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I have been reading some notes on Induction and Strong Induction and fully understand how they work. However I was interested in a formal/mathematical way of expressing their definition and was wondering if it is correct!

For regular induction I have in my notes:

let $p(n)$ be a proposition such that: $p(1)$ holds and for all $n \in N$, $p(n)$ $\implies p(n+1)$. Then $p(n+1)$ holdes for all $n \in N$.

Is this correct? I understand induction but I feel like this definition is wrong :/

Also if it is correct, how can I build on it to make a definition for Strong Induction? Thank you!!

Sniper Clown
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nicefella
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  • To explain why this is the definition: if $p(1)$ is true, then $p(2)$ is true (since $p(n) \Rightarrow p(n+1)$). If $p(2)$ is true, then $p(3)$ is true. If ... – George V. Williams Apr 12 '13 at 01:17
  • in fact it's wrong you feel like this is wrong then you say it is wrong only if you can give the proof – Xiaolang Apr 12 '13 at 01:18
  • @Xiaolang, Im not sure I understand what you mean :/ – nicefella Apr 12 '13 at 01:22
  • i mean you should give more your thought to let us know instead of just say "it's wrong" .we had problem helping you don't know where you are incorrect – Xiaolang Apr 12 '13 at 01:24
  • I am a bit iffy on the formal definition of it and just wanted some confirmation :/ Apologies – nicefella Apr 12 '13 at 01:26
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    i think the definition is natrual and just like what George V.Williams said...it can be expanded to any integer $n \in N$ – Xiaolang Apr 12 '13 at 01:31

2 Answers2

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Ordinary induction need not start at $1$; it can start at any integer, positive, negative, or $0$. It’s the following principle:

Let $n_0$ be any integer, and let $P(n)$ be a proposition (about integers) such that $P(n_0)$ is true, and for each $n\ge n_0$, if $P(n)$ holds, then so does $P(n+1)$; then $P(n)$ holds for all integers $n\ge n_0$.

Here your induction hypothesis is $P(n)$, and the induction step consists in proving that $$P(n)\to P(n+1)\;.$$

Strong induction is the following principle:

Let $n_0$ be any integer, and let $P(n)$ be a proposition (about integers) such that $P(n_0)$ is true, and for each $n\ge n_0$, if $P(k)$ holds for all integers $k$ such that $n_0\le k\le n$, then so does $P(n+1)$; then $P(n)$ holds for all integers $n\ge n_0$.

Here the induction hypothesis is $P(n_0)\land P(n_0+1)\land\ldots\land P(n)$, and the induction step consists in proving that

$$P(n_0)\land P(n_0+1)\land\ldots\land P(n)\to P(n+1)\;.$$

In both cases the conclusion is that $P(n)$ holds for all integers $n\ge n_0$; one cannot conclude anything about the truth or falsity of $P(n)$ for integers $n<n_0$.

The two principles are logically equivalent: each can be proved from the other.

Brian M. Scott
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Formal definitions:

Regular Induction:

$(\forall n \in N, n \geqslant n_0)(h(n)) \iff h(n_0) \land ... \land h(n_1) \land (\forall k \in N, k \geqslant n_1)( h(k) \implies h(k + 1))$

Strong Induction:

$(\forall n \in N, n \geqslant n_0)(h(n))\\ \iff h(n_0) \land ... \land h(n_1) \land (\forall k \in N, k \geqslant n_1)( (\forall i \in [n_1, k])(h(i)) \implies h(k + 1))$

Note:

The above definitions heavily rely on formal logic, its concepts, symbols, and formulae. Formal logic is generally divided into predicate logic and propositional logic. It's effectively the basis language of all mathematical axioms, theorems, and proofs.

Both proofs can have as many basis conditions as they need. So, multiple basis conditions can never be an indication of which of these you should use.

For either regular or strong induction, the inductive condition only has to build from the last of the set of basis conditions.

The strong induction formula acts like a kind of catch-all where any combination of the previous "rungs" up to (and including) the current rung on the ladder can be used to imply that the next rung is true. You usually won't need all those rungs, but it helps if you need some random rung or a couple of rungs besides the current rung.

If you only need the current rung to prove the next rung, then strong induction is a bit overkill.

Though, if you think about it, regular induction is a subset of strong induction where the subset of previous "rungs" up to (and including) the current rung is restricted to only the current rung.

Regular Induction Example:

Prove that $\sum_{i=0}^{n}(i) = \frac{n(n + 1)}{2}$:

1. Preparations:

Let $n_0 = 0$

Let $n_1 = 0$

Let $(\forall n \in N)(f(n) = \sum_{i=0}^{n}{i})$

Let $(\forall n \in N)(g(n) = \frac{n(n + 1)}{2})$

Let $(\forall n \in N)(h(n) \iff f(n) = g(n))$

2. Definition:

$(\forall n \in N, n \geqslant n_0)(h(n)) \iff h(n_0) \land ... \land h(n_1) \land (\forall k \in N, k \geqslant n_1)( h(k) \implies h(k + 1)) $

$(\forall n \in N)(h(n)) \iff h(0) \land (\forall k \in N)( h(k) \implies h(k + 1))$

3. Evaluate the basis condition of $h(0)$:

$h(0) \iff f(0) = g(0) \iff \sum_{i=0}^{0}{(i)} = \frac{0(0 + 1)}{2} \iff 0 = 0 \iff T$

$h(0) \iff T$

4. Build up to the second condition (inductive condition) of the definition:

$(\forall k \in N)( h(k + 1) \iff f(k + 1) = g(k + 1))$

$(\forall k \in N)( h(k + 1) \iff \sum_{i=0}^{k + 1}{i} = \frac{(k + 1)(k + 2)}{2})$

$(\forall k \in N)( h(k + 1) \iff \sum_{i=0}^{k}{i} + (k + 1) = \frac{(k + 1)k}{2} + \frac{(k + 1)2}{2})$

$(\forall k \in N)( h(k + 1) \iff f(k) + (k + 1) = \frac{k(k + 1)}{2} + (k + 1))$

$(\forall k \in N)( h(k + 1) \iff f(k) + (k + 1) = g(k) + (k + 1))$

$(\forall k \in N)( h(k + 1) \iff f(k) = g(k))$

$(\forall k \in N)( h(k + 1) \iff h(k))$

$(\forall k \in N)( h(k) \iff h(k + 1))$

$(\forall k \in N)( h(k) \implies h(k + 1))$

5. Substitute into and work the definition:

$(\forall n \in N)(h(n)) \iff h(0) \land (\forall k \in N)( h(k) \implies h(k + 1))$

$(\forall n \in N)(h(n)) \iff T \land T$

$(\forall n \in N)(h(n)) \iff T$

$(\forall n \in N)(h(n))$

$(\forall n \in N)(f(n) = g(n))$

$(\forall n \in N)(\sum_{i=0}^{n}{i} = \frac{n(n + 1)}{2})$

$\square$

Strong Induction Example:

Prove that $a(n)$ is odd if $a(1) = 1$, $a(2) = 3$, and $a(n \gt 2) = 2a(n - 1) + a(n - 2)$:

1. Preparations:

Let $n_0 = 1$

Let $n_1 = 2$

$(\forall n \in N, n \geqslant n_0)(h(n) \iff a(n)$ is odd $)$

2. Definition:

$(\forall n \in N, n \geqslant n_0)(h(n))\\ \iff h(n_0) \land ... \land h(n_1) \land (\forall k \in N, k \geqslant n_1)( (\forall i \in [n_1, k])(h(i)) \implies h(k + 1))$

$(\forall n \in N, n \geqslant 1)(h(n))\\ \iff h(1) \land h(2) \land (\forall k \in N, k \geqslant 2)( (\forall i \in [2, k])(h(i)) \implies h(k + 1))$

3. Evaluate the basis conditions of $h(1)$ and $h(2)$:

$h(1) \iff a(1)$ is odd $\iff 1$ is odd $\iff T$

$h(2) \iff a(2)$ is odd $\iff 3$ is odd $\iff T$

4. Build up to the second condition (inductive condition) of the definition:

$(\forall k \in N, k \geqslant 2)( (h(k + 1) \iff a(k + 1)$ is odd$)$

$(\forall k \in N, k \geqslant 2)( (h(k + 1) \iff 2a(k + 1 - 1) + a(k + 1 - 2)$ is odd$)$

$(\forall k \in N, k \geqslant 2)( (h(k + 1) \iff 2a(k) + a(k - 1)$ is odd$)$

$(\forall k \in N, k \geqslant 2)( (h(k + 1) \iff a(k - 1)$ is odd$)$

$(\forall k \in N, k \geqslant 2)( (h(k + 1) \iff h(k - 1))$

$(\forall k \in N, k \geqslant 2)( (h(k - 1) \iff h(k + 1))$

$(\forall k \in N, k \geqslant 2)( (h(k - 1) \implies h(k + 1))$

Notice that we only need the "rung" below the current rung to prove the next rung.

Now, let's generalize:

$(\forall k \in N, k \geqslant 2)( ( h(2) \land h(3) \land ... \land h(k - 1) \land h(k) \implies h(k + 1))$

$(\forall k \in N, k \geqslant 2)( ( (\forall i \in [2, k])(h(i)) \implies h(k + 1))$

5. Substitute into and work the definition:

$(\forall n \in N, n \geqslant 1)(h(n))\\ \iff h(1) \land h(2) \land (\forall k \in N, k \geqslant 2)( (\forall i \in [2, k])(h(i)) \implies h(k + 1))$

$(\forall n \in N, n \geqslant 1)(h(n)) \iff T \land T \land T$

$(\forall n \in N, n \geqslant 1)(h(n)) \iff T$

$(\forall n \in N, n \geqslant 1)(h(n))$

$(\forall n \in N, n \geqslant 1)(a(n)$ is odd$)$

$\square$

Gustav Streicher
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