Express a Proposition In Formal Logic I am doing a question where I have to express: There is no largest prime number,
in formal logic.
This is the solution given:
Of course this is a true statement, so it could be expressed by the logically
equivalent formula “1 = 1,” but even if we didn’t know this, we could transcribe the
statement directly as:
$¬(∃p.   IS­PRIME(p) ∧ (∀q. IS­PRIME(q) \implies p ≥ q))$
My first question is in what way does 1 = 1 mean there is no largest prime number?
Secondly, to me the transcribed statement says that for all prime numbers q, there exists a prime number p that is greater than or equal to it (shouldn't it be just greater than?), and then there is the not at the start of the statement, I'm not sure why that is there or if it even makes sense to have it?
 A: One way to express There is no largest prime number in formal logic:
$$\forall p \exists q [\operatorname{prime}(q) \land (p<q)] $$
A: Your formalization of the statement :

There is no largest prime number

is correct; it can be translated also as :


$\forall p[Prime(p) \rightarrow \exists q (Prime(q) \land (p < q))]$


that means :

for every prime number $n$, there is a prime number $q$ such that $p < q$.

The two formulae are equivalent; starting from the formula :


$¬∃p[IS­PRIME(p)∧∀q(IS­PRIME(q) \rightarrow p≥q)]$


we can rewrite is, using first the rules for quantifiers [$\lnot \exists$ is equivalent to $\forall \lnot$] and De Morgan, as :

$∀p[¬ IS­PRIME(p) \lor \lnot ∀q(IS­PRIME(q) \rightarrow p≥q)]$

then, using again the rules for the quantifiers and the equivalence between $\lnot p \lor q$ and $p \rightarrow q$, as :

$∀p[IS­PRIME(p) \rightarrow ∃q \lnot (IS­PRIME(q) \rightarrow p≥q)]$

and finally, using the equivalence between $\lnot (p \rightarrow q)$ and $p \land \lnot q$, as :

$∀p[IS­PRIME(p) \rightarrow ∃q (IS­PRIME(q) \land \lnot (p≥q))]$.

A: The translation mentioned in the post is correct. Your suggested sentence that says there is no prime $p$ such that for all primes $q$ we have $p\gt q$ is trivially true, for given a $p$, take $q=p$. We need to know nothing about primes to assert the truth of that sentence. It is true even if $ISPRIME(k)$ is interpreted as being true only $k=2$, $8$, and $15$.
As to the "not" at the beginning, note that $\lnot(\exists \varphi(p))$ is logically equivalent to $\forall p(\lnot \varphi(p)$. The first was chosen presumably because of its somewhat greater faithfulness to the ordinary mathematical English we started with. I would myself prefer to use a universal quantifier, as suggested by Mauro Allegranza. But, to repeat, the translation of the post is correct, and we cannot replace $\ge$ by $\gt$ keeping the rest unchanged. 
