Prove that $\sqrt{2}>1$

Question

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..

Answer 1

If $0 \leq x \leq 1$, then you also have $0 \leq x^2 \leq 1$. But $\left(\sqrt2\right)^2 = 2 > 1$, so...

Answer 2

$\sqrt2+2>0$ 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 $2>0$.

Answer 3

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.

Answer 4

It's really not possible to give a satisfactory answer to this question without knowing exactly what the 15 axioms are to which the OP refers and which theorems (the OP's "whatnot") have already been proved from them and are at our disposal, but presumably one theorem says that if $0\lt a$ and $b\lt c$, then $ab\lt ac$. From that one can argue that if $0\lt x\lt1$ then $x\cdot0\lt x\cdot x\lt x\cdot1$. The theorems or axioms saying $x\cdot0=0$ and $x\cdot1=x$ turn this into $0\lt x\lt 1\implies0\lt x^2\lt1$. If you now know that $1\lt2$ then you can conclude that $0\lt x\lt1\implies x^2\not=2$. Combining this with the assumption $\sqrt2\gt0$ gives $\sqrt2\ge1$, so all you have to note is that $\sqrt2\not=1$ (since $1^2=1\not=2$). But to do all this, or something like it, formally, you need to be able to cite the exact axioms and whatnot you're using.

Answer 5

You know that $1>0$ (because $1$ is a square), so $3=1+1+1 > 0+0+0=0$. On the other hand, you have $$3=4-1=(\sqrt{2}-1)(\sqrt{2}+1).$$ So you know that $$(\sqrt{2}-1)(\sqrt{2}+1)>0 \ \text{and} \ \sqrt{2}+1>1+0=1>0,$$ hence (by dividing) $$\sqrt{2}-1>0, \ \text{ie.} \ \sqrt{2}>1.$$