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

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The discussion revolves around proving the existence of a rational number y such that x < y and y^2 < 2, given that x is a rational number greater than 1 and less than √2. The original poster presents a proof using the definition of a positive rational number α and subsequently a smaller β to establish the inequalities needed. Other participants suggest simplifying the proof by assuming α is small and avoiding complex expressions, emphasizing the importance of constructing a strictly increasing sequence of rationals approaching √2. The conversation also touches on the implications for Dedekind cuts in rational numbers, highlighting the necessity of proving y^2 < 2 without assuming it initially. Overall, the thread showcases various approaches to the problem while refining the proof's clarity and conciseness.
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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.]
 
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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}##.
 
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.
 
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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##
 
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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: <br /> x + x^2 = x + 2. Now dividing by x + 1 gives us <br /> 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&#039;(\sqrt{2}) \in (-1,0)); but we can fix that by taking two steps of this iteration, yielding <br /> x_{n+1} = (f \circ f)(x_n) = \frac{4 + 3x_n}{3 + 2x_n}. You can then easily verify that if x^2 &lt; 2 then x &lt; (f \circ f)(x) and ((f \circ f)(x))^2 &lt; 2.
 
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.
 
IssacNewton said:
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&lt;2
(2n^2-m^2)a^2-2ma-1&gt;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.
 
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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 ?
 
IssacNewton said:
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:
IssacNewton said:
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## .
 
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IssacNewton said:
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} &lt; \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&lt;2
which is written as the quadratic inequality,
(2n^2-m^2)A^2-2mA-1&gt;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##.
 
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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.
 
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