Question:
There are two positive integers – $x$ and $y$. A function of $x$ and $y$ is defined such that:
$f(0,y) = y + 1$,
$f(x + 1, 0) = f(x, 1)$,
$f(x + 1, y + 1) = f(x, f(x + 1, y))$
What is the value of $f(1, 2)$?
There are two positive integers – $x$ and $y$. A function of $x$ and $y$ is defined such that:
$f(0,y) = y + 1$,
$f(x + 1, 0) = f(x, 1)$,
$f(x + 1, y + 1) = f(x, f(x + 1, y))$
What is the value of $f(1, 2)$?
$f(0,y) = y + 1$,
$f(x + 1, 0) = f(x, 1)$,
$f(x + 1, y + 1) = f(x, f(x + 1, y))$
What is the value of $f(1, 2)$?
Show Hint
When working with recursive functional equations, break them down step-by-step and substitute from the base case upward.
Updated On: Aug 5, 2025
- 2
- 4
- 3
- Cannot be determined
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The Correct Option is B
Solution and Explanation
We use the recursive definition step-by-step:
Step 1: Start with $f(1,2)$
From the 3rd rule:
\[
f(1,2) = f(0, f(1,1))
\]
Step 2: Evaluate $f(1,1)$
Again from the 3rd rule:
\[
f(1,1) = f(0, f(1,0))
\]
Step 3: Evaluate $f(1,0)$
From the 2nd rule:
\[
f(1,0) = f(0,1)
\]
Step 4: Evaluate $f(0,1)$
From the 1st rule:
\[
f(0,1) = 1 + 1 = 2
\]
Step 5: Substitute back
\[
f(1,0) = 2
\]
\[
f(1,1) = f(0, 2) = 2 + 1 = 3
\]
\[
f(1,2) = f(0, 3) = 3 + 1 = 4
\]
\[
\boxed{4}
\]
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