Use the unique factorization of integers theorem to prove the following statement. If n is any positive integer that is not a perfect square, then ✓n is irrational. The following is a proposed proof by contradiction of the statement with at least one incorrect step. 1. Suppose not. Suppose there exists an integer n such that n is not a perfect square and n is rational. 2. By definition of rational number, there exist integers a and b such that - and b0. 3. Without loss of generality, we may assume that a and b have no common divisors. Squaring both sides of the equation gives n = 4, and multiplying by b gives nb = a. 4. By the unique factorization of integers theorem, a, b, and n have representations as products of primes that are unique except for the order in which the prime factors are written down. 5. Since n is not a perfect square, there is a prime factor p in n that occurs an odd number of times. 6. Since nb = a, then p is a factor of a. 7. Thus a and b share the common divisor p. 8. This contradicts the fact that a and b are relatively prime. 9. Hence the supposition is false and the given statement is true. Identify the mistake(s) in the proof. (Select all that apply.) U Statement 7 does not follow from statements 3, 4, 5, and 6. M Statement 6 does not follow from statements 3, 4, and 5. The proof assumes a and b have no common factors, but the statement should apply for all integers a and b. The assumption in Statement 1 should be that n is a perfect square and n is rational. The equation in Statement 2 is not justified by the definition of rational number. The proof excludes the case where b = 0, but the statement should apply for every integer b.