Let \( f : \mathbb{N} \to \mathbb{N} \) be a function such that \( f(m + n) = f(m) + f(n) \) for every \( m, n \in \mathbb{N} \). If \( f(6) = 18 \), then \( f(2) \cdot f(3) \) is equal to :
Show Hint
For any functional equation of the form \( f(x+y) = f(x) + f(y) \), the solution is always a linear function passing through the origin, i.e., \( f(x) = kx \).
This saves time during the exam instead of deriving \( f(1), f(2), \dots \) iteratively.
Step 1: Understanding the Concept:
The given functional equation \( f(m + n) = f(m) + f(n) \) for all \( m, n \in \mathbb{N} \) is a Cauchy functional equation.
For a function defined on natural numbers, this relation implies that \( f(n) \) must be of the form \( f(n) = cn \), where \( c \) is a constant. Step 2: Key Formula or Approach:
We use the given value \( f(6) = 18 \) to find the value of the constant \( c \).
Once \( c \) is known, we can find \( f(2) \) and \( f(3) \). Step 3: Detailed Explanation:
Given \( f(n) = cn \).
Substituting \( n = 6 \) in the expression:
\[ f(6) = c \cdot 6 \]
Since \( f(6) = 18 \), we have:
\[ 6c = 18 \implies c = 3 \]
Now, we calculate \( f(2) \) and \( f(3) \):
\[ f(2) = 3 \cdot 2 = 6 \]
\[ f(3) = 3 \cdot 3 = 9 \]
The product \( f(2) \cdot f(3) \) is:
\[ f(2) \cdot f(3) = 6 \cdot 9 = 54 \] Step 4: Final Answer:
The value of \( f(2) \cdot f(3) \) is 54.
