Show that the function f:N→N, given by f(1)=f(2)=1 and f(x)=x−1 for every x>2, is onto but not one-one.Easy
Question
Show that the function f : N → N, given by
f(1) = f(2) = 1
and
f(x) = x - 1 for every x > 2,
is onto but not one-one.
Solution
Sure, let's break this down:
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A function f: N→N is said to be onto (or surjective) if for every element y in the co-domain N, there is at least one element x in the domain N such that f(x) = y.
For the given function f(x), we can see that for any y in N, we can always find an x in N such that f(x) = y. For y = 1, we have x = 1 or 2. For y > 1, we have x = y + 1. Therefore, the function is onto.
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A function f: N→N is said to be one-one (or injective) if for any two different elements x1 and x2 in the domain N, the images f(x1) and f(x2) are different.
For the given function f(x), we can see that f(1) = f(2) = 1. This means that there are two different elements in the domain N (1 and 2) that have the same image in the co-domain N (1). Therefore, the function is not one-one.
So, the function f: N→N, given by f(1)=f(2)=1 and f(x)=x−1 for every x>2, is onto but not one-one.
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