Prove: $\forall y \in \mathbb{R}^{\geq 0}, \forall n \in \mathbb{Z}^+, n > y \implies (\exists \epsilon \in \mathbb{R}^{\geq 0}, 0 \leq \epsilon < 1 \land y = (n+\epsilon)^2 - n^2)$
What do I do? The thing that confuses me the most is why is $y$ given as arbitrary and why do we have to show that it's equal to something. Wouldn't that mean $y$ isn't arbitary? How do I approach this proof?
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Solve $y=(n+\varepsilon)^{2}-n^{2}$ for $\varepsilon$, the equation has two roots $\varepsilon_1= \sqrt{y+n^{2}}-n$ and $\varepsilon_2=-\sqrt{y+n^{2}}-n$. Then prove that
$0\leq\varepsilon_1<1$ or $0\leq\varepsilon_2<1$. Is easy to show that $\varepsilon_2\leq0$ and for $\varepsilon_1$ note that $$0\leq y<n<2n+1\Rightarrow 0\leq y<2n+1$$ sum $n^{2}$ $$n^{2}\leq y+n^{2}<n^{2}+2n+1\Rightarrow n^{2}\leq y+n^{2}<(n+1)^{2}$$ take square root $$n\leq \sqrt{y+n^{2}}<(n+1)\Rightarrow 0\leq \sqrt{y+n^{2}}-n<1\Rightarrow0\leq\varepsilon_1<1$$.
Let $y\in\mathbb{R}^+$ and $n\in\mathbb{N}$ such that $n>y$. We want to show the existence of $\varepsilon\in[0,1[$ such that $y=(n+\varepsilon)^2-n^2$. Let $f(\varepsilon)=(n+\varepsilon)^2-n^2$ for all $\varepsilon\in[0,1[$. $f$ is a continuous non-decreasing function, thus by the intermediate value theorem, $f([0,1[)=[f(0),f(1)[=[0,(n+1)^2-n^2[=[0,2n+1[$. But $n>y$ so $2n+1>y$ and thus $y\in[0,2n+1[$ and because of what said above, there exists $\varepsilon\in[0,1[$ such that $y=f(\varepsilon)=(n+\varepsilon)^2-n^2$.