Unit 6 Review Exercises

Mathematical Proofs

All proof questions are Paper 1 style. Questions marked AHL are for HL students.

1 State whether each statement is true or false, with a reason.

(a) \(\forall\, x \in \mathbb{R},\; x^2 > 0\)

(b) \(\exists\, n \in \mathbb{Z}\) such that \(n^2 = 2n\)

Solution

(a) False: \(x = 0\) gives \(0^2 = 0\), not \(> 0\).

(b) True: \(n = 0\) or \(n = 2\). See 6.1.

2 Prove that \((x + 1)^3 - x^3 \equiv 3x^2 + 3x + 1\).

Solution

LHS \(= x^3 + 3x^2 + 3x + 1 - x^3 = 3x^2 + 3x + 1 =\) RHS. See 6.2.

3 Prove that the product of two consecutive even numbers is a multiple of 8.

Solution

\(2n(2n + 2) = 4n(n + 1)\). One of \(n\), \(n + 1\) is even, so \(n(n + 1) = 2m\) and the product is \(8m\). See 6.2.

4 AHL Show that “\(n^2 - n + 11\) is prime for every positive integer \(n\)” is false.

Solution

\(n = 11\): \(121 - 11 + 11 = 121 = 11^2\), not prime. See 6.3.

5 AHL Given that, for \(n \in \mathbb{Z}\), if \(n^2\) is a multiple of 5 then \(n\) is a multiple of 5, prove by contradiction that \(\sqrt{5}\) is irrational.

Solution

Assume \(\sqrt{5} = \dfrac{p}{q}\) in lowest terms. Then \(p^2 = 5q^2\), so \(p = 5m\). Then \(25m^2 = 5q^2\), \(q^2 = 5m^2\), so \(q\) is a multiple of 5. Then \(p\) and \(q\) share the factor 5 — a contradiction. See 6.3.

6 AHL Prove by contradiction that there are no integers \(x\) and \(y\) with \(x^2 - 4y = 2\).

Solution

Assume they exist. Then \(x^2 = 4y + 2\) is even, so \(x\) is even, \(x = 2k\), and \(x^2 = 4k^2\) is a multiple of 4. But \(4y + 2\) leaves remainder 2 when divided by 4. Contradiction. See 6.3.

7 AHL Prove by induction that \(\displaystyle\sum_{r=1}^{n} r(r + 2) = \frac{n(n + 1)(2n + 7)}{6}\) for all \(n \in \mathbb{Z}^+\).

Solution

Basis: \(1 \times 3 = 3\) and \(\frac{1 \times 2 \times 9}{6} = 3\). Step:

\[ \frac{k(k + 1)(2k + 7)}{6} + (k + 1)(k + 3) = \frac{(k + 1)\left(2k^2 + 13k + 18\right)}{6} = \frac{(k + 1)(k + 2)(2k + 9)}{6}, \]

which is the formula for \(n = k + 1\). Conclusion as usual. See 6.4.

8 AHL Prove by induction that \(3^{2n} - 1\) is divisible by 8 for all \(n \in \mathbb{Z}^+\).

Solution

Basis: \(9 - 1 = 8\). Step: assume \(3^{2k} = 8M + 1\). Then \(3^{2k+2} - 1 = 9(8M + 1) - 1 = 8(9M + 1)\). Conclusion. See 6.4.

9 AHL A sequence is defined by \(u_1 = 1\) and \(u_{n+1} = u_n + 2n + 1\). Prove by induction that \(u_n = n^2\).

Solution

Basis: \(u_1 = 1 = 1^2\). Step: \(u_{k+1} = k^2 + 2k + 1 = (k + 1)^2\). Conclusion. See 6.4.

10 AHL Prove by induction that \(n! > 3^n\) for all integers \(n \ge 7\).

Solution

Basis: \(7! = 5040 > 2187 = 3^7\). Step: assume \(k! > 3^k\) for some \(k \ge 7\). Then

\[ (k + 1)! = (k + 1)\,k! > (k + 1)\,3^k > 3 \times 3^k = 3^{k+1}, \]

since \(k + 1 > 3\). Conclusion (for \(n \ge 7\)). See 6.4.