If ##x> 1## and ##x^2 <2##, prove ##x < y##, ##y^2<2##

[issacnewton]

Suppose $x \in \mathbb{Q}$ and $x > 1$ and $x^2 < 2$. I need to come up with some $y \in \mathbb{Q}$ such that $x < y$ and $y^2 < 2$. Here is my attempt. Give that $x > 1$ and $x^2 < 2$, I have $(2-x^2) > 0$ and $4x > 0$. Also, $2x >0$. Now define

$$ \alpha = \text{ min} \Bigl\{ \frac{(2-x^2)}{4x}, 2x \Bigl\} $$

It can be seen that $\alpha > 0$. Also, I have

$$ \alpha \leqslant \frac{(2-x^2)}{4x} \cdots \cdots (1) $$
$$ \alpha \leqslant 2x \cdots \cdots (2)$$

Since $x \in \mathbb{Q}$, it can be seen that $\alpha \in \mathbb{Q}$. Since $\alpha > 0$, there exists some $\beta \in \mathbb{Q}$ such that $0 < \beta < \alpha$. From above equations, it follows that

$$ 0 < \beta < \frac{(2-x^2)}{4x} \cdots \cdots (3) $$
$$ 0 < \beta < 2x \cdots \cdots (4)$$

This simplifies to the following

$$ x^2 + 4x \beta < 2 $$
$$ \beta^2 < 2\beta x $$

Now consider $(x + \beta)^2$

$$ (x + \beta)^2 = x^2 + 2\beta x + \beta^2 $$

Using above inequalities, I have

$$ (x + \beta)^2 = x^2 + 2\beta x + \beta^2 < x^2 + 2\beta x + 2\beta x $$
$$ x^2 + 2\beta x + 2\beta x = x^2 + 4 \beta x < 2 $$

Hence $(x + \beta)^2 < 2$. Now, let $y = x + \beta$. Due to closure of $\mathbb{Q}$, it can be seen that $y \in \mathbb{Q}$. Also
$x < y$ and finally $y^2 < 2$.

Is this proof good enough ?
Thanks

[Moderator's note: moved from a technical forum.]

 

[anuttarasammyak]

I take it easy as follows.
1<x and $x^2$<2 imply $1<x<\sqrt{2}$
For all x < y and $y^2$<2 imply　y=$\{\varnothing\}$
For some x < y and $y^2$<2 imply x<y<$\sqrt{2}$.

 

[Office_Shredder]

This is probably intended to prove that $\sqrt{2}$ exists so that's probably not sufficient annuta.

Issac, i think your proof is good but complicated. You can just define $\alpha$ to be 9/10 of your definition, and then you don't need to also define $\beta$.

The bigger win is learning how to avoid having to min multiple complicated expressions. You can assume $\alpha$ is small, which if it's less than 1 $\alpha^2 < \alpha$. So $(x+\alpha)^2 < x^2+(2x+1)\alpha$, which makes picking $\alpha$ really easy, $\frac{9}{10}min(1,(2-x^2)/(2x+1)$.

Starting off by assuming your number will be really small often makes the proof a lot shorter, and more intuitive since the interesting case is going to be when $x^2$ is close to 2.

 

[anuttarasammyak]

For any x such as
$2-x^2:=\epsilon>0$
There exists y such that
$2-y^2=\alpha \epsilon$ where 0<$\alpha$<1
So
$x^2<y^2<2$
thus $1<x<y$ and $y^2<2$

 

[pasmith]

Suppose we wanted to construct a strictly increasing sequence of rationals starting with $x_0 = 1$ such that $x_n \to \sqrt{2}$ as $n \to \infty$. One method is to start with $x^2 = 2$ and add $x$ to both sides: $$x + x^2 = x + 2.$$ Now dividing by $x + 1$ gives us $x = \dfrac{x + 2}{x + 1}$ so a starting point is $$x_{n+1} = f(x_n) \equiv \frac{x_n+2}{x_n+1}.$$ Although $\sqrt{2}$ is a stable fixed point of this iteration, the convergence is not monotonic ($f'(\sqrt{2}) \in (-1,0)$); but we can fix that by taking two steps of this iteration, yielding $$x_{n+1} = (f \circ f)(x_n) = \frac{4 + 3x_n}{3 + 2x_n}.$$ You can then easily verify that if $x^2 < 2$ then $x < (f \circ f)(x)$ and $((f \circ f)(x))^2 < 2$.

 

[issacnewton]

Thanks for replies. Actually, I was trying to prove that $\{x : x^2 <2 \text { or } x < 0 \}$ is a Dedekind left set. So, what I asked here is part of that proof, where I am showing that the above set does not have a maximal element. Since Dedekind cuts are defined in terms of rational numbers, I can't use any non-rational numbers here.

 

[anuttarasammyak]

Since Dedekind cuts are defined in terms of rational numbers, I can't use any non-rational numbers here.

Let
$$x=\frac{m}{n},y=\frac{am+1}{an}$$
where a, m and n are positive integers.
$$y^2<2$$
$$(2n^2-m^2)a^2-2ma-1>0$$
Quadratic equation of a
$$(2n^2-m^2)a^2-2ma-1=0$$
has two real solutions. We know that a, which should be larger than the larger solution, have integer solutions for the inequality.

 

[issacnewton]

Hello anuttarasammyak, we have to prove that $y^2 < 2$. You are assuming it to begin with. Also, I did not understand your last line. Can you explain ?

 

[SammyS]

Hello @anuttarasammyak, we have to prove that $y^2 < 2$. You are assuming it to begin with. Also, I did not understand your last line. Can you explain ?

In the second sentence of the Original Post in this tread, you stated:

I need to come up with some $y \in \mathbb{Q}$ such that $x<y$ and $y^2 < 2$ .

All that @anuttarasammyak has done is to show a way to come up with such a $y$ .

 

[anuttarasammyak]

Hello anuttarasammyak, we have to prove that $y^2 < 2$. You are assuming it to begin with. Also, I did not understand your last line. Can you explain ?

Let 　
$$x=\frac{m}{n} < \frac{Am+1}{An}=y$$
where A, m and n are positive integers and $m^2 < 2n^2$.
We want to prove that there exist A which satisfies
$$y^2<2$$
which is written as the quadratic inequality,
$$(2n^2-m^2)A^2-2mA-1>0$$.
The associated quadratic equation
$$(2n^2-m^2)x^2-2mx-1=0$$
has two real opposite sign solutions, say $\alpha < 0 <\beta$. Any positive integer A > $\beta$ satisfies the inequality thus also satisfies $y^2<2$.

 

[issacnewton]

Ok I see it now. So, we have a parabola in the variable A and since this is upward opening parabola, for any positive integer greater than $\beta$, the inequality would be positive. Great. My solution is actually smaller on paper, but I had to explain to the audience here, so it looks long.