Is this statement correct $\forall x>0 \space \exists M \in \mathbb{N}\space \forall \space m \ge M:\frac{1}{m^2} <x$?

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}

1. Negating the statement:

$$\exists x>0\forall M \in \mathbb{N}\forall m \ge M:\frac{1}{m^2}\ \ge x$$

1. Which statement is correct? My approach: I solved the inequality for $$x$$:

$$\frac{1}{m^2}

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?

• Is there some shortage of letters like $m,n,k,i,j$ or others that are often used to denote natural numbers that we need to resort to $\psi$ and $\Psi$? – Asaf Karagila Nov 1 '18 at 13:35
• @AsafKaragila I don't think so. This is how it is written on my problem sheet. Is this incorrect? – Nullspace Nov 1 '18 at 14:10
• No, it's just very unusual to see $\psi$ and $\Psi$ denote natural numbers. – Asaf Karagila Nov 1 '18 at 14:11
• @AsafKaragila Do you want me to change it to $m$ and $M$? – Nullspace Nov 1 '18 at 14:16
• It's your question, given to you by your teachers. I'm just pointing out that it is quite the odd choice of notation. – Asaf Karagila Nov 1 '18 at 14:17

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.
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 "$$\gt$$.")
• The "ceiling function" was exactly what I was looking for. I was lacking the mathematical vocabulary for saying "take $\frac{1}{\sqrt{x}}$ and go to the next integer/natural number. Thanks for your help! – Nullspace Nov 1 '18 at 15:24
• Come of think of it, $M=\lfloor1/\sqrt x\rfloor+1$, with the "floor" function, would have worked as well. The trick is to get an integer that's strictly larger than $1/\sqrt x$. – Barry Cipra Nov 1 '18 at 16:20