Suppose f :N→N where f(n) ＝ 4n²＋1 is f one-to-one?(explain)

To determine if f is one-to-one, we need to show that for any distinct inputs n₁ and n₂ in the domain N, their corresponding outputs f(n₁) and f(n₂) are also distinct.

Let's assume that there exist two distinct inputs n₁ and n₂ in N such that f(n₁) = f(n₂). Then we have:

4n₁² + 1 = 4n₂² + 1

4n₁² = 4n₂²

n₁² = n₂²

Taking the square root of both sides, we get:

n₁ = ±n₂

Since n₁ and n₂ are distinct, we must have n₁ ≠ n₂. Therefore, the only possibility is that n₁ = -n₂. However, this contradicts the fact that n₁ and n₂ are both in the domain N, which is the set of positive integers. Thus, it is impossible for f(n₁) to be equal to f(n₂) if n₁ and n₂ are distinct, and we can conclude that f is one-to-one.

Therefore, f(n) = 4n² + 1 is one-to-one on the domain N.