Answer
1. \(A(1,1) = 3.\)
2. \(A(2,1) = 5.\)
Work Step by Step
Below is a step‐by‐step derivation of the two requested Ackermann function values, using the classical three‐part definition (for all nonnegative integers \(m,n\)):
\[
A(0, n) \;=\; n+1,
\]
\[
A(m, 0) \;=\; A(m-1, 1) \quad\text{(for \(m>0\))},
\]
\[
A(m, n) \;=\; A\bigl(m-1,\; A(m,\,n-1)\bigr) \quad\text{(for \(m>0,\,n>0\))}.
\]
---
## (a) \(A(1,1)\)
We use the rule for \(m>0\) and \(n>0\):
\[
A(m,n) \;=\; A\bigl(m-1,\; A(m,\,n-1)\bigr).
\]
Hence,
\[
A(1,1)
\;=\;
A\bigl(1-1,\; A(1,\,1-1)\bigr)
\;=\;
A\bigl(0,\; A(1,0)\bigr).
\]
But when \(m>0\) and \(n=0\), we have:
\[
A(m,0) \;=\; A(m-1,1).
\]
So
\[
A(1,0) \;=\; A(0,1).
\]
By the first rule (\(A(0,n)=n+1\)),
\[
A(0,1) \;=\; 1 + 1 \;=\; 2.
\]
Thus \(A(1,0)=2\). Substitute back:
\[
A(1,1) \;=\; A(0,2) \;=\; 2 + 1 \;=\; 3.
\]
**Answer for (a):** \(\boxed{3}.\)
---
## (b) \(A(2,1)\)
Again, for \(m>0\) and \(n>0\):
\[
A(m,n) \;=\; A\bigl(m-1,\; A(m,\,n-1)\bigr).
\]
So
\[
A(2,1)
\;=\;
A\bigl(2-1,\; A(2,\,1-1)\bigr)
\;=\;
A\bigl(1,\; A(2,0)\bigr).
\]
Since \(m>0\) and \(n=0\) implies \(A(m,0)=A(m-1,1)\),
\[
A(2,0) \;=\; A(1,1).
\]
From part (a), we have \(A(1,1)=3\). Hence
\[
A(2,1) \;=\; A(1,3).
\]
Now evaluate \(A(1,3)\). For \(m=1\) and \(n>0\):
\[
A(1,3)
\;=\;
A\bigl(0,\; A(1,2)\bigr).
\]
We need \(A(1,2)\):
\[
A(1,2)
\;=\;
A\bigl(0,\; A(1,1)\bigr).
\]
But we already know \(A(1,1)=3\), so
\[
A(1,2)
\;=\;
A(0,3)
\;=\;
3 + 1
\;=\;
4.
\]
Hence
\[
A(1,3)
\;=\;
A(0,4)
\;=\;
4 + 1
\;=\;
5.
\]
Therefore,
\[
A(2,1)
\;=\;
A(1,3)
\;=\;
5.
\]
**Answer for (b):** \(\boxed{5}.\)
