Réponse:
Given f: N → N, defined by f(x) = x2 + x + 1 Let us prove that given function is one-one Injectivity: Let x and y be any two elements in domain (N), such that f(x) = f(y). ⇒ x2 + x + 1 = y2 + y + 1 ⇒ (x2 – y2) + (x – y) = 0 ` ⇒ (x + y) (x - y) + (x – y) = 0 ⇒ (x – y) (x + y + 1) = 0 ⇒ x – y = 0 [x + y + 1 cannot be zero because x and y are natural numbers ⇒ x = y So, f is one-one. Surjectivity: When x = 1 x2 + x + 1 = 1 + 1 + 1 = 3 ⇒ x + x +1 ≥ 3, for every x in N. ⇒ f(x) will not assume the values 1 and 2. Therefore, f is not ontoRead more on Sarthaks.com - https://www.sarthaks.com/603666/prove-that-the-function-f-n-n-defined-by-f-x-x-2-x-1-is-one-one-but-not-onto
