Q1.
Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression.
Answer
Handle each one the way the book handles 4m + 2q and x² + 2 — either pull out the factor 2, or produce a non-example.
| Expression | Explanation | Examples / non-examples |
|---|---|---|
| 2a + 2b | = 2(a + b), so 2 is a factor | a = 3, b = 5 → 16; a = 4, b = – 7 → – 6. Never odd. |
| 3g + 5h | Odd × odd stays odd; the sum of two odds is even but odd + even is odd | Example g = 1, h = 1 → 8 (even). Non-example g = 1, h = 0 → 3 (odd). |
| 2u – 4v | = 2(u – 2v) | u = 5, v = 1 → 6; u = – 3, v = 2 → – 14. Never odd. |
| 13k – 5k | = 8k = 2(4k) — the letters collect first | k = 1 → 8; k = – 3 → – 24. Never odd. |
| 6m – 3n | = 3(2m – n); 3 is odd, so the parity follows 2m – n, which follows n | Example m = 1, n = 2 → 0 (even). Non-example m = 1, n = 1 → 3 (odd). |
| b² + 1 | b² is even when b is even and odd when b is odd, so b² + 1 flips with b | Example b = 1 → 2 (even). Non-example b = 2 → 5 (odd). |
| 4k × 3j | = 12kj = 2(6kj) | k = 1, j = 1 → 12; k = 2, j = – 1 → – 24. Never odd. |
Why it happens: an expression is even for every integer value exactly when 2 can be taken out as a factor of the whole expression. If it cannot, some choice of the letters will make it odd — and one non-example is enough to settle the matter.