0

I believe that it can be rewritten in the form: "If two integers are positive then their sum is positive.". i.e. A -> B

My math teacher believes that the contradiction is: "There exists two positive integers whose sum is nonpositive.".

But I believe that it is: "The sum of any two positive integers is nonpositive.".

Please note the main difference between my answer and his is the difference between "there exists and any". i.e. $\exists$ and $\forall$

Alexander Williams
  • 11
  • 1
    When you wrote the "contradiction of the statement", does that mean the same thing as the "negation" of the statement? – Joe Oct 21 '21 at 23:58
  • Joe I mean contradiction in the scope of “proof by contradiction”. i.e. the format A -> ~B used to prove a given proposition – Alexander Williams Oct 22 '21 at 00:11
  • Your teacher is right. If you found exactly two integers that summed to a negative number, you would have disproved the proposition. Your version requires a higher burden. – user317176 Oct 22 '21 at 00:33
  • I am not familiar with the "contradiction of a statement", but in a proof by contradiction, the logical form is $(\neg A \implies \mathrm{FALSE}) \implies A$. When you want to show that proposition $A$ holds using a proof by contradiction, you show that the negation of $A$ implies a contradiction, which in turn implies $\neg (\neg A)$, which is $A$. Your teacher's statement is the negation of the original statement. I recommend discussing this with your teacher to see if the negation is what was intended. – Joe Oct 22 '21 at 01:50
  • A suggestion: replace the term “any two” with “every pair of”. For non-native English speakers and students not used to reading about higher mathematics, the term “any” can be confusing since it sometimes means “all” and sometimes means “at least one”. See https://math.stackexchange.com/questions/402020/whats-the-difference-between-any-all-and-some/402073#402073 – KCd Oct 22 '21 at 01:51
  • The formula is: $\forall n,m [(n >0 \land m>0) \to (n+m >0)]$. Its negation will be: $\exists n,m [(n >0 \land m>0) \land (n+m \le 0)]$ because $\lnot (P \to Q)$ is $(P \land \lnot Q)$. – Mauro ALLEGRANZA Oct 22 '21 at 08:11

1 Answers1

1

View your initial argument as "for all positive $a, b$, $a + b$ is positive", then remember that the negation of $\forall$ is $\exists$. To prove by means of contradiction, it will suffice to suppose that there exists a positive pair $a, b$ such that $a + b$ is not positive.

Luis Alexandher
  • 1,260