Classify the function as injection, surjection or bijection f: Z → Z, defined by f(x) = x2 + x

Given f: Z → Z, defined by f(x) = x² + x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x²+ x = y² + y

Here, we cannot say that x = y.

For example, x = 2 and y = – 3

Then,

x² + x = 2² + 2 = 6

y² + y = (−3)² – 3 = 6

So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (Z),

such that f(x) = y for some element x in Z (domain).

f(x) = y

x² + x = y

Here, we cannot say x ∈ Z.

For example, y = – 4.

x² + x = − 4

x² + x + 4 = 0

x =\( \dfrac{(-1 ± \sqrt{-5)}}{2}=\dfrac{-1+i\sqrt{5}}{2}\) which is not in Z.

So, f is not a surjection and f is not a bijection.