Question

Prove the following statement either by contradiction or by contraposition, clearly indicating which method you are using. For all m ∈ℤ, if m² is irrational,

Answer 1

Using the method of contraposition, the statement 'For all m ∈ℤ, if m² is irrational,' is proven by showing that if m² is rational, then m must also be rational, confirming the validity of the contrapositive and original statement.

Step-by-step explanation:

Proof by Contraposition

To prove the given statement, we shall use the method of contraposition. The original statement is 'For all m ∈ℤ, if m² is irrational.' The contrapositive of this statement is 'For all m ∈ℤ, if m² is not irrational (i.e., rational), then m is not irrational (i.e., rational).' By proving the contrapositive, we prove the original statement, as they are logically equivalent.

Let's assume that m² is rational. For any rational number, there exist two integers 'a' and 'b' (where b is not zero), such that m² = a/b and a and b have no common factors other than 1 (in other words, the fraction is in lowest terms). Now, we take the square root of both sides, which gives √m² = √(a/b). Since √m² = m, and the square root of a rational number in lowest terms is either another rational number or an irrational number, in this case, it must be rational, therefore, m is rational.

Since we have demonstrated that if m² is rational, then m must also be rational, by contraposition, if m is irrational, then m² must also be irrational. Thus, the original statement has been proven.