Prove that $\sqrt{2}>1$ I need to prove that $\sqrt{2}>1$, but the initial assumption I am given is that $\sqrt{2}>0$.  I have $\sqrt{2}>0$ so $2>0$ (multiply by $\sqrt{2}$ on each side). I don't know what my next step should be. We are supposed to use axioms and what not. The axioms that we are supposed to use are the 15 basic complete field axioms, such as associativity, commutativity, inverses, identity, etc..

We know, by the usual ordering axioms, that $2 = 1+1>1$.
Now, suppose that $\sqrt{2} \leq 1$.  It would follow that 
$$
\sqrt{2}\cdot \sqrt{2} \leq 1 \cdot 1
$$
since given positive $a,b,c,d$, we have $a\leq c$ and $b\leq d$ implies $ab \leq cd$.  However, this leads to the conclusion
$$
2 \leq 1
$$
Which is a contradiction.
