Analysis proof

Printable View

April 28th 2010, 09:13 PM
tharris

Analysis proof

How would I prove there exists a positive real number x such that x^2=2? I need to use the Archimedean property.
April 28th 2010, 09:16 PM
Drexel28

Quote:

Originally Posted by tharris

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

Suppose there wasn't a root, appeal to the completeness of $\mathbb{R}$ and derive a contradiction using the Archimedean principle.
April 28th 2010, 09:57 PM
tharris

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?
April 28th 2010, 09:59 PM
Drexel28

Quote:

Originally Posted by tharris

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?

Assume that $x^2\ne 2,\text{ }\forall x\in\mathbb{R}$ and think about the set $\left\{x\in\mathbb{R}:x^2>2\right\}$ . By assumption it has a infimum, right? But what then?
April 28th 2010, 10:17 PM
tharris

this implies that the infimum is 0 which is a contradiction since x^2 >2?
April 28th 2010, 10:19 PM
Drexel28

Quote:

Originally Posted by tharris

this implies that the infimum is 0 which is a contradiction since x^2 >2?

Well, the point is that it must have an infimum $\gamma$ . But, $\gamma\in\mathbb{R}$ and so $\gamma^2\ne 0$ and so argue that $\gamma^2>2\implies \gamma^2-2>0$ but Archimedean implies what?
April 28th 2010, 10:52 PM
tharris

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?
April 28th 2010, 10:57 PM
Drexel28

Quote:

Originally Posted by tharris

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?

Think about it like this.

The Archimedean principle says (geometrically) that

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

in the below we can always find some $n\in\mathbb{N}$ such that

0----- $\frac{1}{n}$ -----|-------------

So, now we're claiming that $\gamma$ is the infimum of this set. But, $\gamma^2\ne 2$ and so if you argue correctly we have this scenario

0---- $\gamma^2-2$ -----------

So, we should be able to "fit something" in between zero and $\gamma^2-2$ (by the A.P.)

but, what's the problem with that?
April 28th 2010, 11:12 PM
tharris

Does this imply that that "something" must be between 2 and "lambda" squared which is a contradiction since "lamda" squared is > 2?
April 28th 2010, 11:17 PM
Drexel28

Quote:

Originally Posted by tharris

Does this imply that that "something" must be between 2 and "lambda" squared which is a contradiction since "lamda" squared is > 2?

A) $\gamma=\text{gamma},\lambda=\text{lambda}$

B) No. The point is that you can fit something squared in between $\gamma-2$ and zero and with a little algebra that something squared will be strictly greater than two but less than $\gamma$ .

P.S. I don't know why I've been writing $\gamma^2$ , it should always have been $\gamma$
April 28th 2010, 11:43 PM
tharris

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?
April 29th 2010, 08:49 AM
Drexel28

Quote:

Originally Posted by tharris

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?

Of course it's a square $\frac{1}{n}+2=\sqrt{\frac{1}{n}+2}^2$