While studying more advanced variants of induction (like strong induction), I've been wondering about the possibility to use a variant of induction to prove some statements for all integers.
My principle is the following:
- Prove the statement $P$ is true for a base case $k_0 \in \mathbb{Z}$
- Prove that $P(n+1)\iff P(n)$.
(Or $P(n) \implies (P(n+1) \land P(n-1))$
From which the statement $P$ is true for all $k \in \mathbb{Z} $
As an exemple I'll prove in this way that: $$\forall z \in \mathbb{C^*},\forall k\in\mathbb{Z}: Arg(z^k)\equiv k\cdot Arg(z)\quad[2\pi]$$
Let $z \in \mathbb C^*\quad (\mathbb C^*= \mathbb C\setminus \{0\})$ , we will Prove that $$\forall k\in\mathbb{Z}: Arg(z^k)\equiv k\cdot Arg(z)\quad[2\pi]$$
- For the case $k = 0,\quad z^0=1\quad (z\not=0)$, $Arg(1)\equiv 0 \quad[2\pi],\quad$hence it is true for $k=0$
- Let $k\in \mathbb Z$
$$\begin{align}Arg(z^{k+1})&\equiv Arg(z^k\cdot z)\quad[2\pi]\\
&\equiv Arg(z^k)+ Arg(z)\quad[2\pi]\\
\end{align}
$$
So we can deduce that:
$$Arg(z^k)\equiv k\cdot Arg(z)\quad[2\pi]\\
\iff Arg(z^k)+ Arg(z)\equiv k \cdot Arg(z) + Arg(z)\quad[2\pi]\\
\iff Arg(z^{k+1})\equiv(k+1)\cdot Arg(z)\quad [2\pi]$$ - Conclusion: as $z$ is arbitrary, then our claim is true
Is this Principle in general valid?
if so, is my proof valid?
Best Answer
Yes, your principle is correct and your proof is correct.
If you wish to prove your principle, that can be done simply from normal induction. First argue via normal induction that $P(n+1)\iff P(n)$ and $P(k_0)$ together proves $P(k)$ for all $k\geq k_0$. Then define $Q(n):=P(-n)$ and do the same thing to prove that $Q(n+1)\iff Q(n)$ and $Q(k_0)$ together proves $Q(k)$ for all $k\geq k_0$. This second principle can be shown to be equivalent to the other direction that you need, completing your proof.