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I am trying to prove that following set is a Dedekind left set

$$ A = \{q \in \mathbb{Q} : q^2 < 2 \text{ or } q < 0 \} $$

I have proved most of the properties necessary. Now I am in the last part and I have $ 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, I am trying to prove that $A$ has no maximal element. Here is what I have done so far. Since $ x \in \mathbb{Q}$, there are some $m, n \in \mathbb{Z}$ such that $x = m/n$ with $n \ne 0$. Since, $x > 1$ and $ x^2 < 2$, it will follow that $n < m $ and $m^2 < 2n^2$. Which also means that $ n^2 < m^2 < 2 n^2$. Since $m, n \in \mathbb{Z}$, it must be the case that $ n \geqslant 3$. I am not able to go beyond this. Please help.

user9026
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1 Answers1

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$(x+\epsilon)^2=x^2+2\epsilon x+\epsilon^2< x^2+4\epsilon x$ for all positive $\epsilon<2x$.

All you need is a positive rational value of $\epsilon$ such that $\epsilon<\dfrac{2-x^2}{4x}$ and $\epsilon<2x$.

Chris Sanders
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  • Ok, so let $\delta = \text{min}\left(\frac{2-x^2}{4x}, 2x\right)$. Then, we have $\delta > 0 $. Then , we can choose some rational between $\delta$ and $0$. We let that be $\epsilon$ and so $y = x + \epsilon$. Hence the proof. Thanks Chris. – user9026 Jul 02 '22 at 08:43