Wrong theorem - The number 1 is the largest integer

Question

I have a theorem and its proof. I am trying to find the flaw in this wrong theorem.

Theorem : The number 1 is the largest integer.

Proof -> Suppose the conclusion is false. Then let $ n> 1$ be the largest integer. Multiplying both sides of this inequality by $n$, yields $n^2 > n$. This is contradiction, because $n^2$ is another integer larger than $n$. Thus the theorem is proved.

Now, using the quantifiers, I think theorem will be expressed as follows.

$$ \forall \,x \in \mathbb{Z}\left[ \, x < 1 \, \right] $$

And negation of this will be as follows.

$$ \exists \,x \in \mathbb{Z}\left[ \, x \geqslant 1 \, \right] $$

So, we have $ x \in \mathbb{Z} $ and $ x \geqslant 1$. So, I think that the negation which they have done is wrong. Is my intuition correct ?

I have seen few other posts talking about a relation question. But I could not follow their arguments. Hence the new question.

Answer 1

The claim here is$$(\forall n\in\Bbb Z):n\leqslant1.$$Denying this is to assert that$$(\exists n\in\Bbb Z):n>1.$$And the error lies in assuming that there is a largest integer.

Answer 2

Yes, when you do the negation, what you get is "1 is not the largest integer". Then, the wrong part comes when it says "Then let n>1 be the largest integer", the problem is that that $n$ may not exist.

Answer 3

The negation of "$1$ is the largest integer" should be "there exists some larger integer $n$ than $1$", in lieu of "there exists some largest integer $n$ larger than $1$". So you actually had assumed more than the negation of the statement, which is not allowed.