Question

Let function \( f:N \to Y \) is defined as \( f(x)=4x+3 \) where \( Y=\{y \in N : y=4x+3 \text{ for } x \in N\} \). Prove that f is invertible, also find the inverse of the function f.

Hint:

When a function's codomain is explicitly defined to be its range, as in this problem, the 'onto' property is satisfied by definition. The main task then becomes proving the 'one-one' property and finding the inverse.

Answer 1

Step 1: Understanding the Concept:
A function is invertible if and only if it is bijective, which means it must be both one-one (injective) and onto (surjective). We need to prove these two properties for the given function f and then find its inverse.
Step 2: Proving One-One (Injectivity):
A function is one-one if different inputs from the domain produce different outputs in the codomain.
Let \( x_1, x_2 \in N \) such that \( f(x_1) = f(x_2) \).
\[ 4x_1 + 3 = 4x_2 + 3 \] Subtracting 3 from both sides: \[ 4x_1 = 4x_2 \] Dividing by 4: \[ x_1 = x_2 \] Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function f is one-one.
Step 3: Proving Onto (Surjectivity):
A function is onto if its range is equal to its codomain.
The domain is \( N \) (the set of natural numbers).
The codomain is given as \( Y = \{y \in N : y=4x+3 \text{ for } x \in N\} \).
The range of f is the set of all possible output values, which is \( \{f(x) | x \in N\} = \{4x+3 | x \in N\} \).
By the given definition of the codomain Y, the range of f is exactly equal to the codomain Y.
Therefore, the function f is onto.
Step 4: Finding the Inverse:
Since f is both one-one and onto, it is a bijective function and hence is invertible.
Let \( g: Y \to N \) be the inverse of f. To find the rule for g, we let \( y = f(x) \).
\[ y = 4x + 3 \] We solve this equation for x in terms of y: \[ y - 3 = 4x \] \[ x = \frac{y - 3}{4} \] So, the inverse function, which we can call \( f^{-1} \), is given by: \[ f^{-1}(y) = \frac{y-3}{4} \] Step 5: Final Answer:
The function f is one-one and onto, hence it is invertible. The inverse function is \( f^{-1}: Y \to N \) defined by \( f^{-1}(y) = \frac{y-3}{4} \).