Establish equivalence between the principle of Weak Induction and Strong Induction.

Question

I am looking for less jargons from predicate logic and more prominence on simple language. Also, please hint me as to what approach should I use in establishing the implication both way. I know that's the way to get around it.

Answer 1

The two directions of implication:

If something can be proven using weak induction, it can be proven using strong induction.

This is trivial, since the strong inductive hypothesis implies the weak inductive hypothesis.

If something can be proven using strong induction, it can be proven using weak induction.

Suppose you're trying to prove $P(n)$ for natural numbers $n$. Let $Q(n)$ be the statement "$P(k)$ holds for all natural numbers $0 \leq k \leq n$, i.e., $Q(n)$ means $P(0)$ and $P(1)$ and ... and $P(n)$. Clearly $Q(n)$ implies $P(n)$. Show that the strong inductive hypothesis on $P(n)$ is equivalent to the weak inductive hypothesis on $Q(n)$, so a weak inductive proof of $Q(n)$ is a sufficient proof of $P(n)$.

Answer 2

What more do you need to know?

The principal of induction is that if you can establish that a property being true for one case will imply it must also be true for the next immediately proceeding case. And if it is demonstrably true for a base case, then it must be true for all cases, via an indefinite and infinite running through of all cases one after another and realizing that because it is true for the current case, it must be true for the next proceeding case, and for all subsequent proceeding cases after that.

Weak induction vs. Strong induction is about how one shows that the next immediately proceeding case follows. Weak induction only requires that the current case, and none of the earlier preceeding cases are necessary to show the next proceeding case must be true. Strong induction may require not just the current case but some other previous cases, possibly all, are also true.

If weak induction is possible then strong induction is possible. Both prove all cases. The only difference is weak induction doesn't use all previous cases, just the current one. Strong induction is always available but may not always be necessary.

Frankly, I find the two to be so similar I wish they didn't teach the difference. People hear the terms and think there is some great significant difference between the two when there really isn't.

So for example:

Prove: that $n! \ge 2^{n-1}$ for all $n \ge 2$.

Base case: $n=2$ and $2! = 2*1 = 2 = 2^{2-1}$.

Weak induction: If $n = k\ge 2$ and $k! \ge 2^{k-1}$ then $(k+1)! = k!*k \ge 2^{k-1}*k \ge 2^{k-1}*2 = 2^{(k+1) -1}$.

That proof only relied upon $k! \ge 2^{k-1}$ it did not rely on $2! \ge 2^1$ or $3! > 2^2$ or $4! > 2^3$ or .... or $(k-1)! \ge 2^{k-2}$, all of which are also true but not used.

A strong induction could use those.

Strong induction: Suppose for all $k \le $ then we can assume we know that $k! \ge 2^{k-1}$ Then if $k = 2m + i$ where $i = 0, 1$ then $2^k = (2^m)^2*2^i \le (m!)^2*2^i = (1......m)(1......m)*2^i < (1......m)((m+1)(m+2)......(m+m)*2^i$ if $i=0$ that is equal to $(2m)! = k!$ and if $i=1$ that is equal to $(2m)!*2 < (2m)!(2m+1) = k!$. So $2^k \le k!$.

Okay, that was inefficient and wierd and unnecessarily complicated, but the point is it didn't just use $2^k \le k!$. It used that $m < k$ and $2^m \le m!$.