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..

$\sqrt2+2>0$ since both $2$ and $\sqrt2$ are positive. If we prove that $2>\sqrt{2}$ than we've proven that $\sqrt2>1$ (because from $2>\sqrt{2}$ we have $2\sqrt{2}>2$ or $\sqrt{2}>1$). Let we suppose that $2<\sqrt{2}$, than $\sqrt2-2>0$. Now from $\sqrt2+2>0$ and $\sqrt2-2>0$ we have that $2-4>0$ or $-2>0$ or $2<0$ which is contradictory with $2>0$.
