0

In a usual proof by induction the inductive step involves assuming $P(n)$ is true and then showing that $P(n+1)$ follows.

Are there any other valid inductive steps? That is, can we define an equivalent inductive step $P(q) \implies P(k)$, where $q \neq n$ and $k \neq n+1$ such that proving the implication in the inductive step proves that the given proposition holds true for all $n$

EDIT: This question comes from a problem that I had in my Discrete math class:

Assume we have the statement $P(n)$ which is defined over the set of natural numbers. Is this sufficient to conclude that $P(n)$ is true for all $n \in N$ : \begin{align*} &P(2) \ \text{is true} \\ &\forall n \in N. P(n) \implies P(2n) \\ &\forall n \in N. P(n) \ \wedge n>1 \implies P(n-1) \end{align*}

Karan Mehta
  • 79
  • Not sure what you mean. It's standard, for example, to use "strong" induction, that is $P(≤n)\implies P(n+1)$. But one could imagine something using, say, $P(n-1)≤P(n+1)$, though your base case would need to cover both even and odd numbers. – lulu Mar 02 '23 at 16:15
  • 1
    Other such would almost certainly be able to be reworded as traditional induction. Strong induction for instance might have been worded $P(1)\wedge P(2)\wedge \cdots \wedge P(k)\implies P(k+1)$ but it could have been reworded as $Q(k)\implies Q(k+1)$ where $Q(k)\equiv P(1)\wedge P(2)\wedge \cdots \wedge P(k)$ – JMoravitz Mar 02 '23 at 16:16
  • Typo: in my first comment I meant to write $P(n-1)\implies P(n+1)$. Thus, your induction separates even $n$ from odd. – lulu Mar 02 '23 at 16:26
  • Maybe look up transfinite induction – Alex Mar 02 '23 at 16:33
  • @Shinrin-Yoku What do you mean by "finite in the theory but not the meta theory"? And why would it contradict JMoravitz's observation – Alex Mar 02 '23 at 16:45
  • That Edit is useful! So note: now you don;t have a single conditional that potentially 'replaces' the normal step, but several conditionals. And yes, if you have several conditionals, then there are many ways to make it work. – Bram28 Mar 02 '23 at 16:50
  • With these two properties it should not be too difficult that you indeed have $P(k) \Rightarrow P(k+1)$. – Qise Mar 02 '23 at 17:16
  • One semi-common case shows $P(0)$ and $P(n)\implies P(2n)\land P(2n+1).$ – Thomas Andrews Mar 02 '23 at 18:52
  • See here for the induction method used in your problem. – Bill Dubuque Mar 14 '23 at 10:08

2 Answers2

2

If you can prove multiple conditionals, then yes, there are many (infinitely, really) ways to 'cover' all numbers.

For example, suppose I can show all of:

$P(1)$

$P(2)$

$P(n) \to P(n+2)$

Well, then $P(n)$ should be true for all $n$: Since we have $P(1)$, and since we have $P(n) \to P(n+2)$, we now also have $P(3)$. Applying $P(n) \to P(n+2)$ again, that gives us $P(5)$ .. and then $P(7), P(9)$, ... etc. So, every odd number has then property $P$

But $P(n)$ will also be true for any even number: we have $P(2)$, and therefore $P(4)$, and thus $P(6), P(8)$, ... etc. So, all even numbers have property $P$ as well. Butif all even numbers and all odd numbers have property $P$, then all numbers have property $P$.

So, collectively, these three claims can be used to prove that all numbers have property $P$. And to prove those 3 claims, you are doing some kind of induction: $P(1)$ and $P(2)$ are now two base cases, and $P(n) \to P(n+2)$ is a kind of inductive step to 'get to' all other numbers.

You were probably given the domino metaphor for induction. 'Standard' induction will place all 'dominoes' in one long line:

$1 \to 2 \to 3 \to 4 \to 5 ...$

However, there are many other ways to place all the dominoes so that, when you knock over a initial bunch, they eventually all get knocked over. What I just described above corresponds to having two lines:

$1 \to 3 \to 5 \to ....$

$2 \to 4 \to 6 \to ....$

So, if I knock over $1$ and $2$, will they all get knocked over? Yes.

For your specific case, ask yourself this: can I 'get to' any number $n$, given that I start at number $2$, and from there, either can go to a number twice as large ('double'), or (assuming I am at any number greater than $1$) get to a previous number ('previous')?

For example, can you get to 13? Sure! From $2$ we can 'double' to go to $4$, and then 'double' again to get to $8$. From $8$ we can do 'previous' to get to $7$, then 'double' to get to $14$, and finally 'previous' to get to $13$.

Note that this is not the only way to get to $13$ .. I could do $2 \to 4 \to 8 \to 16 \to 15 \to 14 \to 13$. But being able to get to some number in multiple ways does not take away from the fact that I can get to that number, and that's all you are about.

So again, the question is: can we 'get to' every number by applying the 'double' or 'previous' operation some number of times? And if you think the answer is yes, how would you mathematically prove that?

Bram28
  • 100,612
  • 6
  • 70
  • 118
  • Note that for this example you could also set an other property: $P'(n) = P(n)$ and $P(n+1)$. Then it is trivial to prove $P'(k) \rightarrow P'(k+1)$, so $P'(n)$ is true $\forall n$. Then you just recall that $P(n)$ is true as well, being implied by $P'(n)$. – Qise Mar 02 '23 at 17:14
0

Ok lets try to generalize a bit. We could replace the map $n\mapsto n+1$ with some map $f:\mathbb N \to \mathbb N$. Then we could pose conditions on $f$ so that $$ 0\in N\text{ and } (n\in N \implies f(n)\in N) $$ implies that $N=\mathbb N$ for every subset $N\subset \mathbb N$. The condition on $f$ would be $$ \{f(0),f(f(0)),\dots,f^k(0),\dots\}=\mathbb N. $$

Lets go a step further. Let $S:\mathscr P(\mathbb N)\to\mathscr P(\mathbb N)$. We shall call $S$ inductive if $$ 0\in N\text{ and } (X\subset N\implies S(X)\subset N) $$ implies that $N=\mathbb N$ for every $N\subset \mathbb N$. Then we could have some kind of generalized induction. Show that $0$ staisfies your proposition $P(0)$. Then the induction step would be: If $P$ is satisfied on $X\subset \mathbb N$ then it is also satisfied on $P(S(X))$. This would then imply that the proposition is true on $\mathbb N$, provided that $S$ is inductive.

"Strong induction" would then be given with $S(\{0,\dots,n\})=\{n+1\}$ (and $S=id$ on all other sets)

Now one could try to describe which $S$ are inductive.

Is that what you are looking for?

Alex
  • 934