Show that f: N → N, f(x) = x² + x + 1 is one-one. Is the same rule one-one on R?
How to lay out the answer. Follow the frame, then write it in your own words.
1 · Write what is given and what is to be found. Given: f(x) = x² + x + 1 on N. To show: f is one-one on N. To check: whether the rule is one-one on R.
2 · Write the rule. f is one-one if f(x₁) = f(x₂) ⇒ x₁ = x₂.
3 · Set up the equation. Let f(x₁) = f(x₂): x₁² + x₁ + 1 = x₂² + x₂ + 1.
4 · Factorise.
(x₁² − x₂²) + (x₁ − x₂) = 0
(x₁ − x₂)(x₁ + x₂) + (x₁ − x₂) = 0
(x₁ − x₂)(x₁ + x₂ + 1) = 0
5 · Use the fact that x₁, x₂ are natural numbers. x₁ + x₂ + 1 ≥ 3, so it is not zero. Hence x₁ − x₂ = 0, that is x₁ = x₂. So f is one-one on N.
6 · Test the rule on R. f(0) = 1 and f(−1) = 1 − 1 + 1 = 1, but 0 ≠ −1. So the rule is not one-one on R.
7 · Check and box the answer. f is one-one on N, but the same rule is not one-one on R.
Scoring tip: Always say why the second factor cannot be zero. That one line carries a mark.
