My lecturer defined reducible a element $z \in \mathbb{Z}$ if $\exists x,y \in \mathbb{Z}$ such that $z=xy$ with $x,y \notin \mathcal{U}(\mathbb{Z})$ (the latter being the set of invertible elements of $\mathbb{Z}$. He then proceeds to say that an element is irreducible if $\nexists x,y \in \mathbb{Z}$ such that $z=xy$ with $x,y \notin \mathcal{U}(\mathbb{Z})$, and says that the latter is equivalent to $\forall x,y \in \mathbb{Z}, z=xy \implies (x \in \mathcal{U}(\mathbb{Z})) \vee (x \in \mathcal{U}(\mathbb{Z}))$.

(i) What "such that" and "with" mean logically in this context? I think that the reducible definition means $\exists x,y\in\mathbb{Z}, (z=xy)\wedge(x\notin \mathcal{U}(\mathbb{Z}))\wedge (y\notin \mathcal{U}(\mathbb{Z}))$. Is this correct? Or is there an implication like $\exists x,y\in\mathbb{Z}, (z=xy)\implies ((x\notin \mathcal{U}(\mathbb{Z}))\wedge (y\notin \mathcal{U}(\mathbb{Z})))$?

(ii) I am not convinced by the negation, should't the definition of irreducible be "not reducible" and so $\neg(\exists x,y \in \mathbb{Z}$ such that $z=xy$ with $x,y \notin \mathcal{U}(\mathbb{Z}))$? If …

I'm trying to prove that if $f$ is twice differentiable in $(a,b)$ and there exists $M\ge 0$ such that $|f''(x)| \le M$ for each $x \in (a,b)$, then $f$ is uniformly continuous. Let $x,y \in (a,b)$ and assume $x<y$. By the mean value theorem, there exists $c_{x,y} \in (x,y)$ such that $|f'(x)-f'(y)|=|x-y|\cdot |f'(c_{x,y})| \le M|x-y|$; this proves that $f'$ is Lipschitz continuous in $(a,b)$.

Hence, $f'$ is bounded by some $K\ge 0$. Using again the mean value theorem $|f(x)-f(y)|=|x-y| \cdot |f'(d_{x,y})| \le K|x-y|$ for some $d_{x,y} \in (x,y)$; so, $f$ is Lipschitz continuous as well an…