Let f, g : N → N such that f(n + 1) = f(n) + f(1) ∀ n∈N and g be any arbitrary function. Which of the following statements is NOT true?

Solution

Maths sir · 2 steps

First pin down f: the condition forces f(n) = n f(1), a straight line through the origin.

1. Find f. Let f(1) = k, where k ∈ ℕ. Then f(2) = 2k, f(3) = 3k, and so on, so f(n) = kn for every n.

   f(n) = kn, k = f(1) ≥ 1 ⇒ (f° g)(n) = k g(n)

   Statement	Check with f(n) = kn	Verdict
   (a) g onto ⇒ f° g one-one	Take g(1) = 1 and g(n) = n - 1 for n ≥ 2. It is onto, but g(1) = g(2) = 1, so k g(1) = k g(2).	not true
   (b) f is one-one	kn₁ = kn₂ gives n₁ = n₂ since k ≠ 0.	true
   (c) f onto ⇒ f(n) = n	The range of kn is the multiples of k. It contains 1 only if k = 1.	true
   (d) f° g one-one ⇒ g one-one	If g(m₁) = g(m₂), then f(g(m₁)) = f(g(m₂)), so m₁ = m₂.	true

2. So. Onto does not mean one-one. A g that is onto can still repeat a value, and then f° g repeats it too. The false statement is (a).

The idea to keep. If f° g is one-one, then the inner function g must be one-one. Knowing that g is onto tells you nothing about f° g being one-one, so test such claims with an onto map that repeats a value.

Answer from the NTA final answer key. Solution by PrepWiser.