Exercises 6.12 Exercises

Let \(n \) represent an integer with \(n \ge 2 \text{.}\) Prove that \(n \) is prime if and only if \(n/m \) is not an integer for every integer \(m \) with \(2 \le m \lt \frac{n}{2} \text{.}\)

Let \(n \) represent an integer with \(n \ge 2 \text{.}\) Suppose \(p_1,p_2,\dotsc,p_\ell \) is a complete list of prime numbers which are less than or equal to \(n/2 \text{.}\) Prove that \(n \) is prime if and only if none of the \(p_i \) divide \(n \text{.}\)

Careful. Is the statement actually true in the case \(n=2 \text{?}\) \(n=3 \text{?}\) (Why should these cases be given special consideration?)

Prove directly: The sum of two rational numbers is a rational number.

Prove directly: If \(n \) is even, then \(n^2 \) is divisible by \(4 \text{.}\)

Recall that the triangle inequality states that \(\abs{x+y} \le \abs{x} + \abs{y} \) for all numbers \(x \) and \(y \text{.}\)

Use the triangle inequality to prove directly: \(\abs{x+y+z} \le \abs{x} + \abs{y} + \abs{z} \) for all numbers \(x,y,z \text{.}\)

Prove by reduction to cases: \(n^3 - n \) is always divisible by \(3 \text{.}\)

Prove by proving the contrapositive: if \(2^n - 1 \) is prime, then \(n \) is prime.

Prove by counterexample that the following statement is false.

(See Exercise 6.12.4.)

Prove the biconditional: \(n \) is even if and only if \(n^2 \) is divisible by \(4 \text{.}\)

(See Exercise 6.12.5.)

Prove by contradiction: If \(m \) and \(n \) are integers such that \(11m + 19n \) is odd, then either \(m \) or \(n \) (or both) must be odd.

Prove by contradiction: For \(x,y>0 \text{,}\) \(\sqrt{x+y} \ne \sqrt{x} + \sqrt{y} \text{.}\)

Prove by contradiction: The sum of a rational number and an irrational number is irrational.

(See Exercise 6.12.4 and Exercise 6.12.9.)

Prove that if \(\ell \text{,}\) \(m \text{,}\) and \(n \) are integers such that \(\ell \) divides \(m \) and \(\ell \) divides \(n \text{,}\) then \(\ell \) divides \(mn \text{.}\)

Prove that if \(\ell \text{,}\) \(m \text{,}\) and \(n \) are integers such that \(mn \) divides \(\ell \text{,}\) then both \(m \) and \(n \) divide \(\ell \text{.}\)

Suppose that \(m \) and \(n \) are integers, and \(p \) is a prime number. Prove that if \(p \) does not divide the product \(mn \text{,}\) then \(p \) cannot divide either of \(m \) or \(n \text{.}\)