Answer
See explanation
Work Step by Step
Below is a common proof (sometimes called a “backward induction” or “downward induction” proof) that **McCarthy’s 91 function** forces every positive integer \(n \le 101\) to map to \(91\). Recall its definition:
\[
M(n) \;=\;
\begin{cases}
n - 10, & \text{if } n > 100,\\[6pt]
M\bigl(M(n+11)\bigr), & \text{if } n \le 100.
\end{cases}
\]
We want to show:
\[
\forall\,1 \le n \le 101,\quad M(n) \;=\; 91.
\]
---
## **Proof Idea**
1. **If \(n>100\),** then \(M(n)=n-10\).
In particular, \(M(101)=101-10=91\).
2. **If \(n \le 100,\)** the definition says \(M(n)=M\bigl(M(n+11)\bigr)\).
- If \(n+11>100\), then \(M(n+11)=(n+11)-10=n+1\).
- If \(n+11 \le 100\), then we again apply the same rule to \(M(n+11)\).
Eventually (as we keep adding 11), we pass 100, and the function peels off 10. Tracing back shows the final value collapses to 91.
A succinct way to make this rigorous is by **downward induction** starting from \(n=101\) and working down to \(n=1\).
---
## **Detailed Proof by Downward Induction**
### **Step 1: Base Case** \((n=101)\)
Since \(101>100\), the definition gives
\[
M(101) \;=\; 101 - 10 \;=\; 91.
\]
So \(M(101)=91\) holds.
### **Step 2: Inductive Step**
We assume for some integer \(k\) with \(1 \le k \le 100\) that
\[
\forall m \text{ with } k < m \le 101,\quad M(m)=91.
\]
We need to prove \(M(k)=91\). Two subcases arise:
1. **Case A:** \(k+11 > 100\).
This means \(k \ge 90\). Then
\[
M(k) \;=\; M\bigl(M(k+11)\bigr)
\;=\; M\bigl((k+11)-10\bigr)
\;=\; M(k+1).
\]
But \(k+1 \le 101\). By the inductive hypothesis (since \(k+1>k\)), we have \(M(k+1)=91\). Hence
\[
M(k) = M(k+1) = 91.
\]
2. **Case B:** \(k+11 \le 100\).
Then
\[
M(k) \;=\; M\bigl(M(k+11)\bigr).
\]
But \(k+11 > k\). By the inductive hypothesis, \(M(k+11)=91\). Thus
\[
M(k) \;=\; M(91).
\]
We already know \(91 \le 101\) and by the same inductive reasoning (or by a direct chain evaluation), \(M(91)=91\). Therefore
\[
M(k) = 91.
\]
In both subcases, \(M(k)=91\). This completes the inductive step.
### **Conclusion**
By downward induction, for every \(n\) with \(1 \le n \le 101\), we obtain \(M(n)=91\). Equivalently,
\[
\boxed{M(n)=91 \quad\text{for all positive integers } n \le 101.}
\]
---
## **Intuitive “Chain” Explanation**
Sometimes one sees a more explicit “chain” computation:
- Start with \(n \le 100\). Then \(M(n)=M(M(n+11))\).
- Repeatedly apply “\(+11\)” until you exceed 100. The first time you exceed 100, say \(n+k\cdot 11>100\), the function value becomes \((n+k\cdot 11)-10\).
- That new number is still \(\le 101\), and further applications of \(M(\cdot)\) keep collapsing down until eventually everything maps to 91.
Either way, the key takeaway is that **for all \(n\le101\), McCarthy’s 91 function returns 91.** For \(n>101\), of course, \(M(n)=n-10\).
