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Question
Classify the following function as injection, surjection or bijection :
f : Q → Q, defined by f(x) = x3 + 1
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Solution
f : Q → Q, defined by f(x) = x3 + 1
Injection test :
Let x and y be any two elements in the domain (Q), such that f(x) = f(y).
f(x) = f(y)
x3+1 = y3+ 1
x3 = y3
x = y
So, f is an injection .
Surjection test:
Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q(domain).
f(x) = y
x3+ 1 = y
`x = 3sqrt(y-1) ,` which may not be in Q.
For example, if y= 8,
x3+ 1 = 8
x3= 7
`x = 3sqrt7,`which is not in Q.
So, f is not a surjection and f is not a bijection.
So, f is a surjection and f is a bijection.
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