Short Answer TypeCheck the injectivity and surjectivity of the following functions:
f : N → N given by f(x) : x2
f : N → N given by
f(x) : x2
It is seen that for x, y 
, f(x) = f(y) 
x3 = y3 
x = y
f is injective
Now, 2 
, But there does not exist any element x in domain N such that f(x) = x3=2
f is not surjective
Hence, function f is injective but not surjective.
Check the injectivity and surjectivity of the following functions.
f : Z → Z given by f(x) = x2
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
f : R → R defined by f(x) = 3 – 4x
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
f : R → R defined by f(x) = 1 + x2
Consider the binary operation A on the set {1, 2, 3, 4, 5} defined by a ∧ b = min. {a, b}. Write the operation table of the operation ∧.
Let A = Q x Q. Let be a binary operation on A defined by (a, b) * (c, d) = (ac, ad + b).
Then
(i) Find the identify element of (A, *)
(ii) Find the invertible elements of (A, *)
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