MATH 3150 Class Notes 02-04-2025

notereal-analysis

Theorem

The equation $x^2=2$ has a unique positive solution in $\mathbb{R}$.

Proof

Existence: Let $S=\{x\in \mathbb{R}\mid x\ge 0,x^2\le 2\}$. We want to show that $S$ is bounded above, and that $M=\sup S$ satisfies $M^2=2$. If there exists a solution to $x^2=2$, then $x=\sup S=\max S=\sqrt{2}$ satisfies $x^2=2$.

$\bar{M}=2$ is an upper bound of $S$. If $x>2⟹x^2>4$ so $x\notin S$. Thus, by the completeness axiom, $S$ has a supremum. Let $M=\sup S$. We want to show that $M^2=2$. Assume for the sake of contradiction that $M^2\ne 2$.

Case 1: $M^2<2$. There exists $\epsilon >0$ such that $M^2+\epsilon <2$. For $\delta >0$ to be specified below, let $\bar{M}=M+\delta$, so ${\bar{M}}^2=(M+\delta {)}^2=M^2+2m\delta +{\delta }^2$. Therefore, ${\delta }^2<\delta$ ($\delta <1$), and $2M\delta <4\delta$ because $M<2$ $⟹{\bar{M}}^2=M^2+2m\delta +{\delta }^2<M^2+5\delta <(2-\epsilon )+5\delta$.

For $\delta =\frac{\epsilon }{5}$ it follows that ${\bar{M}}^2<2-\epsilon +5\delta =2$, so $\bar{M}\in S$, which contradicts $M=\sup S$ and $\bar{M}=M+\delta >M$.

Case 2: $M^2>2$. $M^2>2⟹\exists \epsilon >0$ such that $M^2>2+\epsilon$. Since $M=\sup S⟹\forall \delta >0\exists x=x(\delta )\in S$ so that $x\le M\le x+\delta ⟹M^2\le (x+\delta {)}^2=x^2+2x\delta +{\delta }^2\le 2+5\delta$. Therefore, $2+\epsilon <M^2\le 2+5\delta$ contradicts $\delta$ being arbitrary.

Therefore, $M^2=2$.

Infimum

Remark

Any set $S$ bounded below has an infimum.

This follows from the completeness axiom.