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It seems that the contrapositive of a true statement is true. Consider the following:

  1. $A,B,C,K\in \mathbb{N}$
  2. $\exists A,B,C: A^K+B^K=C^K→K≤2$
  3. Take the contrapositive: $$K>2→∀A,B,C:A^K+B^K<C^K∨A^K+B^K>C^K$$

However, can the contrapositive of a true statement be false? I ask because of the following:

  1. $A,B,C,D,K\in \mathbb{N}$
  2. $\exists A,B,C,D: A^K+B^K+C^K=D^K→K≤2$
  3. Take the contrapositive: $$K>2→∀A,B,C,D:A^K+B^K+C^K<D^K∨A^K+B^K+C^K>D^K$$ But this seems false due to the existence of Plato’s Number.

Another example where the contrapositive of a true statement seems false is the following:

  1. $A,B,C,K\in \mathbb{N}$
  2. $\exists A,B,C: A^K+B^K=C^K→K≤1$
  3. Take the contrapositive: $$K>1→A^K+B^K<C^K∨ A^K+B^K>C^K$$ But this seems false too.
qianqian07
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  • The correct contrapositive should be: If $K > 2$, then for all $A, B$, and $C$, $A^K + B^K ≠ C^K$. This is logically equivalent to Fermat's Last Theorem and is also true.
  • The correct contrapositive should be: If $K > 2$, then for all $A, B, C$, and $D$, $A^K + B^K + C^K ≠ D^K$. The truth of this statement is not known, and it is not disproven by the existence of Plato's number, which is not a counterexample for this statement.
    • – rumathe Mar 19 '23 at 11:51
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  • The correct contrapositive should be: If $K > 1$, then for all $A, B$, and $C$, $A^K + B^K ≠ C^K$. This statement is not false; it is actually true for $K > 2$, as it is equivalent to Fermat's Last Theorem.
    • – rumathe Mar 19 '23 at 11:51
  • The first counterexample, I gave is true though. For $1^2+6^2+18^2=19^2$ –  Mar 19 '23 at 11:59
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    You are right. The original statement is false for $K = 2$, as there are non-trivial solutions ($A$, $B$, $C$, and $D$ not being all zeros). Thus, the contrapositive is also false. The lesson here is that the contrapositive is always logically equivalent to the original statement, meaning they are either both true or both false – rumathe Mar 19 '23 at 12:01
  • @rumathe, here's the thing though. I said less than or equal to two though in the first counterexample. Which makes the statement true. –  Mar 19 '23 at 12:08
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    In this case, the original statement and its contrapositive are both true – rumathe Mar 19 '23 at 12:09
  • This seems like a contradiction then. –  Mar 19 '23 at 12:10
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    This doesn't contradict the original statement because the original statement claims that if such a solution exists, $K \leq 2$, which is true in this case. – rumathe Mar 19 '23 at 12:14