Short Answer TypeCheck the injectivity and surjectivity of the following functions:
f : N → N given by f(x) : x2
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, *)
(i) Let (x, y) be the identify element of (A, *).
∴ (a, b) * (x,y) = (a, b) ∀ a, b ∈ Q ⇒ (ax, ay + b) = (a, b)
⇒ ax = a, ay + b = b ⇒ ax = a, ay = 0 ∀ a ∈ Q ⇒ x = 1, y = 0
Now (a, b) * (1,0) = (a, b) ∀ a, b, ∈ Q Also (1,0)* (a, b) = (a, 1. b + 0) = (a, b)
∴ (1, 0) is the identity element of A.
(ii) Let (a, b) ∈ A be invertible ∴ there exists (c, d) ∈ A such that (a, b) * (c, d) = (1, 0) = (c, d) * (a, b)
∴ (ac, ad + b) = ( 1,0) ⇒ ac = 1, ad + b = 0
Now a ≠ 0
∵ if a = 0, then (0, b) is not invertible as (0, b) invertible implies that here exists (c, d)∈ A such that (0, b) * (c, d) = (1, 0) or (0, b) = (1, 0), which is senseless.
invertible elements of A arc (a, b), a 
and (a, b)-1 = 
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