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I am trying to prove the existence of the square root of $2$. I have some steps with a very vague explanation and I would like to clarify.

The proof: Let

$$S=\{x\in\mathbb R\mid x\geqslant 0 \text{ and } x^2<2\}.$$

I understand the proof of LUB, ∝ and so I am at the step where $\alpha^2=2$.

I know that we are to prove by contradiction so we state let $\alpha^2 <2$ and $\alpha^2 >2$. Now my instructor wants us to use the Archimedean Axiom $1/n = \varepsilon$.

$(\alpha^2 + 1/n)^2$ then what.....

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    $\begingroup$ Don't reask a question. Ever. Again. That is a site norm strictly upheld. Most likely you didn't know. Not much harm done. The questions are identical, so I can just merge them. ALL: I deleted the comments, because the merger caused discontinuities. $\endgroup$ – Jyrki Lahtonen Sep 2 '15 at 13:48
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Let $E$ be the set $\{y \in R : y\ge 0 \text{ and } y^2 < 2\}$; thus $E$ is the set of all non-negative real numbers whose square is less than $2$. Observe that $E$ has an upper bound of $2$ (because if $y > 2$, then $y^2>4>2$ and hence $y\in E$). Also, $E$ is non-empty (for instance, $1$ is an element of $E$). Thus by the least upper bound property, we have a real number $x:=\sup(E)$ which is the least upper bound of $E$. Then $x$ is greater than or equal to $1$ (since $1\in E$) and less than or equal to $2$ (since $2$ is an upper bound for $E$). So $x$ is positive. Now we show that $x^2=2$.

We argue this by contradiction. We show that both $x^2<2$ and $x^2>2$ lead to contradictions. First suppose that $x^2<2$. Let $0<\epsilon<1$ be a small number; then we have $$(x+\epsilon)^2=x^2+2\epsilon x+\epsilon^2\le x^2+4\epsilon+\epsilon=x^2+5\epsilon$$ since $x\le2$ and $\epsilon^2\le\epsilon$. Since $x^2<2$, we see that we can choose an $0<\epsilon<1$ such that $x^2+5\epsilon<2$, thus $(x+\epsilon)^2<2$. By construction of $E$, this means that $x+\epsilon\in E$; but this contradicts the fact that $x$ is an upper bound of $E$.

Now suppose that $x^2>2$. Let $0<\epsilon<1$ be a small number; then we have $$(x-\epsilon)^2=x^2-2\epsilon x+\epsilon^2\ge x^2-2\epsilon x\ge x^2-4\epsilon$$ since $x\le2$ and $\epsilon^2\ge0$. Since $x^2>2$, we can choose $0<\epsilon<1$ such that $x^2-4\epsilon>2$, and thus $(x-\epsilon)^2>2$. But then this implies that $x-\epsilon\ge y$ for all $y\in E$. (If $x-\epsilon<y$ then $(x-\epsilon)^2<y^2\le2$, a contradiction.) Thus $x-\epsilon$ is an upper bound for $E$, which contradicts the fact that $x$ is the least upper bound of $E$. From these two contradictions we see that $x^2=2$, as desired.

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  • $\begingroup$ I will look at this and reply with any questions, thank you! $\endgroup$ – David House Aug 30 '15 at 23:02
  • $\begingroup$ @MartinSleziak I don't remember why I deleted it. Thanks for your advice. $\endgroup$ – Cristhian Gz Sep 2 '15 at 12:07
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$\sqrt 2$ is the positive root of $x^2-2$.

Let $f(x)=x^2-2$. Then $f$ is continuous and $f(1)f(2)<0$. Now we can deduce that $f(x)$ has a root betwen $1$ and $2$ by IVT.

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  • $\begingroup$ nice! easy, simple. $\endgroup$ – L.F. Cavenaghi Sep 2 '15 at 20:35
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    $\begingroup$ Is this an actual proof? Isn't saying $\sqrt 2$ is a root of $f$ circular? Also, the fact that there's a root between 1 and 2 doesn't imply that root is $\sqrt 2$. $\endgroup$ – someonewithpc Feb 26 '17 at 18:53
  • $\begingroup$ For the last question. How do you define $\sqrt2$? Positive and a solution of $x^2=2$. $\endgroup$ – vudu vucu Feb 27 '17 at 20:04
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The statement $x \in S$ is equivalent to $x \geq 0 \text{ and } x^2 < 2$. So in particular $x \in S \Rightarrow x^2 < 2$. The contrapositive of this latter statement is $x^2 \geq 2 \Rightarrow x \not \in S$. Does this answer your question?

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  • $\begingroup$ It's a bit clearer, thanks $\endgroup$ – David House Aug 30 '15 at 23:00
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You can use $f(x)=x^{2}-2$ then $f(3)=7>0$ and $f(1)=-1<0$, now use Weierstras theorem for continuous functions. for the method that you want, let $\alpha$ be the Lub for then $b=2-\alpha^2>0$ there is $n_0$ such that $b=2-\alpha^2>\dfrac{1}{n_0}$. now it is may choose $n>n_0$ such that $(\alpha+\dfrac{1}{n})^2<2$ contrary that $\alpha$ is Lub

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It is no contrapositive statement since it is no statement. A set of numbers is defined with such and such conditions, and it is proved this set is bounded from above by $2$. That's all.

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Here is my attempt (I believe some details need still to be fleshed out).

Notice that if $0\le x^2<2$, then $x<2$.

So for any $\varepsilon>0$ we have $(x+\varepsilon)^2=x^2+2\varepsilon x+\varepsilon^2$. If we choose $\varepsilon<1$, then we have $\varepsilon^2<\varepsilon$. Now by using Archimedean axiom it is possible to choose $\varepsilon>0$ such that $$(x+\varepsilon)^2 = x^2+2\varepsilon x +\varepsilon ^2 \le x^2+5\varepsilon \le x^2+5\varepsilon \overset{(*)}\lt 2.$$ (The inequality marked by $(*)$ is the step where Archimedean property is used.)

So we have that if $x\in S$ then there exists a larger number which is still in $S$, which implies that $\alpha\notin S$, i.e. $$\alpha^2 \ge 2.$$

Using similar reasoning we can get that for any $x\ge 0$ and $x^2>2$ there is a smaller number $0\le y<x$ such that still $y^2>2$. This implies that $\alpha>2$ cannot be true, hence we get $$\alpha^2\le2$$

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  • $\begingroup$ let me take a look for a few, thanks! $\endgroup$ – David House Sep 2 '15 at 11:45

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