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

• issacnewton

#### 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.]

Last edited by a moderator:
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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• mfb and 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##

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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: $$x + x^2 = x + 2.$$ 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 $((f \circ f)(x))^2 < 2$.

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.

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.
$$y^2<2$$
$$(2n^2-m^2)a^2-2ma-1>0$$
$$(2n^2-m^2)a^2-2ma-1=0$$
has two real solutions. We know that a, which should be larger than the larger solution, have integer solutions for the inequality.

• 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 ?

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## .

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,
$$(2n^2-m^2)A^2-2mA-1>0$$.
$$(2n^2-m^2)x^2-2mx-1=0$$
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