We compare [[There is a unique real number that is the square root of 2]] and [[Existence & uniqueness nth root of positive reals]].
| Summary | $\sqrt{ 2 }$ | $\sqrt[q]{ y }$ |
| ---------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| Take $y>1$ | We already have $2>1$ | Take $y>1$ |
| Define useful set | $S= \{ x \in \mathbb{R} \mid x^{2} < 2 \}$ | $S= \{ x \in \mathbb{R} \mid x^{q} < y \}$ |
| Show that $1 \in S$ and that $S$ has an upper bound and deduce that it has supremum $r$ that is geq to 1 (since $1 \in S$) by [[Real numbers]] q| $x^{2}<2<4 \implies x<2$ | For $y>1$, $x^{q}<y<y^{q}\implies x <y$ |
| Show that $(r+\epsilon)$ is bounded above for $\epsilon<1$ | $(r+\epsilon)^{2} = r^{2} + 2\epsilon r + \epsilon^{2} < r^{2}+\epsilon(2r+1)$ since $\epsilon<1 \implies \epsilon^{2}< \epsilon$ | Using binomial expansion, $r \geq 1$ and $\epsilon<1$ deduce that $(r+\epsilon)^{q} \leq r^{q}(1+\epsilon2^{q})$ |
| Choose upper bound for $\epsilon$ so that $(r+\epsilon)^{q}<y$ | $0<\epsilon< \frac{r^{2}-2}{2r+1} \implies (r+\epsilon)<2$ | $0<\epsilon < (yr^{-q}-1)/2^{q}\implies (r+\epsilon)^{q}\leq y$ using above result |
| Deduce that $r+\epsilon \in S$ so $r$ cannot be an upper bound for $S$ | | |
| Show that $(r-\epsilon)$ is bounded below by an expression containing $-\epsilon$ for $\epsilon<1$ | $(r-\epsilon)^{2}= r^{2}-2\epsilon r+\epsilon^{2} \geq r^{2}-2\epsilon r$ since $\epsilon^{2}>0$ | For $-\frac{\epsilon}{r} > -1$, uysing [[Bernoulli's Inequality]], $(r- \epsilon)^{q} = r^{q}\left( 1- \frac{q\epsilon}{r} \right)^{q} \geq r^{q} \left( 1- \frac{q\epsilon}{r} \right)$ |
| Choose upper bound for $\epsilon$ (lower bound for $-\epsilon$) so that $(r-\epsilon)^{q}>y$ | $0<\epsilon <\min(1, (r^{2}-2)/2r)\implies(r-\epsilon)^{2}>2$ | $0<\epsilon<\min(1,rq^{-1}(1-yr^{-q})) \implies (r-\epsilon)^{q}>y$ |
| Deduce that this contradicts the assumption that $r$ is the least upper bound for $S$ | | |
| Deduce by trichotomy that $r^{q}=y$ since both $r^{q} < y$ and $r^{q}>y$ are false. | | |
| Prove uniqueness using *trichotomy* again | | |