1) There exists an integer n such that for every integer m, we have m ≤ n
2) If n^2 is even, then n is even
3) Pigeonhole Principle
Reference: Different Websites
3 proofs.pdf
1) There exists an integer n such that for every integer m, we have m ≤ n
2) If n^2 is even, then n is even
3) Pigeonhole Principle
Reference: Different Websites
3 proofs.pdf
The first of those, that "there exist an integer n such that for every m we have m<n" is simply NOT TRUE! For any integer, n, m= n+ 1 does not satisfy "m< n". So of course your "proof" is incorrect. The error is where you say "There exists an integer m such that m > n. Since for any m, we have that m > n is true". You have jumped from
"there exists and integer m" to "for any m" and the second does NOT follow from the first.