# Mathematical induction from $n=1$

In T.Tao's Analysis 1 book I've read the following:

Axiom 2.5 (Principle of mathematical induction):

Let $P(n)$ be any property pertaining to a natural number $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n+1)$ is also true. Then $P(n)$ is true for every natural number $n$.

Proposition 2.2.14 (Strong principle of induction):

Let $m_0$ be a natural number, and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $m\geq m_0$ we have the following implication: If $P(m')$ is true for all natural numbers $m_0\leq m'<m$ then $P(m)$ is also true. Then $P(m)$ is true for all numbers $m\geq m_0$.

Since I've seen in many proofs the principle of induction (not the strong version) used with the base case $n=1$ I want to prove the following:

"Let $P(m)$ be a property pertaining to a natural number $m$ such that $P(1)$ is true and whenever $P(m)$ is true, $P(m+1)$ is also true. Then $P(m)$ is true $\forall m\in\mathbb{N}, m\geq 1$."

Here's what I've done:

Let $Q(n)$ be the property that $P(m)$ is true for $m=n+1$; then $Q(0)$ is true by hypothesis and if we suppose that $Q(n')$ is true for all natural numbers $0\leq n'<n$ then this means in particular that $Q(n-1)$ is true thus $P(n)$ is true so by hypothesis we have also that $P(n+1)$ is true thus $Q(n)$ is true and so by the strong principle of induction $Q(n)$ is true $\forall n\in\mathbb{N}$ and (being $\{m\in\mathbb{N}:m=n+1, n\in\mathbb{N}\}=\{m\in\mathbb{N}:m\geq 1\}$) we can conclude that $P(m)$ is true $\forall m\in\mathbb{N}, m\geq 1$, as desired.

a) Is my proof correct?

b) Is there any way to make it simpler/better?

c) Is there a way to prove the statement using only the axiom of mathematical induction?

Best regards,

lorenzo

As you say, let $Q(n) = P(n+1)$. Then $P(1) = Q(0)$, and $$P(n)\Rightarrow P(n+1), n\ge 1 \equiv Q(n-1)\Rightarrow Q(n), n\ge 1 \equiv Q(n)\Rightarrow Q(n+1), n\ge 0.$$ So together, $P(0)$ and $P(n)\Rightarrow P(n+1)$, $n\ge 1$ are equivalent to $Q(0)$ and $Q(n)\Rightarrow Q(n+1)$, $n\ge 0$. Thus $Q(n)$ holds for all $n\ge 0$; i.e., $P(n+1)$ holds for all $n\ge 0$; i.e., $P(n)$ holds for all $n\ge 1$.
But really, the "names" of the numbers $0$ and $1$ don't matter: just squint so you can't see the labels on the number line. It's the fact that the set of integers you're concerned with has a lower bound that matters. You can equally well prove induction where the base case is, say $P(17)$, or $P(-12)$.