Answer :
To answer the question regarding the function [tex]\( f: N \rightarrow E \)[/tex] defined by [tex]\( f(x) = 2x \)[/tex] for all [tex]\( x \in N \)[/tex], we need to determine the properties of the function, specifically whether it is one-one (injective) and/or onto (surjective).
Let's break this down step-by-step:
### 1. Check if the function is One-One (Injective)
A function is one-one (injective) if for every pair of different elements in the domain, their images are different. This means [tex]\( f(a) = f(b) \implies a = b \)[/tex].
For the given function [tex]\( f(x) = 2x \)[/tex]:
- Suppose [tex]\( f(a) = f(b) \)[/tex]
- This implies [tex]\( 2a = 2b \)[/tex]
- Dividing both sides by 2, we get [tex]\( a = b \)[/tex]
Since [tex]\( f(a) = f(b) \)[/tex] implies [tex]\( a = b \)[/tex], the function [tex]\( f(x) = 2x \)[/tex] is injective.
### 2. Check if the function is Onto (Surjective)
A function is onto (surjective) if for every element in the codomain, there is a preimage in the domain. This means that every element in the set [tex]\( E \)[/tex] (the codomain) should have a corresponding element in the set [tex]\( N \)[/tex] (the domain).
For the given function [tex]\( f(x) = 2x \)[/tex]:
- The codomain [tex]\( E \)[/tex] is the set of all even natural numbers.
- An even natural number can be written as [tex]\( 2k \)[/tex] where [tex]\( k \)[/tex] is a natural number.
For any [tex]\( y \in E \)[/tex] (where [tex]\( y \)[/tex] is an even natural number):
- We can write [tex]\( y = 2k \)[/tex] where [tex]\( k \)[/tex] is a natural number.
- Therefore, [tex]\( y \)[/tex] can be obtained by [tex]\( f(k) \)[/tex], where [tex]\( k \in N \)[/tex].
Since every even natural number [tex]\( y \in E \)[/tex] can be expressed as [tex]\( 2k \)[/tex], and [tex]\( k \)[/tex] is an element of [tex]\( N \)[/tex], the function [tex]\( f(x) = 2x \)[/tex] is surjective.
### Conclusion
Since the function [tex]\( f: N \rightarrow E \)[/tex] defined by [tex]\( f(x) = 2x \)[/tex] is both injective and surjective, it is a one-one onto function.
Therefore, the correct answer is:
[tex]\[ \text{b) One - One onto} \][/tex]
Let's break this down step-by-step:
### 1. Check if the function is One-One (Injective)
A function is one-one (injective) if for every pair of different elements in the domain, their images are different. This means [tex]\( f(a) = f(b) \implies a = b \)[/tex].
For the given function [tex]\( f(x) = 2x \)[/tex]:
- Suppose [tex]\( f(a) = f(b) \)[/tex]
- This implies [tex]\( 2a = 2b \)[/tex]
- Dividing both sides by 2, we get [tex]\( a = b \)[/tex]
Since [tex]\( f(a) = f(b) \)[/tex] implies [tex]\( a = b \)[/tex], the function [tex]\( f(x) = 2x \)[/tex] is injective.
### 2. Check if the function is Onto (Surjective)
A function is onto (surjective) if for every element in the codomain, there is a preimage in the domain. This means that every element in the set [tex]\( E \)[/tex] (the codomain) should have a corresponding element in the set [tex]\( N \)[/tex] (the domain).
For the given function [tex]\( f(x) = 2x \)[/tex]:
- The codomain [tex]\( E \)[/tex] is the set of all even natural numbers.
- An even natural number can be written as [tex]\( 2k \)[/tex] where [tex]\( k \)[/tex] is a natural number.
For any [tex]\( y \in E \)[/tex] (where [tex]\( y \)[/tex] is an even natural number):
- We can write [tex]\( y = 2k \)[/tex] where [tex]\( k \)[/tex] is a natural number.
- Therefore, [tex]\( y \)[/tex] can be obtained by [tex]\( f(k) \)[/tex], where [tex]\( k \in N \)[/tex].
Since every even natural number [tex]\( y \in E \)[/tex] can be expressed as [tex]\( 2k \)[/tex], and [tex]\( k \)[/tex] is an element of [tex]\( N \)[/tex], the function [tex]\( f(x) = 2x \)[/tex] is surjective.
### Conclusion
Since the function [tex]\( f: N \rightarrow E \)[/tex] defined by [tex]\( f(x) = 2x \)[/tex] is both injective and surjective, it is a one-one onto function.
Therefore, the correct answer is:
[tex]\[ \text{b) One - One onto} \][/tex]
