Araceli Soto

2022-04-22

Proof Confusions: How do I approach this?

$\mathrm{\forall}y\in {\mathbb{R}}^{\ge 0},\mathrm{\forall}n\in {\mathbb{Z}}^{+},n>y\Rightarrow (\mathrm{\exists}\u03f5\in {\mathbb{R}}^{\ge 0},0\le \u03f5<1l{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}y={(n+\u03f5)}^{2}-{n}^{2})$

amisayq6t

Beginner2022-04-23Added 18 answers

Step 1

Let $y\in {\mathbb{R}}^{+}$ and $n\in \mathbb{N}$ such that $n>y$. We want to show the existence of $\epsilon \in [0,1[$ such that $y={(n+\epsilon )}^{2}-{n}^{2}$

Let $f\left(\epsilon \right)={(n+\epsilon )}^{2}-{n}^{2}$ for all $\epsilon \in [0,1[$ f is a continuous non-decreasing function, thus by the intermediate value theorem,

$f([0,\text{}1[)$

But

$f([0,1\left[\right)=[f\left(0\right),f\left(1\right)[=[0,{(n+1)}^{2}-{n}^{2}[=[0,2n+1[n>y\mathrm{so}2n+1y$ and thus $y\in [0,2n+1[$ and because of what said above, there exists $\epsilon \in [0,1[$ such that $y=f\left(\epsilon \right)={(n+\epsilon )}^{2}-{n}^{2}$

Jonas Dickerson

Beginner2022-04-24Added 22 answers

Step 1

Solve$y={(n+\epsilon )}^{2}-{n}^{2}$ sor $\epsilon$ , the equation has two roots ${\epsilon}_{1}=\sqrt{y+{n}^{2}}-n$ and ${\epsilon}_{2}=-\sqrt{y+{n}^{2}}-n$ . Then prove that $0\le {\epsilon}_{1}<1$ or $0\le {\epsilon}_{2}<1$ . Is easy to show that ${\epsilon}_{2}\le 0$ and for $\epsilon}_{1$ note that

$0\le y<n<2n+1\Rightarrow 0\le y<2n+1$

sum$n}^{2$

$n}^{2}\le y+{n}^{2}<{n}^{2}+2n+1\Rightarrow {n}^{2}\le y+{n}^{2}<{(n+1)}^{2$

take square root

$n\le \sqrt{y+{n}^{2}}<(n+1)\Rightarrow 0\le \sqrt{y+{n}^{2}}-n<1\Rightarrow 0\le {\epsilon}_{1}<1$

Solve

sum

take square root

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