How would I prove there exists a positive real number x such that x^2=2? I need to use the Archimedean property.
Think about it like this.
The Archimedean principle says (geometrically) that
0--------|------------------
in the below we can always find some $\displaystyle n\in\mathbb{N}$ such that
0-----$\displaystyle \frac{1}{n}$-----|-------------
So, now we're claiming that $\displaystyle \gamma$ is the infimum of this set. But, $\displaystyle \gamma^2\ne 2$ and so if you argue correctly we have this scenario
0----$\displaystyle \gamma^2-2$-----------
So, we should be able to "fit something" in between zero and $\displaystyle \gamma^2-2$ (by the A.P.)
but, what's the problem with that?
A) $\displaystyle \gamma=\text{gamma},\lambda=\text{lambda}$
B) No. The point is that you can fit something squared in between $\displaystyle \gamma-2$ and zero and with a little algebra that something squared will be strictly greater than two but less than $\displaystyle \gamma$.
P.S. I don't know why I've been writing $\displaystyle \gamma^2$, it should always have been $\displaystyle \gamma$