Consider the set $T$={$t\in\Bbb{R}:t^2<2$}

Let $\alpha=supT$ which implies $\forall\epsilon>0\exists t\in T$ such that $\alpha-\epsilon<t$.

Suppose $\alpha^2<2$ $\Rightarrow$ $\alpha^2=2-\epsilon_0$, for some $\epsilon_0\in\Bbb{R}$. By the Archimedian Property, for every $\epsilon_0$ there exist $n\in\Bbb{N}$ such that $\frac{1}{n}<\epsilon_0$. Let us let $\epsilon_1=\frac{1}{n_1}<\epsilon_0$ for some $n_1\in\Bbb{N}$. Then there exist $t_1\in T$ such that $t_1^2=2-\epsilon_1$ which implies that $t_1^2=2-\epsilon_1>\alpha^2=2-\epsilon_0$ . By property of a supremum, this cannot be because $t_1\in T$. Hence, a contradiction. Therefore, it cannot be that $\alpha^2<2$

Suppose $\alpha^2>2$ $\Rightarrow$ $\alpha^2=2+\epsilon_2$ where $\epsilon_2>0$. Again by the Archimedean property, we can find a real number $\epsilon_3=\frac{1}{n_3}$ for some $n_3\in\Bbb{N}$ such that $\epsilon_3<\epsilon_2$, $\forall\epsilon_2\in\Bbb{R}$. Let $q=\epsilon_2 - \epsilon_3$. Then $\alpha^2-q=2 + \epsilon_3$ which means $(supT)^2-q>t, \forall t\in T$. Thus contradicting what a supremum is.

Therefore, $\alpha^2$ cannot be greater or less than $2$. Meaning $\alpha^2=2$

  • 1
    $\begingroup$ The third line of your argument refers to the number $\sqrt{2}$. But that's the number you're trying to prove the existence of. You have to rephrase your logic (which I haven't read beyond this point) to avoid that contradiction. $\endgroup$ Jan 17, 2018 at 13:40
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    $\begingroup$ The method of proof will depends on how you define the real numbers. For instance, if you use dedekind cuts then it is a one-liner. I presume you're using the complete ordered field axioms? $\endgroup$ Jan 17, 2018 at 13:43
  • $\begingroup$ Damn. I need to start over. thanks! $\endgroup$ Jan 17, 2018 at 13:43
  • $\begingroup$ Assuming $\alpha^2<2$ show that there is a number $x\in T$ which is greater than $\alpha$. This is a contradiction. Similar contradiction can be derived with assumption $\alpha^2>2$. $\endgroup$
    – Paramanand Singh
    Jan 17, 2018 at 13:44
  • $\begingroup$ @dbx i haven't reach dedekind cuts yet. I'm using abbott's analysis, this is a theorem with a proof provided in the preliminaries. $\endgroup$ Jan 17, 2018 at 13:44

1 Answer 1


Consider taking the set $S = \{r \in \mathbb{R} | r^2 > 2\}$

Then show that $\alpha = \sup(T)$ is not in $S$ or $T$ which means $x^2$ is neither greater than or less than than $2$.

To finish the proof, first, assume $\alpha \in S$. Then $\alpha ^2 > 2 \implies (\alpha-\epsilon)^2 >2$ for sufficiently small $\epsilon \implies t \leq \alpha-\epsilon < \alpha$ for $t \in T$ which contradicts $\sup$.

Assume $\alpha \in T$. Then $\alpha^2 < 2 \implies (\alpha + \epsilon)^2 < 2$ for sufficiently small $\epsilon \implies \alpha + \epsilon \in T$ but $\alpha < \alpha + \epsilon$ which is a contradiction.

  • $\begingroup$ You need to define $T$ as the complement of $S$. $\endgroup$ Jan 17, 2018 at 14:49
  • $\begingroup$ Well, to get around doing that entirely I guess this could be done without the creation of a second set by just saying that $\alpha^2$ has to be less than, equal to, or greater than to 2. And show the contradictions that way. $\endgroup$ Jan 17, 2018 at 15:02
  • $\begingroup$ That would be fine, too. I just note that $T$ is not defined anywhere. I started my earlier comment by suggesting that it should be $r \lt 2$ in the definition of $S$ but then noted you were working with the $\sup$ of $T$. Either way would make it correct. $\endgroup$ Jan 17, 2018 at 15:06

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