Question -
Answer -
Givenf: N → N, defined by f(x) = x2 + x +1
Now we have to provethat given function is one-one
Injectivity:
Let x and y be any two elements in the domain (N), suchthat 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 isone-one.
Surjectivity:
When x = 1
x2 + x+ 1 = 1 + 1 + 1 = 3
⇒ x2 +x +1 ≥ 3, for every x in N.
⇒ f(x) will notassume the values 1 and 2.
So, f is not onto.