Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.6 - Page 206: 31

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### Part (a): Proof by Contraposition **Statement:** For all positive integers \(n\), \(r\), and \(s\), if \[ r \cdot s \le n, \] then \[ r \le \sqrt{n} \quad \text{or} \quad s \le \sqrt{n}. \] **Contrapositive Statement:** The contrapositive is logically equivalent to the original statement. Its negation is: \[ \text{If } r > \sqrt{n} \text{ and } s > \sqrt{n}, \text{ then } r \cdot s > n. \] **Proof of the Contrapositive:** Assume that \(r > \sqrt{n}\) and \(s > \sqrt{n}\). Then \[ r \cdot s > \sqrt{n} \cdot \sqrt{n} = n. \] Thus, if both \(r\) and \(s\) exceed \(\sqrt{n}\), their product exceeds \(n\). This completes the proof of the contrapositive, and therefore the original statement is true. --- ### Part (b): Direct Proof **Statement:** For all integers \(n > 1\), if \(n\) is not prime (i.e. \(n\) is composite), then there exists a prime number \(p\) such that \[ p \le \sqrt{n} \quad \text{and} \quad p \text{ divides } n. \] **Proof:** Suppose \(n > 1\) is not prime. Then \(n\) is composite, so there exist integers \(r\) and \(s\) with \[ 1 < r \le s < n \] such that \[ n = r \cdot s. \] By part (a) (applied to \(r\) and \(s\) with \(r \cdot s = n\)), we know that at least one of the factors must be no larger than \(\sqrt{n}\). Without loss of generality, assume \[ r \le \sqrt{n}. \] If \(r\) is prime, then we have found our prime \(p = r\) satisfying \(p \le \sqrt{n}\) and \(p \mid n\). If \(r\) is not prime, then \(r\) itself is composite. Hence, \(r\) has a prime factor \(p\) with \[ p \le r \le \sqrt{n}. \] Since \(p\) divides \(r\) and \(r\) divides \(n\), it follows that \(p\) divides \(n\). Thus, in either case, there exists a prime \(p \le \sqrt{n}\) such that \(p\) divides \(n\). --- ### Part (c): Contrapositive of Part (b) **Original Statement of Part (b):** For all integers \(n > 1\), if \(n\) is not prime, then there exists a prime number \(p\) such that \[ p \le \sqrt{n} \quad \text{and} \quad p \mid n. \] **Contrapositive Statement:** The contrapositive is: For all integers \(n > 1\), if for every prime number \(p\) with \(p \le \sqrt{n}\) it holds that \(p\) does not divide \(n\), then \(n\) is prime. In words, if no prime number less than or equal to \(\sqrt{n}\) divides \(n\), then \(n\) must be prime.