What is the flaw of this proof (largest integer)?

Question

Let $n$ be the largest positive integer. Since $n ≥ 1$, multiplying both sides by $n$ implies that $n^2 ≥ n$. But since $n$ is the biggest positive integer, it is also true that $n^2 ≤ n$. It follows that $n^2 = n$. Dividing both sides by $n$ implies that $n = 1$.

The goal is to find the flaw in the reasoning of the proof rather than find a proof that proves it wrong. Here's where I think the problem is:

It was stated that $n^2 ≥ n$ after multiplying both sides of the inequality by $n$. But then because $n$ is the largest possible integer, $n^2 ≤ n$. This is where I think the flaw is, because if $n$ is the single largest integer, then we'd get $n^2 < n$ rather than $n^2 ≤ n$, so writing the latter is incorrect.

Answer 1

In fact, you have given a valid proof of a true theorem.

Theorem. If $n$ is the largest positive integer, then $n=1$.

You would start a proof of this theorem exactly the way you did start: "Let $n$ be the largest positive integer." So your proof is perfectly valid. But it doesn't prove that $1$ is the largest positive integer; that would be a different theorem: There exists a largest positive integer, and it is equal to $1$. To prove that stronger theorem, you'd first have to prove existence, which of course you cannot do.

The theorem statement you did prove is an example of a mathematical statement that is "vacuously true." This means it is true because its hypothesis is always false. If you look at the truth table for the implication $P\implies Q$, you'll see that in all cases in which $P$ is false, the implication is true. So you proved a true (but entirely uninteresting) result!

Answer 2

You've assumed what you wanted to prove: this would work as a proof that there is no largest integer (assuming there is an reaching a contradiction).

Regarding your answer, think about this: if $2>1$ then it is true that $2\ge 1$ (why?).

Answer 3

There is no flaw in the argument. You have proved the statement

If the largest positive integer $n$ exists, then $n=1$.

This statement is true, indeed you have proved it. Note that also the statement

If the largest positive integer $n$ exists, then $n=42$

is true, because there's no largest positive integer.

Answer 4

There's no flaw in the reasoning!! The thing is that, when you reach that point : $n=1$ then you should say "which is an absurd since for instance $2>1$ "(if you are trying to prove that there's no largest integer).

Maybe you could've discarded that case before: You know that ,supposing you have $n$ the largest positive integer, then $n\neq 1$, which is the only positive integer such that $n^2=n $, so you KNOW that when you multiply by $n$ , you have $n^2>n$ , but $n$ was the largest, so $n\geq n^2$ ABS! (again)

Answer 5

This seems like a proof by contradiction, i.e. in order to prove that there is no largest integer, make the assumption that there IS one and poke a hole in it. Since the flawless logic indicates that the largest integer would be 1, and we can clearly indicate that it is not, you have a contradiction and a proof that there is no single largest integer.

Answer 6

The flaw in the logic occurs here:

Let $n$ be the largest positive integer.

And this statement should be corrected:

Dividing both sides by $n$ implies that $n=1$.

The correct statement should be:

It follows that $n^2=n$. Since this is only true when $n=1$ and there exist integers larger than $1$, n is not the biggest positive integer.

You have a proof that there is not a biggest positive integer.

EDIT:

This is the inverse of Proof by infinite Descent - maybe it should be called "Proof by Infinite Ascent" since you can always have a larger $n$

Answer 7

The alternative flaw is with squaring. When you take $n^2$, you implicitly assume that there exists a number $x$ with $x=n^2$. If $n$ is the largest positive integer (and naturally you must be working with a number system where such a largest positive integer exists) then this is incompatible with the assumption that successors, sums, products, squares, etc. necessarily exist. If we have a set of positive numbers $X$ such that, whenever $x$ is in this set, then so is $x+1$, and $x$ never equals $x+1$, then necessarily $X$ has no greatest element. Likewise if we assume we can always square. To assume successors and products always exists and that there is a greatest positive integer is the problem: you can have one, but not the other.