# Statement is true but contrapositive is false?

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: $$K>2→∀A,B,C:A^K+B^KC^K$$

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: $$K>2→∀A,B,C,D:A^K+B^K+C^KD^K$$ 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: $$K>1→A^K+B^KC^K$$ But this seems false too.
• 1. 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. 2. 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. Commented Mar 19, 2023 at 11:51
• 3. 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. Commented Mar 19, 2023 at 11:51
• 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 Commented Mar 19, 2023 at 12:01
• In this case, the original statement and its contrapositive are both true Commented Mar 19, 2023 at 12:09
• 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. Commented Mar 19, 2023 at 12:14

Too long to post as a comment, but note that in the below, (1) and (1C) are logically equivalent, as are (2) and (2C).

$$∃A,B,C,D: A^K+B^K+C^K=D^K→K≤2\tag1$$

$$∃A,B,C,D\:( A^K+B^K+C^K=D^K)→K≤2\tag1$$ and $$∃A,B,C,D\:( A^K+B^K+C^K=D^K→K≤2)$$ are logically inequivalent, so please replace that confusing colon with parentheses. Fixing this and making explicit your implicit universal quantification of $$K:$$ $$\forall A,B,C,D,K\:( A^K+B^K+C^K=D^K→K≤2).\tag1$$

Contrapositive: $$\forall A,B,C,D,K\:( K>2\to A^K+B^K+C^K\ne D^K).\tag{1C}$$

$$∃A,B,C: A^K+B^K=C^K→K≤1\tag2$$

$$\forall A,B,C,K\:(A^K+B^K=C^K→K≤1)\tag2$$

Contrapositive: $$\forall A,B,C,K\:(K>1→A^K+B^K\ne C^K).\tag{2C}$$

• 1. I did not downvote you. 2. I'm not sure that downvoters are obligated to [blah blah]. Employing one of the site mechanisms (up/down votes and accepting answers) for organising this site is actually being quite considerate, haha. 3. You're welcome for my edits and long comment/answer. Commented Mar 19, 2023 at 15:15
• Thank you! I appreciate it
– user1148275
Commented Mar 19, 2023 at 15:16
• @AutistandProud 1.You're welcome; do you understand my answer (incl the part where I flipped your ∃A,B,C to ∀A,B,C? $\quad$ 2. What does edited OP mean? The edits have merely improved your post. Commented Mar 19, 2023 at 15:19
• @AutistandProud No, just logical equivalence. Commented Mar 19, 2023 at 15:22
• Got it. Thank you!
– user1148275
Commented Mar 19, 2023 at 15:22