How would I prove there exists a positive real number x such that x^2=2? I need to use the Archimedean property.

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- Apr 28th 2010, 09:13 PMtharrisAnalysis proof
How would I prove there exists a positive real number x such that x^2=2? I need to use the Archimedean property.

- Apr 28th 2010, 09:16 PMDrexel28
- Apr 28th 2010, 09:57 PMtharris
Do I start with: suppose x^2 does not equal 2? Or do I begin with assuming either a lower bound or an upper bound?

- Apr 28th 2010, 09:59 PMDrexel28
- Apr 28th 2010, 10:17 PMtharris
this implies that the infimum is 0 which is a contradiction since x^2 >2?

- Apr 28th 2010, 10:19 PMDrexel28
- Apr 28th 2010, 10:52 PMtharris
I guess I don't see the connection. The arcimedian principle states that if x>0, then there exists a positive integer n such that 1/n <x.

How does this relate to the infimum? - Apr 28th 2010, 10:57 PMDrexel28
Think about it like this.

The Archimedean principle says (geometrically) that

0--------|------------------

in the below we can always find some such that

0----- -----|-------------

So, now we're claiming that is the infimum of this set. But, and so if you argue correctly we have this scenario

0---- -----------

So, we should be able to "fit something" in between zero and (by the A.P.)

but, what's the problem with that? - Apr 28th 2010, 11:12 PMtharris
Does this imply that that "something" must be between 2 and "lambda" squared which is a contradiction since "lamda" squared is > 2?

- Apr 28th 2010, 11:17 PMDrexel28
- Apr 28th 2010, 11:43 PMtharris
Sorry, my greek symbols are rusty. You have been a great help. Does this stuff get more understandable? I hope so. So far I have that:

0 < 1/n < gamma - 2

If I add 2 to everything I get:

2 < 1/n + 2 < gamma

How did I show that 1/n + 2 is a square? - Apr 29th 2010, 08:49 AMDrexel28