Propositional logic : commas and implication confusion

Question

Im trying to transalate this sentence into propositional logic:

The square of a nonzero integer is positive.

The solution of my book states the following: $$ \forall x \in \mathbf{Z}, (x \neq 0) \implies ( x^2 > 0) $$

However i did this:

$$ (\forall x \in \mathbf{Z} \land x \neq 0 ) ( x^2 > 0) $$ The reason why i didn't use implication was because in the same book for this sentence:

For all natural numbers n, $n^2 + n + 41$ is prime.

The book translates it like this

$$(\forall n \in N)(n^2 + n + 41 \text{ is prime}) $$

and the implication was not used which left me confused

My question is:

When is necessary to use an implication , and it is correct to use an "and" rather than a comma?

Answer 1

Perhaps $(\forall x \in \mathbb Z \land x \neq 0 ) ( x^2 > 0) $ may sound intuitive to you, but is not the correct way to convey what you want. You want every integer that is not $0$ to have a positive square. The way you wrote it is syntactically incorrect (rigurously, it is ambiguous) , but even if it was written as $\forall x \in \mathbb Z (x \neq 0 \land x^2 > 0) $, it would say that every integer is not $0$ AND has a positive square, both of these assertions are false. You want the integers which are not $0$, to have a positive square, so in this case, in order to "check" if some $x \in \mathbb Z$ has a positive square, you first need to "check" if said $x$ is not $0$, hence $\forall x \in \mathbb Z ( x \neq 0 \Rightarrow x^2 > 0)$ . As for the latter proposition abount $n^2 + n +41$, there is nothing to check a priori, since the wording of the sentence simply states "for all natural numbers". This last statement is actually false by the way.

Answer 2

The square of a nonzero integer is positive.

The correct form for your attempt would be: $$\forall x~.((x\in\Bbb Z\wedge x\neq 0)\to(x^2>0))$$

and the implication was not used which left me confused

Universal statements with forms like $\forall x~. (x\in\mathcal S\to \varphi(x))$ may be shortened by abbreviating the set inclusion with the declaration of the bound variable — to $\forall x\in\mathcal S~.\varphi(x)$ —, but this is only allowed for set inclusions.

So, since $\Bbb Z{\smallsetminus}\{0\}$ is the set of non-zero integers, you might represent the statement as: $$\forall x\in \Bbb Z{\smallsetminus}\{0\} ~. x^2>0 $$

Alternatively you may notice that, by the exportation equivalence, your attempt may also be written as: $$\forall x~. (x\in\Bbb Z\to(x\neq 0\to x^2>0))$$

Thus this may be abbreviated to the answer's form:$$\forall x\in\Bbb Z~.(x\neq 0\to x^2>0) $$

Or with bracketing around the variable binding instead of the punctuation. $$(\forall x\in\Bbb Z)~(x\neq 0\to x^2>0)$$