When solving exercises related to functions, a common type of problem is to find out if what we are given is actually a function or not. As with almost everything in mathematics, we need to start with the definition. Definition: “Let A and B be nonempty sets. A function f from A to B is […]
Why is f not a function from R to R if 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 »
To determine whether an argument is valid or not, we should prove that it is impossible that all the premises are true and the conclusion is false. The way we do that proof is by applying one or more rules of inference on the premises and obtaining the conclusion. Like in other exercises, it is
Like in other exercises related to rules of inferences, you first should know the basic rules of inferences before you can apply them to draw conclusions from certain premises. If you still don’t know the rules of inference, make sure you open your textbook and have rules at hand so you can apply them. Notice
For each of these arguments, explain which rules of inference are used for each step 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 »
Direct proof is one the easiest method to construct proofs. To use this method, we must prove that a theorem, stated as a conditional statement p-> q is true. To this, we assume that p is true, and we apply rules of inference, axioms, definitions and previously proved theorems to prove that q is also
Use a direct proof to show that the sum of two odd integers is even Read More »
As with the previous examples, there are two main things we need to solve this exercise. First, we need to know the rules of inference so we can apply them. Second, we need to identify whether we must use propositional variables or predicates. The first step will always be to translate the argument using propositional
