Answer
See explanation
Work Step by Step
Below is a standard proof (by induction on \(n\)) of each of the three identities for the classical Ackermann function
\[
A(m,n)=
\begin{cases}
n+1 & \text{if }m=0,\\[6pt]
A(m-1,1) & \text{if }m>0\text{ and }n=0,\\[6pt]
A\bigl(m-1,A(m,n-1)\bigr) & \text{if }m>0\text{ and }n>0.
\end{cases}
\]
We show, for all nonnegative \(n\):
1. \(A(1,n) = n+2.\)
2. \(A(2,n) = 2n+3.\)
3. \(A(3,n) = 2^{\,n+3} - 3.\)
---
## 1. **\(A(1,n) = n + 2\)**
We prove by induction on \(n\).
- **Base Case** \((n=0)\):
By definition for \(m>0\) and \(n=0\),
\[
A(1,0) \;=\; A(0,1).
\]
Since \(A(0,k)=k+1\), we get \(A(0,1)=1+1=2\).
On the right‐hand side, \(n+2 = 0+2=2\).
Thus \(A(1,0)=2\) matches \(0+2\).
- **Inductive Step**: Assume \(A(1,n)=n+2\) holds for some \(n\). We show it for \(n+1\).
By definition (\(m=1>0\), \(n+1>0\)):
\[
A(1,\,n+1) \;=\; A\bigl(0,\;A(1,n)\bigr).
\]
By the inductive hypothesis, \(A(1,n)=n+2\). Hence
\[
A(1,\,n+1)
= A\bigl(0,\;n+2\bigr)
= (n+2) + 1
= n+3
= (n+1) + 2.
\]
This completes the induction.
Thus for all \(n\ge0\),
\[
\boxed{A(1,n)=n+2.}
\]
---
## 2. **\(A(2,n) = 2n + 3\)**
Again we do induction on \(n\).
- **Base Case** \((n=0)\):
By definition (\(m=2>0\), \(n=0\)):
\[
A(2,0) \;=\; A(1,1).
\]
From part 1 (just proven), \(A(1,1)=1+2=3\).
On the right side, \(2n+3 = 2\cdot0+3=3\).
So \(A(2,0)=3\) matches \(2\cdot0+3\).
- **Inductive Step**: Assume \(A(2,n)=2n+3\). We prove \(A(2,n+1)=2(n+1)+3\).
By definition (\(m=2>0\), \(n+1>0\)):
\[
A(2,n+1) \;=\; A\bigl(1,\;A(2,n)\bigr).
\]
By the inductive hypothesis, \(A(2,n)=2n+3\). So
\[
A(2,n+1)
= A\bigl(1,\;2n+3\bigr).
\]
From part 1, \(A(1,k)=k+2\). Substituting \(k=2n+3\) gives
\[
A(1,\,2n+3) = (2n+3)+2 = 2n+5 = 2(n+1)+3.
\]
Hence \(A(2,n+1)=2(n+1)+3\), completing the induction.
Therefore for all \(n\ge0\),
\[
\boxed{A(2,n)=2n+3.}
\]
---
## 3. **\(A(3,n) = 2^{n+3} - 3\)**
Once more, prove by induction on \(n\).
- **Base Case** \((n=0)\):
By definition (\(m=3>0\), \(n=0\)):
\[
A(3,0) \;=\; A(2,1).
\]
From part 2, \(A(2,1)=2\cdot1+3=5\).
On the right side, \(2^{\,0+3}-3=2^3-3=8-3=5\).
So \(A(3,0)=5\) matches \(2^{3}-3\).
- **Inductive Step**: Assume \(A(3,n)=2^{n+3}-3\). We show \(A(3,n+1)=2^{(n+1)+3}-3\).
By definition (\(m=3>0\), \(n+1>0\)):
\[
A(3,n+1) \;=\; A\bigl(2,\;A(3,n)\bigr).
\]
By the inductive hypothesis, \(A(3,n)=2^{n+3}-3\). Thus
\[
A(3,n+1)
= A\bigl(2,\;2^{n+3}-3\bigr).
\]
From part 2, \(A(2,k)=2k+3\). Substituting \(k=2^{n+3}-3\) gives
\[
A\bigl(2,\;2^{n+3}-3\bigr)
= 2\bigl(2^{n+3}-3\bigr)+3
= 2^{n+4} -6 +3
= 2^{n+4} -3
= 2^{(n+1)+3}-3.
\]
This completes the induction step.
Hence for all \(n\ge0\),
\[
\boxed{A(3,n)=2^{\,n+3}-3.}
\]
---
## **Summary**
Using induction on \(n\) and the basic definition of Ackermann’s function, we obtain:
1. \(A(1,n) = n+2\).
2. \(A(2,n) = 2n+3\).
3. \(A(3,n) = 2^{n+3} \;-\; 3.\)
