Question

check the injective and surjective


1. f:N - N given by f (x) = x^3​

Answer 1

Answer:

yes injective but not surjective

Step-by-step explanation:

injective is one to one function and surjective in onto function here the range is not equal to co-domain of function

Answer 2

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To check the injectivity and surjectivity of the function f: N -> N defined as f(x) = x^3, we need to examine the properties individually.

1. Injectivity (One-to-One):

To prove that f is injective, we need to show that if f(x1) = f(x2), then x1 = x2 for any x1, x2 ∈ N.

Let's assume f(x1) = f(x2):

x1^3 = x2^3

Taking the cube root of both sides, we get:

x1 = x2

Since x1 = x2, we can conclude that the function f is injective.

2. Surjectivity (Onto):

To prove that f is surjective, we need to show that for every y ∈ N, there exists an x ∈ N such that f(x) = y.

Let's consider an arbitrary y ∈ N. To find an x such that f(x) = y, we need to solve the equation x^3 = y.

Since we are dealing with natural numbers, we can take the cube root of y to find x:

x = ∛y

Note that ∛y may or may not be a natural number for all y ∈ N. For example, if y is not a perfect cube, there will be no natural number x such that x^3 = y.

Hence, we cannot guarantee the existence of an x for every y in N, and therefore the function f is not surjective.

In summary, the function f: N -> N defined as f(x) = x^3 is injective (one-to-one) but not surjective (onto).

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