Let’s refresh the concept of direct proof. “A direct proof of a conditional statement p → q is constructed when the first step is the assumption that p is true; subsequent steps are constructed using rules of inference, with the final step showing that q must also be true. A direct proof shows that a […]
“A direct proof of a conditional statement p → q is constructed when the first step is the assumption that p is true; subsequent steps are constructed using rules of inference, with the final step showing that q must also be true. A direct proof shows that a conditional statement p → q is true
6- Use a direct proof to show that the product of two odd numbers is odd Read More »
Let’s refresh the concept of direct proof. “A direct proof of a conditional statement p → q is constructed when the first step is the assumption that p is true; subsequent steps are constructed using rules of inference, with the final step showing that q must also be true. A direct proof shows that a
7. Use a direct proof to show that every odd integer is the difference of two squares. Read More »
Sometimes, we must attempt different proof methods to prove a certain theorem or complete a certain proof. This is what you will find usually in practice. Think about it, you are doing some research, you think something is true under certain conditions, you create a conjecture, and now you have to prove it. No one
First, let’s refresh the concept of direct proof. “A direct proof of a conditional statement p → q is constructed when the first step is the assumption that p is true; subsequent steps are constructed using rules of inference, with the final step showing that q must also be true. A direct proof shows that
3. Show that the square of an even number is an even number using a direct proof Read More »
The following will serve as a guide to answer this exercise: a) ∀x(C(x)→F(x)) All comedians are funny. b) ∀x(C(x)∧F(x)) Every person is funny and a comedian. A shortened way is every person is a funny comedian. c) ∃x(C(x)→F(x)) There exists at least one person that, if that person is a comedian, then the person is
De Morgan’s Laws: ¬(p∨q)≡¬p∧¬q ¬(p∧q)≡¬p∨¬q As in a previous exercise, if we use propositional variables, we will not make mistakes while solving this type of exercise. a) Kwame will take a job in industry or go to graduate school Let p=”Kwame will take a job in industry” and q=”Kwame will go to graduate school”. ¬(p∨q)≡¬p∧¬q,
8. Use De Morgan’s laws to find the negation of each of the following statements. Read More »
See below the truth table for a conditional statement, and conjunction and disjunctions of two propositions. We will use them to solve this exercise. a) (p∧q)→p From the truth table for the conjunction, we know that if p^q is true, then p and q must be true. If p is true, then (p^q)->p is also
De Morgan’s Laws: ¬(p∨q)≡¬p∧¬q ¬(p∧q)≡¬p∨¬q a) Jan is rich and happy The way to solve this type of exercise and make sure not to make a mistake is straightforward: Let p=”Jan is reach” and q=”Jan is happy” The negation of p^q is as follows: ¬(p∧q)≡¬p∨¬q, by the De Morgan Law. Now, we can write it
7. Use De Morgan’s laws to find the negation of each of the following statements Read More »
Relevant definitions: “A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a ∈ A to denote that a is an element of the set A. The notation a∉A denotes that a is not an element of the set A.”
8. For each of the sets in Exercise 7, determine whether {2} is an element of that set Read More »
