Pi as summation

May 2010
14
0
So I've seen that pi can be written as a series:

http://upload.wikimedia.org/math/4/b/3/4b325f4142cab62b1786d8be1ac3be60.png

\(\displaystyle pi = 4 * series(((-1)^k)/2k+1)\)

So pi = 4/1 - 4/3 + 4/5 - 4/7 + ...

Now I know that Q is an ordered field so if a and b are in Q, if I add them, the result a+b is another element in Q.

But if I sum up infinitely many rational numbers I can get an irrational? Wondering how this works.
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
So I've seen that pi can be written as a series:

http://upload.wikimedia.org/math/4/b/3/4b325f4142cab62b1786d8be1ac3be60.png

\(\displaystyle pi = 4 * series(((-1)^k)/2k+1)\)

So pi = 4/1 - 4/3 + 4/5 - 4/7 + ...

Now I know that Q is an ordered field so if a and b are in Q, if I add them, the result a+b is another element in Q.

But if I sum up infinitely many rational numbers I can get an irrational? Wondering how this works.
Is this really that surprising? Isn't every irrational number the limit of a sequence of rational approximations, which can be represented as the infinite sum of rationa numbers.
 
May 2010
14
0
I know, but when you take the limit of a sum of a bunch of rationals you get an irrational, even though the rationals are closed under addition? That's what's strange about it.
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
I know, but when you take the limit of a sum of a bunch of rationals you get an irrational, even though the rationals are closed under addition? That's what's strange about it.
Closed under finite addition.
 
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May 2010
14
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So when you take a limit of something you can converge to something completely different?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
So when you take a limit of something you can converge to something completely different?
I have no real idea what that means, but I suppose the answer is yes.
 
May 2010
14
0
I mean having a summation converging to a value that is not within the original set.

Ok, thank you.