Proof: (by contradiction) Suppose $\lnot(\forall \epsilon\in\Bbb{Q^{+}}\exists x\in\Bbb{Q^{+}}(x^2<2 \land 2<(x+\epsilon)^2))$, which is equivalent to the following propositions: \begin{align} &\Longleftrightarrow \exists\epsilon\in \Bbb{Q^+}(\lnot\exists x\in\Bbb{Q^+}(x^2<2 \land 2<(x+\epsilon)^2))\\ &\Longleftrightarrow \exists\epsilon\in\Bbb{Q^+}\forall x\in\Bbb{Q^+}(x^2\geq2\lor 2\geq (x+\epsilon)^2)\\ &\Longleftrightarrow \exists\epsilon\in\Bbb{Q^+}\forall x\in\Bbb{Q^+}(x^2<2\Rightarrow 2\geq (x+\epsilon)^2)\\ &\Longleftrightarrow \exists\epsilon\in\Bbb{Q^+}\forall x\in\Bbb{Q^+}(x^2<2\Rightarrow 2> (x+\epsilon)^2), &\text{Since $\lnot\exists q\in\Bbb{Q^+}, q^2=2$.} \end{align}

We compute for $x$, to examine the inequality $(x+\epsilon)^2<2$ such that $x^2<2$, closer. Note that the inequality $(x+\epsilon)^2<2$ $\Longleftrightarrow$ $(x+\epsilon)^{2}-2<0$ is satisfied if and only if $\vert\epsilon^{2}+2\epsilon\vert <\vert x^{2}-2\vert$. By property of absolute values, we have

\begin{align} -(x^2-2)<\epsilon^2 +2\epsilon<x^2-2\\ -(x^2-2<\epsilon^2+2\epsilon)\qquad \land\qquad \epsilon^2+2\epsilon<x^2-2\\ x^2-2>\epsilon^2+2\epsilon\qquad \land\qquad\epsilon^2+2\epsilon+2<x\\ x^2>\epsilon^2+2\epsilon+2\qquad \land\qquad x^2>\epsilon^2+2\epsilon+2\\ x>\sqrt{\epsilon^2+2\epsilon+2} \end{align} But $x,\epsilon\in\Bbb{Q^+}$ implies that $\epsilon^2+2\epsilon+2>2\Longrightarrow \sqrt{\epsilon^2+2\epsilon+2}>\sqrt{2}$. Thus, by transitivity of order in $\Bbb{Q}$, $\sqrt{2}<\sqrt{\epsilon^2+2\epsilon+2}<x$. Therefore, a contradiction.


1 Answer 1


Your proof looks fine.

I would have done it as follows. If $n\in\mathbb N$,$$\left\lfloor2^n\sqrt2\right\rfloor<2^n\sqrt2<\left\lfloor2^n\sqrt2\right\rfloor+1$$and therefore$$\frac{\left\lfloor2^n\sqrt2\right\rfloor}{2^n}<\sqrt2<\frac{\left\lfloor2^n\sqrt2\right\rfloor}{2^n}+2^{-n}.$$So, given $\varepsilon\in\mathbb{Q}^+$, I would choose $n\in\mathbb N$ such that $2^{-n}<\varepsilon$ and choose $x=2^{-n}\left\lfloor2^n\sqrt2\right\rfloor$.

  • $\begingroup$ cool proof! thank you! Can you teach me how I could possibly arrive at a proof like yours, independently? Like the general strategy you used. $\endgroup$ Mar 31, 2018 at 8:23
  • 1
    $\begingroup$ @TheLastCipher What you were trying to prove was that we can always find rational numbes $x$ and $y$ such that $x<\sqrt2<y$ and that $x$ and $y$ are arbitrarily close. So, I condered the sequences of rational numbers$$\left(\frac{\left\lfloor2^n\sqrt2\right\rfloor}{2^n}\right)_{n\in\mathbb N}\text{ and }\left(\frac{\left\lfloor2^n\sqrt2\right\rfloor}{2^n}+2^{-n}\right)_{n\in\mathbb N},$$both of which converge to $\sqrt2$, one from below and the other one from above. $\endgroup$ Mar 31, 2018 at 8:30
  • $\begingroup$ thank you again! :) $\endgroup$ Mar 31, 2018 at 8:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .