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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

Stu
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  • 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”. – egreg Sep 27 '18 at 22:07
  • 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 – Stu Sep 27 '18 at 22:09
  • 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. – egreg Sep 27 '18 at 22:10
  • so that is wrong, thanks – Stu Sep 27 '18 at 22:12
  • yes it's what I told them, that is formally wrong to say that!! – Stu Sep 27 '18 at 22:13
  • Only one more question, why the "there exists n\ge 1" annoy you? It is just the induction hypotesis, how would you reformulate that? – user Sep 28 '18 at 00:13
  • Let any $n\ge 1$, and we assume $P(n)$ holds – Stu Sep 28 '18 at 00:20
  • to prove $\forall n$, you take an unspecified $n$ then we assume that $P(n)$ holds and we show $P(n+1)$ from $P(n)$, if it works for an unspecified $n$ it works for all $n$ – Stu Sep 28 '18 at 00:26
  • if you say we assume there exists n such that P(n), it means we assume $P(n)$ holds for specifics $n$ and not for any – Stu Sep 28 '18 at 00:28
  • so we can't conclude $\forall n\quad P(n)\implies P(n+1)$ but $\exists n \quad P(n)\implies P(n+1)$ – Stu Sep 28 '18 at 00:30
  • @Stu The part $\exists n$ such that $P(n)$ holds is just the induction hypothesis (it suffices one value for n). From here by the induction step we show that $P(n)\implies P(n+1)$ and of course that fact needs to be true for all $n\ge n_1$. Now, with reference to the base case $P(n_0)$, if $n_0\ge n_1$ we are done otherwise we need to prove the base case for some $n\ge n_1$. – user Sep 28 '18 at 01:13

1 Answers1

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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.

user
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