Suppose A and B are nonempty sets and f : A → B is a function.
If a function is one-to-one and onto, then it is invertible.
Here , A={a,b,c} and B={1,2,3}.
Given that -> f(a) = 2 , f(b) = 3 , f(c) = 1
Now here, f maps every element of A to a unique element of B, so f is one-to-one.
Also every element in B has a pre-image in A , so f is onto.
Thus, by definition f is invertible.
A function f-1 : B → A is called an inverse function for f if it satisfies the following condition:
For every x ∈ A and y ∈ B, f(x) = y if and only if f-1(y) = x.
So, here "f(a)=2\\implies f"-1"(2)=a\n\\\\"
"f(b)=3 \\implies f"-1(3) = b
"f(c)=1 \\implies f"-1(1)"\\;=c"
Comments
Dear Olagidi Joshua, thank you for leaving comments.
It is invertible since the function is injective(one to one) and bijective(onto), then there is a inverse
It's invertible
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