Show that the rational number (a + b)2 lies between the rational numbers a and b.
Step-by-step solution
Idea: Start from a < b and build the inequality step by step; adding the same number to both sides, or halving both sides, keeps the inequality the same way round.
- Let a and b be different rational numbers, and name them so that a < b. (a + b2 is rational, since rationals are closed under addition and division by 2.)½ mark
- Add a to both sides of a < b: a + a < a + b, i.e. 2a < a + b.1 mark
- Add b to both sides of a < b: a + b < 2b.½ mark
- So 2a < a + b < 2b. Divide by 2 (positive, so the signs stay): a < a + b2 < b. Hence a + b2 lies between a and b.1 mark
Check: a = −34, b = 12: a + b2 = −18, and −0.75 < −0.125 < 0.5 ✓.
Answer to write in the exam
Let a < b
a + a < a + b ⇒ 2a < a + b
a + b < b + b ⇒ a + b < 2b
2a < a + b < 2b
∴ a < a + b2 < b
Common mistakes that cost marks
- Showing only one inequality (e.g. a + b2 > a) and stopping.
- Using a numerical example as the proof.
- Not stating which of a, b is smaller before comparing.
How this can come in the exam
The rational number exactly halfway between 29 and 49 is
- 13
- 23
- 318
- 69
Show answer
(A) 13
(29 + 49) ÷ 2 = 69 ÷ 2 = 39 = 13.
Find the rational number halfway between −56 and 14 and verify it lies between them.
Show answer
(−1012 + 312) ÷ 2 = −724 (1 mark). −2024 < −724 < 624 ✓ (1 mark).Try one yourself
Show that a + 2b3 also lies between a and b when a < b.
Show answer
a + 2a < a + 2b < 3b (using a < b twice), so 3a < a + 2b < 3b and a < a + 2b3 < b.
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