I've got a question about mathematical induction.

$\bigg(P(0) \land \big[\forall n\in \mathbb{N},\;\; P(n)\implies P(n+1)\big]\bigg)\implies \forall n\in \mathbb{N}, \; P(n)$

for the inductive step usually we say

"Let any $n\in \mathbb{N}$ and we assume that $P(n)$ holds and we will show that P(n+1) holds"

But today I have seen this on an another forum

"We assume there exists $n$ such that $P(n)$ holds..."

I told them that is formally wrong, because we will prove that $P(n)\implies P(n+1)$ only for a specific $n$

What is your point of view, Am I right?

Example we want to prove that $\forall n\ge 1, \quad 2^n>n$

Induction step ($\forall n\ge 1,\;\; P(n)\implies P(n+1)$)

(*) Let any $n\ge 1$ and we assume that $2^n>n$ holds

$2^{n+1}>n+1\iff 2^n+2^n>n+1$ as $2^n>n$ and $2^n>1$, we conclude that$2^{n+1}>n+1$ holds

So we have proved that $\forall n\ge 1,\;\; P(n)\implies P(n+1)$ which means $\forall n\ge 1,\;\; P(1)\implies P(2)\implies \cdots \implies P(n)\implies P(n+1)\cdots$

It remains the base step because we know noting about $P(1)$, $P(2)$ and ...

$P(1) :2^1>1$ so $P(1)$ is true

So we can conclude that $\forall n\ge 1, \quad 2^n>n$

If somone says for the induction step (*)

*"We assume there exists $n\ge 1$ such that $2^n>n$ holds..."*

this "there exists" annoys me

  • $\begingroup$ If you make no special assumption on $n$, then you actually proved that “if $P(n)$ then $P(n+1)$” is true for every $n$. It's just a sloppy way to say “let $n$ be any natural number”. $\endgroup$ – egreg Sep 27 '18 at 22:07
  • $\begingroup$ my question is, if someone says for the induction step "We assume there exists n such that P(n) holds and we will show that P(n+1) holds" is it wrong or not please? that is my question $\endgroup$ – Stu Sep 27 '18 at 22:09
  • $\begingroup$ It's a sloppy way to express the usage of induction. Better not using it. I guess you can reformulate that proof in a more rigorous language. $\endgroup$ – egreg Sep 27 '18 at 22:10
  • $\begingroup$ so that is wrong, thanks $\endgroup$ – Stu Sep 27 '18 at 22:12
  • $\begingroup$ yes it's what I told them, that is formally wrong to say that!! $\endgroup$ – Stu Sep 27 '18 at 22:13

After the base case, that is $\exists n_0$ such that $P(n_0)$ holds, for the induction step we assume as induction hypothesis that for some $n$ $P(n)$ holds and we need to show that $P(n) \implies P(n+1)$.

Note that when we prove the induction step we need to check that $P(n) \implies P(n+1)$ holds for $n\ge n_0$ otherwise we need to check again the base case.

For example when we prove by induction that $2^n\ge n^2$ we have that for the base case $n=0$ works but the induction step holds only for $n\ge 3$ then we need to prove the base case for $n\ge 3$ otherwise the proof by induction is wrong.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Sep 28 '18 at 1:03
  • $\begingroup$ @AloizioMacedo You are right sorry for that, even if our comments were completely inherent to the object raised by the OP. I would have deleted that after a while. Thanks $\endgroup$ – gimusi Sep 28 '18 at 1:16

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