**Proposition** $\exists! r \in \mathbb{R}\;(r^{2}=2)$. See similar proof [[Existence & uniqueness nth root of positive reals]] and [[Comparison of square root of 2 and nth root of y proofs]]. **Remark** This is a simple consequence of the [[Intermediate Value Theorem (IVT)]] i.e given continuous function $f: x \mapsto x^2$ (as shown here [[Examples 4 Draft#^67955b]]) there is a solution for $f(x) = 2$ which is clearly unique. The proof of IVT is similar to the one above since they both use LUBA. **Proof** Consider the set $S:= \{ x \in \mathbb{R} \mid x^{2}<2 \}$. Note that $1 \in S$ (so it's non-empty) and is bounded above since if $x>2$ then $x^{2} > 4$ so $x \not \in S$: it follows that $x\in S \implies x \leq 2$. By [[Real numbers]] the set $S$ has a supremum. Define $r:= \sup \{ x \in \mathbb{R} \mid x^{2} <2 \}.$ By [[Real numbers#^2d1ba9|trichotomy]], there are three possibilities (1) $r^{2}<2$; (2) $r^{2}=2$; (3) or $r^{2}>2$. Suppose (1) holds and $r^{2}<2$. Consider $(r+\epsilon)^{2}= r^{2}+2\epsilon r+\epsilon^{2}.$If $\epsilon<1$ then $\epsilon^{2}<\epsilon$ so in this case $(r+\epsilon)^{2} < r^{2}+2\epsilon r+\epsilon=r^{2}+\epsilon(2r+1).$Now if we choose $\epsilon<\min\left( 1, \frac{2-r^{2}}{2r+1} \right)$ then $(r+\epsilon)^{2}<r^{2}+\frac{2-r}{2r+1}(2r+1)=2$so $r+\epsilon \in S$. But $r+\epsilon>r$ so $r$ cannot be an upper bound for $S$: possibility (1) cannot be true. Now suppose (3) holds and $r^{2}>2$. Consider $(r-\epsilon)^{2}= r^{2}-2r\epsilon+\epsilon^{2} > r^{2}-2r\epsilon.$since $\epsilon^{2}>0$. If we take $0<\epsilon < \frac{r^{2}-2}{2r}$ then $(r-\epsilon)^{2}>2.$It follows that $r-\epsilon$ is an upper bound for $S$, since $x \in S$ we have $x^{2}<2<(r-\epsilon)^{2}$this implies that $x<r-\epsilon$. But contradicts the fact that $r= \sup S$. So case (2) must hold and $r^{2}=2$. To show that there is no other real number with this property, suppose $s^{2}=2$. Then either (i) $s<r$, or (ii) $s=r$, or (iii) $s>r$. If $s<r$ then $s^{2}<r^{2}=2;$ if $s>r$ then $s^{2}>r^{2}=2$; so we must have $s=r$.