides can never be obtained in such a way, neither can [pi], thus it is irrational.
 
  • #7
Hurkyl
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That reasoning really isn't valid. One of the things I like to do when I have an argument I'm not really sure about is to try to apply it to something where the argument should obviously fail, if correct.


In this case, something akin to Zeno's paradox shows that it shouldn't hold. Suppose you are trying to figure out if the number 1 is irrational or not. Well, you can start building line segment of length 1/2. Then you can add to it another line segment of length 1/4, then 1/8, et cetera. You'll be able to get arbitrarily close to the number 1, but you can never actually obtain a line segment of length 1, so 1 must be irrational!
 
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