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Question
Classify the following function as injection, surjection or bijection :
f : Z → Z, defined by f(x) = x2 + x
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Solution
f : Z → Z, defined by f(x) = x2 + x
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 =` (-1 ±sqrt-5)/2 = (-1 ±isqrt5)/2` which is not in Z.
So, f is not a surjection and f is not a bijection.
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