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 »
In this case, we will attempt a proof by contradiction. Remember that to use proof by contradiction to prove a conditional statement(p->q), we do the following: This type of proofs is valid because p->q ≡ (p^q)-> F. p= “mn is even” q= “m is even or n is even” ¬q = “m and n are
Prove that if m and n are integers and mn is even, then m is even or n is even Read More »
This type of exercise tends to be easier, as it tells you and advance what method you should use. Remember the beauty of proving a theorem, is that you have an arsenal of methods to use and most of the time, your skill identifying what method to use is what will take you to a
One of the easiest methods to proof that conjecture is false is to use a counterexample. So, in this case, we will try to find a counterexample. We know that √2 is an irrational number. √2 x √2 = 2 The number 2 is a rational number. Therefore, the product of two irrational numbers is
Prove or disprove that the product of two irrational numbers is irrational Read More »
As in previous cases, we start by expressing the conjecture as a conditional statement. If m is irrational and n is rational, then m+n is irrational. In this case, we must use proof by contradiction. Remember that a contradiction is a compound proposition that is always false. To give a proof by contradiction of a
To use this method, we must prove that a theorem, stated as a conditional statement p-> q is true. In this case, we can restate the conjecture as follows: If m and n are even integers, then m+n is even. If we assume that p is true (m and n are evens), then we have
Use a direct proof to show that the sum of two even integers is even Read More »
