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Cauchy's induction principle states that:

The set of propositions $p(1),...,p(n),...$ are all valid if:

  1. $p(2)$ is true.
  2. $p(n)$ implies $p(n-1)$ is true.
  3. $p(n)$ implies $p(2n)$ is true.

How to prove Cauchy's induction principle? Can we use it to prove what we can prove with weak and strong induction?

If yes how to prove using Cauchy's induction principle

$$ 1+2^1+2^2+...+2^n=2^{n+1}-1 $$

hjpotter92
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Abdelrhman Fawzy
  • 456

2 Answers2

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Let $\tilde p(n):=p(2^n)$. Then $\tilde p(1)=p(2)$ is true, and from $2^{n+1}=2\cdot 2^n$ it follows from your axiom $(3)$ by ordinary induction that $\tilde p(n)$ is true for all $n\geq 1$.

Let an $n\geq1$ be given. It is a fundamental principle of mathematics that $n<2^n$, i.e., $n=2^n-j$ for some $j\geq1$. Let $p'(k):=p(2^n-k)$ when $k<2^n$ and $:={\tt true}$ when $k\geq2^n$. Then $p'(0)=p(2^n)=\tilde p(n)$ is true, hence $p'(k)$ is true for all $k\geq0$ by your axiom $(2)$, in particular for $k=j$.

Christian Blatter
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If $R$ denotes a well ordening on $\mathbb N$ then $R$ can be used for induction.

There will be an $n_0\in\mathbb N$ that serves as minimal element, and there will be a successor function $S:\mathbb N\to\mathbb N$ in the sense that $S(n)$ is according to $R$ the smallest element of $\mathbb N$ with $(n,S(n))\in R$.

Then a set of propositions $p(1),p(2),\dots$ is true if $p(n_0)$ is true and if $p(n)\implies p(S(n))$ is true.

Now observe that $\mathbb N$ can be well ordered according to:$$2,1,4,3,6,5,8,7,10,9,12,11,\dots$$

So we have $n_0=2$, $S(2)=1$, $S(1)=4$ et cetera.

Then what is stated by Cauchy's induction principle is equivalent with:

  • $p(n_0)$ is true
  • $p(n)\implies p(S(n))$ is true.

So this shows that Cauchy's induction principle is indeed valid.

drhab
  • 151,093