Thread: 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.

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 and derive a contradiction using the Archimedean principle.

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?

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 and think about the set . By assumption it has a infimum, right? But what then?

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

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 . But, and so and so argue that but Archimedean implies what?

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?

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 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?

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

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)

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


P.S. I don't know why I've been writing , it should always have been

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?

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