[proofplan] We construct the desired number as the [supremum](/pages/1203) of the [real numbers](/pages/1303) set $S := \{x \in \mathbb{R} : x^2 < 2\}$. The set is non-empty and bounded above, so the [Least Upper Bound Property of the Real Numbers](/theorems/737) gives a real number $c = \sup S$. We then rule out $c^2 < 2$ by producing a larger element of $S$, and rule out $c^2 > 2$ by producing an upper bound for $S$ smaller than $c$. The only remaining possibility is $c^2 = 2$. [/proofplan] [step:Construct the supremum of the set of real numbers whose square is less than $2$] Define the set \begin{align*} S := \{x \in \mathbb{R} : x^2 < 2\}. \end{align*} The set $S$ is non-empty because $1 \in \mathbb{R}$ and $1^2 = 1 < 2$, so $1 \in S$. The number $2$ is an upper bound for $S$: if $x \in S$ and $x \geq 2$, then $x^2 \geq 4 > 2$, contradicting $x^2 < 2$; hence every $x \in S$ satisfies $x < 2$. Since $S \subset \mathbb{R}$ is non-empty and bounded above, the Least Upper Bound Property of the [Real Numbers](/page/Real%20Numbers) gives a real number $c \in \mathbb{R}$ such that \begin{align*} c = \sup S. \end{align*} Because $1 \in S$, every upper bound of $S$ is at least $1$, so $1 \leq c$. Because $2$ is an upper bound of $S$ and $c$ is the least upper bound, $c \leq 2$. Therefore \begin{align*} 1 \leq c \leq 2. \end{align*} [/step] [step:Rule out the possibility that the supremum has square less than $2$] Assume, for contradiction, that $c^2 < 2$. Define \begin{align*} t := \min\left\{\frac{1}{2}, \frac{2 - c^2}{10}\right\}. \end{align*} Then $0 < t < 1$ and $5t < 2 - c^2$. Since $1 \leq c \leq 2$, we have $2c \leq 4$. Hence \begin{align*} (c+t)^2 &= c^2 + 2ct + t^2 \\ &< c^2 + 2ct + t \\ &\leq c^2 + 5t \\ &< 2. \end{align*} Thus $c+t \in S$. But $t > 0$, so $c+t > c$, contradicting that $c$ is an upper bound for $S$. Therefore $c^2 < 2$ is impossible. [guided] Assume, for contradiction, that $c^2 < 2$. The goal is to use the positive gap $2 - c^2$ to move slightly to the right of $c$ while keeping the square below $2$. Define \begin{align*} t := \min\left\{\frac{1}{2}, \frac{2 - c^2}{10}\right\}. \end{align*} Because $c^2 < 2$, the number $(2 - c^2)/10$ is positive, so $t > 0$. Also $t \leq 1/2$, hence $t < 1$, and $t \leq (2 - c^2)/10$, hence $5t \leq (2 - c^2)/2 < 2 - c^2$. We estimate the square of $c+t$. Since $t < 1$, we have $t^2 < t$. Since $c \leq 2$, we have $2c \leq 4$. Therefore \begin{align*} (c+t)^2 &= c^2 + 2ct + t^2 \\ &< c^2 + 2ct + t \\ &\leq c^2 + 4t + t \\ &= c^2 + 5t \\ &< 2. \end{align*} This proves $c+t \in S$ by the definition of $S$. But $t > 0$, so $c+t > c$. This contradicts the fact that $c$ is an upper bound for $S$, because an upper bound must be at least every element of $S$. Hence the assumption $c^2 < 2$ cannot hold. [/guided] [/step] [step:Rule out the possibility that the supremum has square greater than $2$] Assume, for contradiction, that $c^2 > 2$. Define \begin{align*} t := \min\left\{\frac{1}{2}, \frac{c^2 - 2}{8}\right\}. \end{align*} Then $0 < t < 1$, $t < c$, and $4t < c^2 - 2$. Since $c \leq 2$, we have $2c \leq 4$. Hence \begin{align*} (c-t)^2 &= c^2 - 2ct + t^2 \\ &\geq c^2 - 2ct \\ &\geq c^2 - 4t \\ &> 2. \end{align*} We claim that $c-t$ is an upper bound for $S$. Let $x \in S$. If $x < 0$, then $x < c-t$ because $t < c$. If $x \geq 0$, then $x^2 < 2 < (c-t)^2$ and $c-t > 0$, so $x < c-t$. Thus every $x \in S$ satisfies $x < c-t$, so $c-t$ is an upper bound for $S$. Since $t > 0$, we have $c-t < c$, contradicting the fact that $c$ is the least upper bound of $S$. Therefore $c^2 > 2$ is impossible. [guided] Assume, for contradiction, that $c^2 > 2$. Now the gap is above $2$, and the strategy is to move slightly to the left of $c$ while keeping the square still greater than $2$. Define \begin{align*} t := \min\left\{\frac{1}{2}, \frac{c^2 - 2}{8}\right\}. \end{align*} Because $c^2 > 2$, the number $(c^2 - 2)/8$ is positive, so $t > 0$. Also $t \leq 1/2 < 1 \leq c$, so $t < c$ and therefore $c-t > 0$. Finally, $t \leq (c^2 - 2)/8$, so $4t \leq (c^2 - 2)/2 < c^2 - 2$. We estimate the square of $c-t$. Since $t^2 \geq 0$ and $c \leq 2$, we have \begin{align*} (c-t)^2 &= c^2 - 2ct + t^2 \\ &\geq c^2 - 2ct \\ &\geq c^2 - 4t \\ &> 2. \end{align*} We now prove that $c-t$ is an upper bound for $S$. Let $x \in S$. If $x < 0$, then $x < c-t$ because $c-t > 0$. If $x \geq 0$, then $x^2 < 2 < (c-t)^2$ and both $x$ and $c-t$ are non-negative; taking the order-compatible square root comparison for non-negative real numbers gives $x < c-t$. Thus in all cases $x < c-t$. Therefore $c-t$ is an upper bound for $S$. But $t > 0$, so $c-t < c$. This contradicts the defining property of $c = \sup S$, namely that no upper bound of $S$ is strictly smaller than $c$. Hence the assumption $c^2 > 2$ cannot hold. [/guided] [/step] [step:Conclude that the supremum is a real square root of $2$] The real number $c$ satisfies exactly one of the three alternatives $c^2 < 2$, $c^2 = 2$, or $c^2 > 2$ by the trichotomy property of the order on $\mathbb{R}$. The previous two steps ruled out the first and third alternatives. Therefore \begin{align*} c^2 = 2. \end{align*} Since $c \in \mathbb{R}$, taking $x := c$ proves that there exists $x \in \mathbb{R}$ with $x^2 = 2$. [/step]