Question

Let p: If x is an integer and \[{{x}^{2}}\]is even, then x is even. Using the method of contrapositive, prove that p is true.

Answer 1

Hint: Here, the given question is about to prove that p is true using the method of contrapositive, given statement is Let p: If x is an integer and \[{{x}^{2}}\]is even, then x is even. Based on the given statement, we need to prove that p is true, and also, we are going to discuss the types of if-then statements and also what is the contraceptive statement.

Complete step-by-step solution:
Given an if-then statement like ‘if p, then q.’ we can explain three statements.
A conditional statement consists of two elements, a hypothesis in the ‘if’ clause and a conclusion in the ‘then’ clause. For example, ‘if its rains, then they cancel going out.’
Hypothesis is ‘it rains’.
Conclusion is ‘they cancel going out.’
This is the conditional statement, we can form converse, inverse and contrapositive statements of the conditional statements.
To create a converse of the conditional statement, interchange the hypothesis and conclusion.
Example: The converse of ‘if it rains, then they cancel going out’ is ‘if they cancel going out, then it rains.’
To create the inverse of the conditional statement, taking the negation of both the hypothesis and the conclusion.
The inverse of ‘if it rains, then they cancel going out’ is ‘if it doesn’t rain, then they don’t cancel going out.’
To create the contrapositive of a conditional statement, interchanging the hypothesis and the conclusion of the inverse statement is essential.
The contrapositive of ‘if it rains, then they cancel going out’ is ‘if they don’t cancel going out, then it doesn’t rain.’
According to the given question:
p: If x is an integer and \[{{x}^{2}}\]is even, then x is even.
Let q: x is an integer and \[{{x}^{2}}\] is even
r: x is even
to prove that p is true by contrapositive method we assume that r is false and prove that q is also false.
Let x is not even
To prove that q is false, it has to be proved that x is not an integer or \[{{x}^{2}}\]is even
x is not even implying that \[{{x}^{2}}\]is not even
Therefore, statement q is false.
Thus, the given statement p is true.

Note: Before solving these kinds of questions, we need to learn about the conditional statement’s basics, it can help us to solve the contrapositive statements easily.
Statement: if a, then b.
Converse: if b, then a.
Inverse: if not a, then not b.
Contrapositive: if not b, then not a.