Comment Countersignature Proof with airSlate SignNow

Comment countersignature proof

Welcome to an example on proof by counter example. Earlier, we proved, for all integers a and b, if a + b is odd, then a is odd or b is odd. Is the converse true? Recall if the implication is if P then Q, the converses if Q then P. The converse of the given statement is for all integers a and b, if a is odd or b is odd, then a + b is odd. However, this is false. How do we prove this is false? We need to prove the negation of the converse is true, and therefore, the converse is false. Let's look at this in terms of symbols. Begin represent the converse for all integers a and b, if a is odd or b is odd, then a + b is odd using the symbols shown here below. Again, we have for all integers a and b, if a is odd or b is odd, then a + b is odd. Since this is the converse, we now need the negation of this statement, and then prove that it's true to prove the converse is false. So here is a negation of the converse. Simplifying using the rules from the previous sections, we have there exists integers a and b such that a is odd or b is odd and a + b is not odd. Let's go over this negation in more detail in the next slide. As a negation passes by the quantifiers, they change from for every to their exist. And then we negate the implication. Remember, the negation of the implication, if P then Q is logically equivalent to P and not Q. This gives us the conjunction a is odd or b is odd and a + b is not odd. So again, we can state the negation of the converse as their exists integers a and b such that a is odd or b is odd and a + b is not odd. And now we know what to do to prove that the converse is false, we need to find two integers a and b, so that a is odd or b is odd, but a + b is not odd, meaning it's even. And this is pretty easy to do because we have an or statement between a is odd or b is odd. Either a is odd, b is odd, or they're both odd. So we can let a equal one, b equal three, and notice one plus three is equal to four, and four is not odd. This shows the negation of the commerce is true, and proves that the converse statement, for all integers a and b, if a is odd or b is odd, then a + b is odd is false. I hope you found this helpful.