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..

That's the definition of $\sqrt{x}$ - it is the positive value $y$ such that $y^2=x$. — Thomas Andrews, Mar 02 '14 at 15:16
@ThomasAndrews It's the non-negative value, otherwise $\sqrt{0}$ wouldn't be defined. Just nitpicking. — Michael Hoppe, Mar 02 '14 at 15:18
I'm pretty sure you mean ordered field, not complete field there... — fgp, Mar 02 '14 at 15:32

score 5 · Answer 1 · answered Mar 02 '14 at 15:26

5

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.

answered Mar 02 '14 at 15:26

Ben Grossmann

225,327

What if we are not working by Peanos axioms? What if all that we know is that $\sqrt2>0$? – Emo Mar 02 '14 at 15:46
@Emin do you mean to say that $1>0$ might not be given? – Ben Grossmann Mar 02 '14 at 15:53
I mean what if it is not given that 2>1 (but I think it is equivalent with your question). – Emo Mar 02 '14 at 15:55
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@Emin the step I've skipped for brevity is $$1>0 \implies 1+1 > 1+0$$ that $1>0$ is usually one of the axioms given for an ordered field. – Ben Grossmann Mar 02 '14 at 15:56
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@Emin sometimes you are equivalently given $a>0$ and $b>0 \implies ab>0$. In this case, we may state $1$ is positive. Proof: either $1$ is positive or $−1$ is positive. If $−1$ is positive, then $(−1)(−1) = 1$ is positive, which is a contradiction. – Ben Grossmann Mar 02 '14 at 16:00
@Emin: What do the Peano axioms have to do with that? – Asaf Karagila Mar 02 '14 at 16:30
By Peano axioms we prove that 1+1=2 which was used in his prove! – Emo Mar 02 '14 at 16:32
@Emin once you've defined "$1$" and "$+$", $2$ is defined by $2 = 1+1$, and arbitrary integers are defined by $$ n = \overbrace{1 + \cdots + 1}^n$$ regardless of which axioms it took to get to that point. – Ben Grossmann Mar 02 '14 at 16:37
@Omnomnomnom 1 and + are defined, but 2=1+1 is not a definition! It is a theorem. It is a consequence from Peanos axioms! – Emo Mar 02 '14 at 16:40
The only difference with Peano's axioms is that we define (and yes, it most certainly is a definition) $2 = S(1) = 1+1$, where $a+1:=S(a)$. – Ben Grossmann Mar 02 '14 at 16:42
We define 2 as a successor of 1, but we don't define 2 as 2=1+1. We prove that 2=1+1. – Emo Mar 02 '14 at 16:44
If that's your only hangup, then all you need to say is $a + 1 = a + S(0) = S(a + 0) = S(a)$, per the recursive definition of addition. However, this question makes no mention of Peno's axioms, and it is not generally required that one appeal to them in dealing with ordered fields. – Ben Grossmann Mar 02 '14 at 16:49
Now I agree with you. – Emo Mar 02 '14 at 17:02

score 1 · Answer 2 · answered Mar 02 '14 at 15:23

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

score 0 · Answer 3 · answered Mar 02 '14 at 15:19

0

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

answered Mar 02 '14 at 15:19

fgp

21,050

@ThomasAndrews Yes, but at the time it didn't include the part about field axioms... – fgp Mar 02 '14 at 15:27
i said we are supposed to use axioms when I first asked the question, the basic ones are pretty universal to begin with – Jack Armstrong Mar 02 '14 at 15:32
@jack If you want a formal proof that start from just the axioms, you'll have to state the axioms. There are different ways to write some of the axioms, and while the gist of the proof will always be the same, the details may vary. Also, as I pointed out in another comment, if you start from just the axioms, $2$ isn't even defined... So your question did, in multiple ways, implie that you're not looking for a completely formal proof... – fgp Mar 02 '14 at 15:35

Barry Cipra · Answer 4 · 2014-03-02T15:49:49.797

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.

score 0 · Answer 5 · answered Mar 02 '14 at 16:26

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) $$\sqrt{2}-1>0, \ \text{ie.} \ \sqrt{2}>1.$$

Prove that $\sqrt{2}>1$

5 Answers5