Prove that, for every $\epsilon > 0$, there exists $x>0$ such that $2 − \epsilon

Question

Prove that, if $S = \{$real numbers $x > 0 : x^2 < 2\}$, then for every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.

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My Proof

Proposition: If $S = \{$real numbers $x > 0 : x^2 < 2\}$, then for every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.

A (Hypothesis): $S = \{$real numbers $x > 0 : x^2 < 2\}$.

B (Conclusion): For every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.

A1: Let $\epsilon > 0$ be a real number.

B1: There is an element $x \in S$ such that $x^2 > 2 - \epsilon$.

A2: Let $x \in S$.

B2: $x^2 > 2 - \epsilon$

$\implies x > \sqrt{2 - \epsilon} > 0$ and $x > 0 > -\sqrt{2 - \epsilon}$

A3: $ 2 > x^2 > 0$

$\implies 2 > x > 0$ since $x > 0$

A4: $2 > \epsilon^2 > 0$

$\implies 2 > \epsilon > 0$ since $\epsilon > 0$

$\implies 2 - \epsilon > 0 > -\epsilon$

$\implies \sqrt{2 - \epsilon} > 0 > -\epsilon$

$\implies -\sqrt{2 - \epsilon} < 0 < \epsilon$

A5: $2 > x > 0 > -\sqrt{2 - \epsilon}$

$\implies 4 > x^2 > 0 > 2 - \epsilon$

$\therefore x^2 > 2 - \epsilon$ $Q.E.D.$

$\implies x > \sqrt{2 - \epsilon}$

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I would greatly appreciate it if people could please take the time to review my proof to ensure that it is correct.

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EDIT

After reading the responses, I have attempted to fix my proof.

A (Hypothesis): $S = \{$real numbers $x > 0 : x^2 < 2\}$.

B (Conclusion): For every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.

A1: Let $\epsilon > 0$ be a real number.

B1: There is an element $x \in S$ such that $x^2 > 2 - \epsilon$.

A2: Let $x \in S$.

B2: $x^2 > 2 - \epsilon$

$\Leftarrow x > \sqrt{2 - \epsilon} > 0$ where $2 > \epsilon > 0$

A3: $2 > x^2 > 0$

$\implies 2 > \sqrt{2} > x > 0$ since $x > 0$.

A4: $2 > \epsilon^2 > 0$

$\implies 2 > \sqrt{2} > \epsilon > 0$ since $\epsilon > 0$.

$\implies 2 > 2 - \epsilon > 0$

$\implies \sqrt{2} > \sqrt{2 - \epsilon} > 0$

A5: Let $\sqrt{2} > x > \sqrt{2 - \epsilon} > 0$

Now we have to show that $x^2 > 2 - \epsilon$ (B2).

$\implies 2 > x^2 > 2 - \epsilon > 0$

$\therefore x^2 > 2 - \epsilon$ $Q.E.D.$

Answer 1

The idea lurking behind this question might be to use an iteration procedure $x\to A(x)$ converging monotonically to $\sqrt2$, for example,

$$A(x)=\frac{4x}{2+x^2}$$

That is, define $x_0=1$ and, for every $n$, $$x_{n+1}=A(x_n)$$ Since every $x_n$ is positive, the goal becomes to show that, for every $\epsilon>0$, there exists some $n$ such that $$2-\epsilon<x_n^2<2$$ To prove this, note that the function $A$ is increasing on $[1,\sqrt2]$, with $A(x)>x$ for every $x$ in $[1,\sqrt2)$ and $A(\sqrt2)=\sqrt2$.

Thus, $x_n<\sqrt2$ for every $n$ and $x_n\to\sqrt2$ when $n\to\infty$. In particular, for every $\epsilon>0$, choosing $n$ large enough yields $x_n$ such that $2-\epsilon<x_n^2<2$.

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More quantitatively, note that $2-x_0^2=1$ and that, for every $x<\sqrt2$, $$2-A(x)^2=\frac{2(2-x^2)^2}{(2+x^2)^2}<\frac12(2-x^2)$$ Thus, for every positive $n$, $$2-x_n^2<\frac1{2^n}$$ which indicates that every $n\geqslant1-\log_2\epsilon$ yields $x_n$ such that $2-\epsilon<x_n^2<2$.

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Remarks: (1) Every $x_n$ is a rational number hence this actually proves the result for the smaller set $S'=\{x\in\mathbb Q\mid x>0,x^2<2\}$.

(2) No square root is involved in the procedure or in the proof, only rational numbers and rational functions.

(3) The convergence $x_n\to\sqrt2$ is much faster than geometric, since $$\lim_{n\to\infty}\frac{2-x_{n+1}^2}{(2-x_n^2)^2}=\frac18$$ hence each iteration roughly doubles the number of exact digits of $\sqrt2$ in $x_n$.

(4) Other iteration schemes are available, for example, the function $$\bar A(x)=\frac{3x+4}{2x+3}$$ starting from $\bar x_0=1$, yields an increasing sequence $(\bar x_n)$ converging to $\sqrt2$, but "only" at a geometric rate since $$\lim_{n\to\infty}\frac{\sqrt2-\bar x_{n+1}}{\sqrt2-\bar x_n}=(3-2\sqrt2)^2$$ For example, $$x_4=\frac{941664}{665857}\approx\mathbf{1.41421356237}15\qquad\bar x_4=\frac{1393}{985}\approx\mathbf{1.414213}1979695$$ and $$\sqrt2\approx1.4142135623731$$

Answer 2

A5 is definitely wrong.

You begin with:

$$2 \gt x \gt 0 \gt -\sqrt{2 - \epsilon}$$

and then claim that this implies:

$$4 \gt x^2 \gt 0 \gt 2 - \epsilon$$

and so:

$$x^2 \gt 2-\epsilon$$

However, $0\gt -\sqrt{2-\epsilon}$ doesn't imply $0\gt 2-\epsilon$, because when we multiply an inequality by a negative number, the inequality reverses direction.

Answer 3

Most of this post was a discussion about how to properly go from a solution to a sufficient condition that will prove a result, and how not to assume the conclusion to verify a hypothesis. And the result was how to back track from a conclusion to find a correct range of values to test.

But none of my previous answer was address the gyst of the question.

The question is asking to show for any positive value $\epsilon > 0$ we can find an $x$ so that $2 - \epsilon < x^2 < 2$ or in other worde there is always an $x$ so that $0 < 2-x^2 < \epsilon$.

This is really a fundamental idea of continuity and/or the archimedian property of reals so that we can always find "arbitrarily small" values, which is an idea that will become indispenible. For example, we will need to be able to show we can always find $n \in \mathbb Z$ so that $0 < 1/n < \epsilon$ or $1/n^2 < \epsilon$ or in this case $0 < 2- x^2 < \epsilon$.

$0 < 2 - x^2 < \epsilon \implies -\epsilon < x^2 - 2 < 0 \implies 2-\epsilon < x^2 < 2 \implies \max(2-\epsilon) < x^2 < 2 \implies \sqrt{\max(2-\epsilon,0)} < x < \sqrt{2}$ which makes the conclusion easy. All we have to do is pick an $x$ that is between $\sqrt{\max(2-\epsilon,0)}$ and $\sqrt{2}$.

And we are done. Nothing left to show.

But this is actually a cop-out. To assume that there are numbers $k$ and $m$ so that $k^2 = \max(2-\epsilon,0)$ and $m^2 = 2$ and to assume that $k < m$ and to assume that we can pick $x| k < x < m$ are all assumptions that rely upon the very idea that the reals have continuity and the archimedean principal applies, which is what this exercise is trying to illustrate.

So how can we show that the is an $x$ so that $0 < 2 - x^2 < \epsilon$ directly?

I'll get to that later.

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I'm going to actually delete this whole thing in a few hours because: We can't really assume that the numbers $\sqrt{2}$ and $\sqrt{2 - \epsilon}$ exist.

I really wanted this be be a quick excercise in picking values. [If $\sqrt{2 - \epsilon} < x < \sqrt2$ than $2-\epsilon < x^2 < 2$] But it obviously failed.

I'll leave it up for a for hours to finish the discussion.

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Its important when we are working back from a conclusion that our implications go back from our statement to our conclusions. We can't go from our conclusion to prior statements.

For example we can't say $y^2 > x^2 > 0 \implies y > x > 0$ but we can say $y^2 > x^2 > 0 \Leftarrow y> x > 0$

That way we can do a proof starting at the conclusion and work back but: In a "forward" proof each step implies the next. In a backwords proof each step is implied by the next.

So Conclusion $x^2 + \epsilon > 2$

$\Leftarrow x^2 > 2- \epsilon \Leftarrow$ (this is actually an $\iff$ statment)

$2 > x^2 > 2-\epsilon \Leftarrow$

$2 > x > \sqrt{2-\epsilon} > 1$ (this is the first step that absolutely can not go the other way)

$\Leftarrow \exists x| 2> x \sqrt{2-\epsilon} > 1$

$\Leftarrow 2 > \sqrt{2-\epsilon} > 1$

$\Leftarrow 2 > 2-\epsilon > 1$

$\Leftarrow 0 < \epsilon < 1$

However this isn't our Hypothesis. Our hypothesis is $0 < \epsilon$. What if $\epsilon \ge 1$.

So we do it again with $\epsilon \ge 1$ in mind

$x^2 + \epsilon > 2 \Leftarrow \epsilon \ge 1$ and $x^2 > 1$

$\Leftarrow \epsilon \ge 1$ and $2 > x^2 > 1$

$\Leftarrow \epsilon \ge 1$ and $\sqrt{2} > x > 1$.

$\Leftarrow \epsilon \ge 1$ and $\exists x| 1 < x <\sqrt{2}$.

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Of course that is taking working backwards to an absurd and really hard to follow degree.

It be better to present to stop at the first few steps, like so:

To conclude there is an $x^2 + \epsilon > 2$ it is sufficient to show $2>x^2 > 2 - \epsilon$ and if we can assume $2-\epsilon > 1$ (or $0 < \epsilon < 1$) it is sufficient to show $\sqrt 2 > x > \sqrt{2-\epsilon} > 1$.

So if $0 < \epsilon < 1$ then $1< \sqrt{2-\epsilon} < \sqrt{2}$ it is enough to select any $x$ so that $1< x< \sqrt{2-\epsilon} < \sqrt{2}$. As $0 < x < \sqrt{2}$ we know $x^2 < 2$ so $x \in S$ and we know from the first paragraph that $x^2 + \epsilon > 2$, and we are done.

If $\epsilon \ge 1$ then $x^2 + \epsilon > 2 \Leftarrow x^2 + 1 > 2 \Leftarrow x^2 > 1$ so pick any $x$ so that $1 < x < \sqrt2$. Then $x^2 < 2$ so $x \in S$ and $x^2 + \epsilon > 2$.

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But to be more efficient (an much easier to follow) an effective prove would go like this:

$0 < \epsilon$. If $\epsilon < 1$ then $1 < 2-\epsilon < 2$ so $1 < \sqrt{2-\epsilon} < \sqrt{2}$.

Select $x$ so that $1 < x <\sqrt{2-\epsilon}$. Then $x^2 < 2-\epsilon < 2$ so $x \in S$ and $x^2 + \epsilon > 2$. So the statement is true for any real $\epsilon$ so that $0 < \epsilon < 1$.

If $\epsilon \ge 1$ then we can just pick another $\delta$ so that $0 < \delta < 1 \le \epsilon$. Thus there is an $x\in S$ so that $ x^2 + \delta > 2$. And as $\delta < \epsilon$, $x^2 + \epsilon > x^2 + \delta > 2$. So we are done.

Answer 4

Such questions arise when we prove from the axioms that $\sqrt{2}$ exists. So I would avoid using the existence of square root.

We need some $x>0$ with $2-\varepsilon<x^2<2$.

Construct a sequence of intervals $I_n=[a_n,b_n]$ by $$ I_0=[1,2]; \quad I_{n+1}=\begin{cases} \left[a_n,\frac{a_n+b_n}2\right] & \text{if } \left(\frac{a_n+b_n}2\right)^2\ge2 \\ \left[\frac{a_n+b_n}2,b_n\right] & \text{if } \left(\frac{a_n+b_n}2\right)^2<2. \\ \end{cases} $$ Then we have $a_n^2<2\le b_n$ for every $n$.

Since $b_n-a_n=\frac1{2^n}<\frac1n$, by the Archimedean axiom there is an index $n$ with $n>\frac3\varepsilon$, so $b_n-a_n<\frac\varepsilon3$. Then $$ 0 < 2-a_n^2 \le b_n^2-a_n^2 = (b_n+a_n)(b_n-a_n) < 3\cdot \frac\varepsilon3 = \varepsilon. $$ So, $x=a_n$ is an appropriate choice.

$\textbf{Remark.}$ This proof works in every Archimedean (ordered) field. In non-Archimedean fields the statement is not necessarily true. For example, extend $\mathbb{Q}$ with a new element $X$ which is greater than all rationals. The resulting field $\mathbb{Q}(X)$ consists of (equivalence classes) of rational functions with rational coefficients. That field does contain any element $r$ with $2-\frac1X<r^2<2$.