I want to check if this statement is true and then negate it.
$$\forall x>0 \space \exists M \in \mathbb{N}\space \forall \space m \ge M:\frac{1}{m^2} <x$$
- Negating the statement:
$$\exists x>0\forall M \in \mathbb{N}\forall m \ge M:\frac{1}{m^2}\ \ge x$$
- Which statement is correct? My approach: I solved the inequality for $x$:
$$\frac{1}{m^2}<x \\ \iff \frac{1}{\sqrt{x}}<m\\ \implies \frac{1}{\sqrt{x}}<M\le m$$
If my reasoning is correct, no matter what $x$ I am given, I can always find a $M \in \mathbb{N}$ that satisfies the condition. Also, once I have that $M$ all $m \ge M$ will also satisfy that inequality. Is my logic correct here? My reasoning sounds very "handwavy". Is there a way I can formalize this?
When you negative a logical statement that starts with a string of $\forall$'s and $\exists$'s, each $\forall$ becomes a $\exists$ and each $\exists$ becomes a $\forall$. So the "$\forall x\gt0\exists M\in\mathbb{N}\forall m\ge M$" should become "$\exists x\gt0\forall M\in\mathbb{N}\exists m\ge M$." You left the final qualifier ("$\forall m\ge M$") unchanged.
Your proof that the statement (as opposed to its negation) is true is basically correct, but would benefit from a cleaner exposition. (If I were your instructor, I would ask you to redo it.) Here's one way to put it:
For $x\gt0$, let $M=\lceil1/\sqrt x\rceil+1$ (where $\lceil\cdot\rceil$ is the "ceiling" function). It follows that
$$m\ge M\implies m\ge\lceil1/\sqrt x\rceil+1\gt1/\sqrt x\implies\sqrt x\gt1/m\implies x\gt1/m^2$$
I'll leave it as an exercise why it's necessary to let $M=\lceil1/\sqrt x\rceil+1$ instead of just $\lceil1/\sqrt x\rceil$. (Hint: there's a difference between "$\ge$" and "$\gt$.")