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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.

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    $\begingroup$ The assumption that there is a largest positive integer is the flaw. $\endgroup$ Feb 9, 2016 at 22:37
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    $\begingroup$ You haven't stated the claim you are trying to prove. There is no flaw in your argument: it shows that if $n$ is the largest positive integer, then $n = 1$, which is a valid, since the premiss is false. $\endgroup$
    – Rob Arthan
    Feb 9, 2016 at 22:42
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    $\begingroup$ When Oskar Perron (Zur Existenzfrage eines Maximums oder Minimums, Jahresber. Deutsch. Math.-Verein. 22 (1913) 140-144) wanted to explain why a uniqueness proof of the isoperimetric inequality is not enough (Jakob Steiner had shown that any simple closed curve which is not a circle can be improved) he gave the above proof of the maximality of $1$ as an analogy (every integer that is not $1$ can be "improved" (made larger) by squaring). $\endgroup$ Feb 10, 2016 at 7:15
  • $\begingroup$ This thread has better answers than the recent Prove 1 is not the largest integer? $\endgroup$ Feb 10, 2016 at 7:19
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    $\begingroup$ Possible duplicate of Could you explain Perron's paradox to me, please? $\endgroup$
    – Watson
    Feb 10, 2016 at 10:29

7 Answers 7

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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!

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    $\begingroup$ @Chris: not exactly. In fact, if you assume a false statement, you can prove anything you like (because $\langle \text{false}\rangle \implies Q$ no matter what $Q$ is). But what it really says is that a valid proof of Theorem X can only start by assuming whatever is explicitly stated in the hypothesis of Theorem X. If you assume something else (such as existence of a largest integer), then you have a circular argument. Or, at best, a proof of a different theorem (i.e., Theorem X with an additional hypothesis added). $\endgroup$
    – Jack Lee
    Feb 10, 2016 at 1:10
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    $\begingroup$ So, the proof "1 is the largest integer if $n$ is the largest positive integer" is true but "1 is the largest integer" with no hypotheses is false? $\endgroup$
    – Chris
    Feb 10, 2016 at 1:51
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    $\begingroup$ That's correct, Chris. $\endgroup$ Feb 10, 2016 at 2:11
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    $\begingroup$ If you instead applied the successor operation to $n$, you would have a very nice proof that there is no largest integer (by reductio ad absurdum), which is actually a moderately interesting result in the foundations of mathematics. $\endgroup$
    – Kevin
    Feb 10, 2016 at 3:30
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    $\begingroup$ Incidentally, this is called Principle of Explosion. If you accept a false premise, you can "prove" anything as a consequence. $\endgroup$ Feb 10, 2016 at 23:15
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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.

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    $\begingroup$ Interesting choice (42).In The Hitch-Hiker's Guide To The Galaxy it is revealed that 6x9=42. That's not a typo. $\endgroup$ Feb 11, 2016 at 6:43
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    $\begingroup$ @user254665 And works in base 13 ... which is how we know God has 13 fingers... $\endgroup$
    – TripeHound
    Feb 11, 2016 at 12:02
  • $\begingroup$ Amazing! LOL.... $\endgroup$ Feb 11, 2016 at 15:50
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    $\begingroup$ @TripeHound 13 fingers on each hand or on both hands? Wait... does God have two hands? $\endgroup$
    – Earthliŋ
    Feb 11, 2016 at 16:37
  • $\begingroup$ @Earthliŋ We (supposedly) use base10 because we have 10 fingers in total; whether God has one tridecaphalangic hand, one 6-fingered hand and one 7-fingered one or more than two hands I don't know. $\endgroup$
    – TripeHound
    Feb 11, 2016 at 17:11
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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.

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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 $n$

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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?).

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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)

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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.

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  • $\begingroup$ I was disappointed to see that somebody had voted this answer down. This answer should not have been voted down. Thanks for leaving it up. +1 $\endgroup$
    – user729424
    Feb 2, 2020 at 20:04

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