I want to check if this statement is true and then negate it.

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

  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 $x$:

$$\frac{1}{m^2}<x \\ \iff \frac{1}{\sqrt{x}}<m\\ \implies \frac{1}{\sqrt{x}}<M\le m$$

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?

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    $\begingroup$ 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 $\Psi$? $\endgroup$ – Asaf Karagila Nov 1 '18 at 13:35
  • $\begingroup$ @AsafKaragila I don't think so. This is how it is written on my problem sheet. Is this incorrect? $\endgroup$ – Nullspace Nov 1 '18 at 14:10
  • $\begingroup$ No, it's just very unusual to see $\psi$ and $\Psi$ denote natural numbers. $\endgroup$ – Asaf Karagila Nov 1 '18 at 14:11
  • $\begingroup$ @AsafKaragila Do you want me to change it to $m$ and $M$? $\endgroup$ – Nullspace Nov 1 '18 at 14:16
  • $\begingroup$ It's your question, given to you by your teachers. I'm just pointing out that it is quite the odd choice of notation. $\endgroup$ – 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.

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 "$\gt$.")

  • $\begingroup$ 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! $\endgroup$ – Nullspace Nov 1 '18 at 15:24
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    $\begingroup$ 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 $1/\sqrt x$. $\endgroup$ – Barry Cipra Nov 1 '18 at 16:20

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